LABORATORY MANUAL CHEMISTRY 121 2013
LABORATORY MANUAL CHEMISTRY 121 EIGHTH EDITION 2013 Dr. Steven Fawl Science, Mathematics, and Engineering Division Napa Valley College Napa, California 1 Table of Contents Page 3 Preface Laboratory Safety Rules 4 Lab Reports 5 Sample Lab Report………………………………………………………………………………. 6 EXPERIMENT ONE ................................................................................................................... 11 Kinetics of the acid hydrolysis of trans-[Co(en)2Cl2]Cl EXPERIMENT TWO .................................................................................................................. 15 Determination of the t½ of a Radioactive Isotope EXPERIMENT THREE.............................................................................................................. 18 Chemical equilibria and Le Chatelier's principle EXPERIMENT FOUR................................................................................................................. 23 Hydrogen ion concentration and pH of aqueous solutions EXPERIMENT FIVE ...................................................................................................................31 Neutralization and hydrolysis EXPERIMENT SIX ..................................................................................................................... 36 Carbonic acid and its salts EXPERIMENT SEVEN............................................................................................................... 41 Titration Curve for KHP and Determination of pKa1 and pKa2 EXPERIMENT EIGHT............................................................................................................... 46 Determination of The Equilibrium Constant For FeSCN2+ EXPERIMENT NINE...................................................................................................................50 Determination of the Heat of Reaction EXPERIMENT TEN .............................................................................................55 Hess’ Law - Heats of Solution EXPERIMENT ELEVEN ............................................................................................................60 Electrochemical cells EXPERIMENT TWELVE .......................................................................................................... 66 Mystery Experiment 2 PREFACE Chemistry is an experimental science. Thus, it is important that students of chemistry do experiments in the laboratory to more fully understand applications of the theories they study in lecture and how to critically evaluate experimental data. The laboratory can also aid the student in the study of the science by clearly illustrating the principles and concepts involved. Finally, laboratory experimentation allows students the opportunity to develop techniques and other manipulative skills that students of science must master. The faculty of the Napa Valley College clearly understands the importance of laboratory work in the study of chemistry. The Department is committed to this component of your education and hopes that you will take full advantage of this opportunity to explore the science of chemistry. A unique aspect of this laboratory program is that a concerted effort has been made to use environmentally less toxic or non-toxic materials in these experiments. This was not only done to protect students but also to lessen the impact of this program upon the environment. This commitment to the environment has presented an enormous challenge, as many traditional experiments could not be used due to the negative impact of the chemicals involved. Some experiments are completely environmentally safe and in these the products can be disposed of by placing solids in the wastebasket and solutions down the drain. Others contain a very limited amount of hazardous waste and in these cases the waste must be collected in the proper container for treatment and disposal. The Department is committed to the further development of environmentally safe experiments which still clearly illustrate the important principles and techniques. The sequence of experiments in this Laboratory Manual is designed to follow the lecture curriculum. However, instructors will sometimes vary the order of material covered in lecture and thus certain experiments may come before the concepts illustrated are covered in lecture or after the material has been covered. Some instructors strongly feel that the lecture should lead the laboratory while other instructors just as strongly believe that the laboratory experiments should lead the lecture, and still a third group feel that they should be done concurrently. While there is no "best" way, it is important that you carefully prepare for each experiment by reading the related text material before coming to the laboratory. In this way you can maximize the laboratory experience. In conclusion, we view this manual as one of continual modification and improvement. Over the past few years many improvements have come from student comments and criticisms. We encourage you to discuss ideas for improvements or suggestions for new experiments with your instructor. Finally, we hope you find this laboratory manual helpful in your study of chemistry. 3 LABORATORY SAFETY RULES Your participation in this laboratory requires that you follow safe laboratory practices. You are required to adhere to the safety guidelines listed below, as well as any other safety procedures given by your instructor(s) in charge of the course. You will be asked to sign this form certifying that you were informed of the safety guidelines and emergency procedures for this laboratory. Violations of these rules are grounds for expulsion from the laboratory. Note: You have the right to ask questions regarding your safety in this laboratory, either directly or anonymously, without fear of reprisal. Goggles must be worn at all times while in lab. You must purchase a pair of goggle for yourself and you may store them in your locker. You will be advised of the appropriate goggles to be purchased. Locate the emergency evacuation plan posted by the door. Know your exit routes! Locate emergency shower, eyewash station, fire extinguisher, fire alarm, and fire blanket. Dispose of all broken glassware in the proper receptacle. Never put broken glass in the trashcan. Notify you instructor immediately if you are injured in the laboratory; no matter how slight. Never pipette fluids by mouth. Check odors cautiously (i.e. wafting). Never taste a chemical. Shoes must be worn in the laboratory. These shoes must fully enclose your foot. Long hair must be tied up in a bun during lab work. Loose long sleeves should be avoided in the lab. Children and pets are not allowed in the laboratory. Eating or drinking in the lab is prohibited. Do not drink from the laboratory taps. Wash your hands before and after working in the lab. Turn off the Bunsen burner when you are not using it. If any reagents are spilled, notify your instructor at once. Follow the instructor’s directions for disposal of chemicals. Only perform the assigned experiment. No unauthorized experiments are allowed. Every chemical in a laboratory must be properly labeled. If a label is unclear, notify your instructor. Use the proper instrument (eye-dropper, scoopula, etc.) to remove reagents from bottles. Never return unused chemicals to the original container. Do not cross contaminate reagents by using the same instrument for 2 different reagents. (e.g. don’t use the mustard knife in the mayonnaise jar) Material Safety Data Sheets (MSDS) are available for your reference. These contain all known health hazards of the chemicals used in this course. In addition, there is information concerning protocols for accidental exposure to the chemical. You are advised to inspect this binder. 4 LAB REPORTS Every lab has a lab report worksheet for you to fill in and give to your instructor. A more detailed explanation of what is expected in a lab report will be given to you by your instructor but some general considerations are given below. Only turn the lab worksheets into your instructor, not the entire lab. Make sure that your name and date is on every page of the lab. Write in pen and never erase or white-out a result. A simple line through bad data is sufficient. Do not write in pencil and then copy the information into your lab in pen. Always write every digit given to you by a balance. DO NOT ROUND. Pay attention to significant figures. It is usually safe to report an answer to 4 sig.figs. When drawing graphs, always use Guggenheim notation. Never connect the dots or label points when drawing the line on a graph. A best fit line should be drawn to indicate the trend in the data. Labs should be formatted with your name, date, title, objective, procedure, data, calculations, results, conclusions, sources of error and any questions found at the end of a lab must be answered. A more detailed view of what a lab report should look like follows on the next page. 5 NAME (on each page) DATE (on each page) TITLE OBJECTIVES- One or two sentences about the purpose of the experiment. Cute remarks will not be tolerated. PROCEDURE- This is to be a CONCISE OUTLINE of the experimental procedure. It is not necessary to restate your handout. You do not need to write out concentrations and amounts used of reagents or types of glassware and other equipment used. The exception to this rule is whenever you do something not written in the lab instructions then (and only then) you must give a complete description of everything you did. DATA- Must be written in non-erasable ink. It is especially important that your data be written neatly and directly into the table. DO NOT write on scratch or any other paper with the intention of transferring the information your data table later. Any mistakes you make should be corrected by drawing a single line through the incorrect information and the correct information written above. Put your data in tabular form (in tables). There is only one exception to this rule and that is when a single piece of data is taken and it applies to all the rest of the data. It is very important that your data be written neatly and logically and in a table, an example follows from you first experiment Beaker Measurements: Beaker size/mL Circumference/cm (y axis) Diameter/cm (x axis) Your data section should also include any Constants that you plan on using in your calculations examples include R (the gas law constant) and atomic mass. CALCULATIONS- This includes one set of every calculation done to go from your raw data to your final result. The last number you get is usually your result. Every single calculation must be shown even if all you do is a simple addition. Also be very careful with your units. They should be written every time they apply. It is not necessary to show every repeated calculation. Calculations must be neat and easy to follow; units must be used every time they apply. You may add paper to your report, if you need additional room for your calculations. (Hint: do your calculations on scratch paper first, and then copy to report.) It is very important that you do not use a number in your calculations that is not explained in your data section or in an earlier calculation. 6 GRAPHS- Graphs are frequently needed for a proper interpretation of your data and are often an integral part of your calculations. There is a standard method for making graphs that will be outlined here. Always use a separate 8 1/2 x 11 sheet of graph paper for each graph. Always draw your graph such that the x-axis is along the 11 inch side of the paper (turn the paper side-ways). Draw in the axes using a pen and mark the axes so that your graph will fill the page. It is not necessary to start at zero on every axis. Label the axes accordingly giving the units that were used. When labeling the axes use the following format (Guggenheim Notation), Time/sec or Temperature/K or Distance/cm This notation will be explained further in class. Always give a title to your graph., This title should tell you what the axes are and which system was used. For example for your first lab you could label your graph, Circumference vs. Radius for Several Beakers Always include your name and date on your graph. Perhaps the two most common problems when dealing with graphs are in the drawing of the best line and measurement of the slope which represents your data. NEVER CONNECT DOTS. Always draw a smooth line through the data putting as many dots above as below the line. NEVER FIND A SLOPE USING YOUR DATA POINTS. Always use points on your line, as far apart as they can be, to measure your line. Do not indicate the slope and intercept right on your graph. These belong in your calculations section. You may indicate the points along the line that you used to determine the slope on the graph, but make sure they are also in the Calculations section. Look at the graph on the next page as an example of what your graph should look like. RESULTS- A result differs from a conclusion in that a result usually reports numbers, but it may also require reporting the presence or absence of chemical compounds based on the outcome of some qualitative reactions. Do not assume that your data table suffices as a report for your results. Every lab should have a separate result section indicating any unknown number (if required), and the final answers obtained from your data tables, calculations, or graphs. Most labs indicate what you need to have in your results. A result should look like, The formula of unknown #5 was ZnS. The tablets contained 85.5 ± 0.3% aspirin The percentage of oxygen in the air was 20.7% 7 Circumference/cm 8 5 10 15 20 25 30 35 40 45 50 2.5 Name 4.5 6.5 Diameter/cm 8.5 10.5 12.5 Circumference vs. Diameter for Several Beakers 14.5 Date Sample Graph CONCLUSIONS – This is an extension of what you have learned. It is a general statement related to your experiment but giving a large view of what happened. "Models are used in science to propose workable, testable methods from which information can be obtained. These models are considered valid until more information reveals errors whereupon the original model is either modified or completely abandoned for a new model which describes the system more fully." Or, "When two elements combine the product has properties which are different from either of the reactants. This indicates that a new compound has been formed." DO NOT repeat your results or data here. There is always a conclusion that can be formed from your data. A conclusion describes what were you supposed to learn from your experiment? The answer should not be, I learned that my unknown was CaCl2 (which is a result), rather, you might have learned that compounds can be identified by determining how each of the ions reacts with other known compounds. SOURCES OF ERROR- In all experiments there are probable sources of error. In this portion of the lab write up you may discuss the possible reasons why the result differs from the expected result. Note- "Human Error" or "I did not weigh out enough sulfur" are not acceptable. It is assumed that you did the experiment properly. If you can identify error that you could not control it belongs in this section. QUESTIONS- Most experiments will include some questions which must be answered. Not all of the questions will be graded, but all of them must be done. All work must be shown and yes/no or multiple choice answers must be explained. OTHER CONSIDERATIONS- Neatness does not count, but neither will excessive eye-strain be tolerated. In most cases suggestions will be made on how to improve the quality of the lab report as the quarter progresses. Do not write on the back of your report, anything written there will not be graded. When you are finished collecting your data, the I.A. or Instructor will sign it. Data must not be altered. 9 LABORATORY MANUAL GENERAL CHEMISTRY 121 EIGHTH EDITION SPRING 2014 10 EXPERIMENT KINETICS OF THE ACID HYDROLYSIS OF trans-[Co(en)2Cl2]Cl 1 INTRODUCTION In aqueous solution the green complex trans-dichlorbis (ethylenediamine) cobalt(III) chloride dissociates into chloride ion and trans-[Co(en)2Cl2]+. This cobalt complex ion then reacts in acid solution to yield a mixture of the cis and trans forms of [Co(en)2(H2O)Cl]2+. The progress of this reaction can be followed visually because the products are red while the reactant is green. The purpose of this exercise is to determine the rate law for the hydrolysis of trans-[Co(en)2Cl2]+ and to determine the activation energy for the hydrolysis reaction by carrying out the reaction at several different temperatures from 45°C to 85°C. The equation for the hydrolysis reaction is, trans-[Co(en)2Cl2]+ + H2O → cis/trans-[Co(en)2(H2O)Cl]+ + Cl(dark green) (burgundy to red violet) The reactant is green while the product is supposed to be burgundy red to violet, but because of problems we have had in recent years, the product is sometimes yellow or orange. To determine the rate law and rate constants for this reaction, we shall measure the half-life for each experiment. When 50% of the reactant has been converted to product, the mixture (50% green and 50% red) has a characteristic color best described as "gun-metal gray", but other colors are possible and your instructor will inform you of them.. EXPERIMENT: You may work in pairs. Weigh out two samples (one about 0.12 g and one 0.02 g) of trans-[Co(en)2Cl2]Cl, and dissolve them in separate test tubes each containing 8.0 mL of icecold 1 M sulfuric acid, then place these and all following solutions in an ice bath. With a pipet transfer exactly 2 mL (save the rest for the next part) from each of these solutions to separate test tubes, and heat these in boiling water for 5 minutes to completely hydrolyze the complex to trans[Co(en)2(H2O)Cl]+. The solutions should be red, but may be yellow or orange. Cool both the tubes to room temperature, and add 2 mL of the corresponding green unhydrolyzed complex. You should now have two pairs of solutions that are gun metal gray, or some other color. The exact color is not important as long as you have a color representing the half-way point in the reaction. Keep these tubes in an ice bath so that they do not undergo additional reaction during the laboratory period. Adjust the temperature of a large beaker of water to 55 +/- 1°C. One student should place the tubes which contain the remaining unhydrolyzed green complex into the hot water while a second student records the time. As the solutions change color a comparison should be made to our standards prepared previously. When the colors match, the time and temperature should be recorded. From this you should be able to determine the order of the reaction and the rate law. Prepare a solution of 0.15 g of trans-[Co(en)2Cl2]Cl in 30 mL of 1 M H2SO4, transfer 2 mL to a test tube and boil it for 5 minutes. As before, cool it and add 2 mL of unreacted solution to make 4 mL of half converted standard. Using 4 mL portions measure the half life of the reaction at 85, 75, 65, and 45°C. Record the times and temperatures in your notebook. Be sure to record t1/2, k, ln k and 1/T in your data table. 11 Name ______________________ Date _______________ EXPERIMENT ONE KINETICS OF THE ACID HYDROLYSIS OF trans-[Co(en)2Cl2]Cl OBJECTIVE: PROCEDURE: DATA Concentration Time Rate Law = Temperature Time Rate Const. k ln k 1/Temp 45°C = 318 K 55°C = 328 K from above 65°C = 338 K 75°C = 348 K 85°C = 358 K Attach Graph RESULTS: Slope =_____________________________________ Ea = __________________________ Intercept = _________________________________ A = ___________________________ 12 CALCULATIONS (You may use this sheet in your lab report) 1) Write down the rate law for the acid hydrolysis of trans- [Co(en)2Cl2]Cl. What is the order of the reaction? 2) The relationship between the rate constant k and the half-life t1/2 for first and second order reactions is given by, k = 0.693/t1/2 (First order) or k = 1/([A]t1/2) (Second order) Using the order obtained in (1) and the proper half-life equation calculate k at each temperature of your reaction. Show one example here; include all the k's in your data table. 3) Make a plot of ln(k) vs. 1/T, where k is the rate constant at each temperature and T is the temperature in Kelvin. The relationship between the rate constant and a quantity called the activation energy is given by the Arrhenius equation, ln(k) = -Ea/RT + ln(A) Where R = 8.314 J/mol-K, A is called a pre-exponential term (a constant), Ea is the activation energy, and k is the rate constant. This equation is of the general form y = mx + b Where m is the slope and b is the intercept of a line. Your plot of ln(k) vs. 1/T should yield a slope of -Ea/R and an intercept of ln(A). From the slope and intercept calculate Ea and A. Ea and A are your results. 13 PROBLEMS 1) In this experiment you assumed that the t1/2 was very slow at ice bath temperatures of about 0°C. Use your data to calculate the t1/2 at this temperature. Was the assumption valid? 2) While doing a different experiment a student found that the half-life of his reaction (not this experiment) decreased as the concentration of his reacting species increased. What if anything was wrong? 3) It takes 3 minutes to soft-boil an egg at 100°C. Assuming a soft boiled egg represents the halflife of the denaturation of egg albumin, comment on why most people add salt to the water when boiling an egg. (What does the salt do to the water?) 14 EXPERIMENT DETERMINING THE HALF-LIFE OF AN ISOTOPE 2 INTRODUCTION One type of nuclear reaction is called radioactive decay, in which an unstable isotope of an element changes spontaneously and emits radiation. The mathematical description of this process is shown below. Af = Aie-kt In this equation, k is the decay constant, commonly measured in sec-1 (or another appropriate unit of reciprocal time) similar to the rate law constant, k, in kinetics analyses. Ai is the activity (rate of decay) at t = 0. The SI unit of activity is the bequerel (Bq), defined as one decay per second. This equation shows that radioactive decay is a first-order kinetic process. One important measure of the rate at which a radioactive substance decays is called half-life, or t1/2. Half-life is the amount of time needed for one half of a given quantity of a substance to decay. Half-lives as short as 10-6 second and as long as 109 years are known. In this experiment, you will use a source called an isogenerator to produce a sample of radioactive barium. The isogenerator contains cesium-137, which decays to produce barium-137. The newly made barium nucleus is initially in a long-lived excited state, which eventually decays by emitting a gamma photon and becomes stable. By measuring the decay of a sample of barium-137, you will be able to calculate its half-life. EXPERIMENT Each group should have a computer interfaced to a radiation monitor and have the Vernier “Lifetime” program running. You will also be given a shallow aluminum cup that will hold your radioactive sample. Place the radiation monitor on top of, or adjacent to, the cup to get a maximum rate of sample detection. Your instructor will deliver a small sample of radioactive Barium-137 to each group. As soon as delivery is complete, click the Collect button in the upper right corner of the data window. Collect data for thirty minutes. Do not move the radiation monitor or the cup during the data collection. Be careful not to spill the solution in the cup. When the data collection is complete, you may dispose of the barium solution by pouring it down the sink. 15 CALCULATIONS The solution you obtained from the isogenerator may contain a small amount of long-lived cesium in addition to the barium. To account for the counts due to any cesium, as well as for counts due to cosmic rays and other background radiation, you can determine the background count rate from your data. By taking data for 30 minutes, the count rate should have gone down to a nearly constant value, aside from normal statistical fluctuations. The counts during each interval in the last five minutes should be nearly the same as for the 20 to 25 minute interval. If so, you can use the average rate at the end of data collection to correct for the counts not due to barium. BACKGROUND RADIATION Select the data on the graph between 25 and 30 minutes by dragging across the region with your mouse. Click on the statistics button on the toolbar. Read the average counts during the intervals from the floating box, and record the value in your data table as the average background counts. Use the corrected count rates to derive an exponential function for the first fifteen minutes of the data collection. DETERMINING VALUES FOR k and Ai Click and drag the computer mouse across the region between 0 and 15 minutes. Make sure that all of the data points in this region are in the shaded area. Click the curve-fit icon on the tool bar. Select natural exponent from the equation list. In the "Coefficients" box, type in the background radiation value as the "B" value. This will account for the background radiation inherent in the experiment. Click Try Fit. The function will be shown and plotted along with your data. Click to see a full graph of the function and data. Record the fit parameters Ai, k, and B in your data table. From the fit parameters, determine the half-life t1/2 for 137Ba and place it in your data table 16 Name__________________ Date____________________ DETERMINING THE HALF-LIFE OF AN ISOTOPE OBJECTIVE: PROCEDURE: DATA TABLE Experimental Data Fit to Af = Ai exp(- kt) + B Ai = k= B= t1/2 = CRC t1/2 for 137 Ba = PROBLEMS 1) What fraction of the initial activity of your barium sample would remain after 25 minutes? 2) Using your results calculate the counts after 25 minutes. Compare this to the actual value you measured. Was it a good assumption that the counts in the last five minutes would be due entirely to non-barium sources? 3) Would any of your t1/2 or k values change if you had been given more or less sample? Explain. 17 EXPERIMENT CHEMICAL EQUILIBRIA AND LE CHATELIER'S PRINCIPLE 3 INTRODUCTION All chemical reactions proceed toward an equilibrium position, some more rapidly than others. In this experiment we shall consider only reactions that occur fairly rapidly, reaching equilibria in a few minutes or less. Such reactions are said to be rapid and reversible. The equilibrium position varies for different reactions. For example acetic acid dissociates only a small extent. CH3COOH ↔ H+ + CH3COOAt equilibrium in 0.1 M acetic acid, only about 1% of the acetic acid molecules are dissociated into hydrogen ions and acetate ions. We say that this equilibrium "lies far to the left." The equilibrium position is the same whether one starts with reactants or products. For example, when equivalent amounts of H+ and CH3CO2- are mixed, they combine to a large extent to form CH3COOH molecules, and at equilibrium only about 1% of the H+ and CH3COO- remain unreacted. The final concentrations of H+, CH3COO-, and CH3COOH are exactly the same as in a solution of acetic acid of the same concentration. LE CHATELIER'S PRINCIPLE The equilibrium position may be shifted to the left or to the right by changing experimental conditions such as concentration, pressure or temperature. A very useful chemical principle, called Le Chatelier's Principle, enables one to predict in which direction the equilibrium will shift. The principle states that; "If a change is made in any of the factors influencing a system at equilibrium, reaction will occur in the direction which tends to counteract the change made." To illustrate this principle, we shall examine the effect of increasing the concentration of CH3COO- on the equilibrium of the reaction above. If this change causes the equilibrium to shift to the left, then the concentration of H+ will decrease; if the equilibrium shifts to the right, then the concentration of H+ will increase. EQUILIBRIUM SHIFT WHEN A CONCENTRATION IS CHANGED EXPERIMENT: Place a 5 mL portion of 1M acetic acid in each of two test tubes, a 5 mL portion of 0.1 M acetic acid in a third test tube, and 5 mL of de-ionized water in a fourth. Add 2 drops of methyl orange indicator to each test tube and record in your laboratory notebook the color of the contents of each test tube. To one of the test tubes containing 1 M acetic acid, add 2 mL of 4 M sodium acetate (CH3COONa) dropwise. Observe and record the changes in color that occur. 18 EQUILIBRIUM BETWEEN A SOLID SALT AND A SOLUTION As a second example of a rapid reversible reaction we shall investigate the equilibrium between a solid salt and a saturated solution of the salt. The solubility of solid silver acetate (CH3COOAg) is 0.06 mole per liter at room temperature, and the solubility product is therefore (0.06)2 or 3.6 x10-3. CH3COOAg (s) ↔ Ag+ + CH3COO- Ksp = 3.6 x 10-3 EXPERIMENT: Prepare some solid silver acetate as follows: Place 5 mL of 0.1 M silver nitrate (AgNO3) in a centrifuge tube and add 2 mL of 4 M sodium acetate. Stir the mixture thoroughly using a glass rod. Record the color of the precipitate that is formed. Place the centrifuge tube in the centrifuge and balance the centrifuge by placing a second centrifuge tube containing 7 mL of water opposite the first tube. Turn on the centrifuge for about 30 seconds. When the centrifuge has stopped, remove the tubes then decant the liquid and discard it. To the solid which remains, add 3 mL of de-ionized water. Place the tube in a beaker of boiling water and note the amount of solid in the tube. Cool the tube to room temperature by placing it in a beaker of tap water for 5 minutes. Again note the amount of solid. Centrifuge the mixture as above and decant the liquid into a clean test tube for use in another experiment. To the remaining solid add 2 mL 6 M HNO3. Stir with a glass rod until all of the solid dissolves. THE COMMON ION EFFECT EXPERIMENT: To the test tube containing the saturated silver acetate solution add 2 mL of 4 M sodium acetate. Stir the mixture with a glass rod. Observe any reaction that occurs. Write a net reaction to account for the observed change. In your conclusion state briefly how this experiment illustrates Le Chatelier's Principle. This application of Le Chatelier's principle is called the common ion effect. 19 Name ______________________ Date _______________ EXPERIMENT THREE LeChatelier’s Principle OBJECTIVE: PROCEDURE: DATA: Equilibrium Shift When Concentration is Changed Solution Color w/ Methyl Orange 1.0 M Hac 0.1 M HAc Pure Water Volume of NaAc added to HAc Color 0.0 mL 0.5 mL 1.0 mL 1.5 mL 2.0 mL Equilibrium between Solid and Solution & Common Ion Effect Procedure Result (amount) AgNO3 + NaAc Before Boiling AgNO3 + NaAc After Washing and Boiling AgAc(s) + 2 mL of HNO3 AgAc solution + 2 mL NaAc Net Ionic Reaction = 20 PROBLEMS 1) What happens to the H+ concentration in acetic acid when sodium acetate is added? What is the reaction taking place to account for this? Will the [H+] be higher or lower after the sodium acetate/acetic acid system has come to equilibrium? 2) What substance disappeared when HNO3 was added to the wet solid formed by adding AgNO3 to sodium acetate? Write the reaction that has occurred (net ionic equation). 3) How would the solubility of AgCl be affected by addition of 6 M HNO3? Why do AgCl and AgAc behave differently with addition of H+? 4) How would the equilibrium have shifted in the common ion experiment upon addition of Silver Nitrate? of Sodium Nitrate? 21 5) Excess solid Ca(OH)2 is in equilibrium with its ions, Ca(OH)2 ↔ Ca2+ + 2 OHState whether the amount of Ca(OH)2 increases, decreases, or stays the same upon addition of; NaOH KNO3 HCl Ca(OH)2 CaCl2 6) Does the solubility of solid Ca(OH)2 in water depend on the amount of solid in contact with the saturated solution? 7) The solubility of Ca(OH)2 in water is 0.165 g per 100 mL of water. Calculate the solubility product of Ca(OH)2. 22 EXPERIMENT HYDROGEN ION CONCENTRATION AND pH OF AQUEOUS SOLUTIONS; DISSOCIATION CONSTANTS OF ACETIC ACID AND AMMONIA SOLUTIONS 4 INTRODUCTION Water is a weak electrolyte. It dissociates slightly to give H+ and OH-. H2O ↔ H+(aq) + OH-(aq) In pure water [H+] = [OH-] = 1.0 x 10-7 M and thus the amount of dissociation is extremely small. The dissociation constant for water can be written, Kw = [H+] [OH-] / [H2O] = 1.0 x 10-14 @ 25°C Chemists use activities in the formulation of Kw instead of concentrations. By convention the activity of water is set equal to one and the activity of dilute solutions is set equal to the concentrations of the ion. Therefore our dissociation constant can be written as, Kw = [H+] [OH-] = 1.0 x 10-14 We can use the dissociation constant of water to calculate the concentration of H+ or of OH- in any aqueous solution if the concentration of the other ion is known. THE USE OF pH TO MEASURE ACIDITY It is generally easier to express [H+] in terms of pH. We shall define pH = -log[H+]. Thus the pH of pure water would be pH = -log[H+] = -log[10-7] = 7.0. Acidic solutions have a pH less than 7.0 and basic solutions have a pH larger than 7.0. THE LOWER THE pH THE HIGHER THE ACIDITY. DETERMINATION OF HYDROGEN ION CONCENTRATION AND pH In this section you will determine the [H+] and pH of two unknown aqueous solutions. In the back of your Lab Manual you will find a table that will tell you the relationship between color and pH for several different indicators. While the data in this table are useful as a general guide, it should be realized that different individuals respond differently to the same color, and they may also use different words to describe a particular color. For the accurate determination of pH with an indicator, one must always make a direct comparison of the color of the unknown solution with the colors of solutions of known pH values. Furthermore, the solutions being compared should be as similar as possible in factors that may affect the color such as size of test tube, volume of solution, and number of drops of indicator. It is usually helpful to observe the colors by placing the test tubes side by side on a piece of white paper, and then looking down through the solutions from the top. 23 EXPERIMENT: Obtain your unknowns from the instructional assistant. Test 2mL your first unknown with 2 drops of bromothymol blue to ascertain whether it is acidic or basic. Then, proceed with the appropriate indicators to determine the approximate pH. Record the name of each indicator. Use its color with your unknown and the pH range indicated into your data table. For these tests use 2 mL of your unknown solution and 2 drops of indicator. When you have decided the approximate pH, you will need to verify it. To do this you must prepare a solution of this pH and test it with the appropriate indicators. If your solution is in the 1 M to 10-3 M H+ concentration range (0 to 3 pH), prepare it from 6 M HCl. Solutions in the range from 10-4 to lO-11 M H+ (pH 4 to 11) are available as buffers from the instructional assistant. Solutions in the 10-11 to 10-14 M range of H+ concentration range (pH 11 to 14 ) are prepared from 6 M NaOH. Use serial dilution to prepare solutions from HCl and NaOH. For these final critical comparisons use 5 mL of both the standard and the unknown (in different test tubes) and 3-4 drops of indicator. Compare the color of the standard solution to the color of your unknown with at least 2 and preferably 3 different indicators. If they are the same, then you know the pH and [H+] of your unknown solution. AMMONIA SOLUTIONS Ammonia is a gas at room temperature that is very soluble in water. The resulting solution is basic because of the reaction, NH3(g) + H2O ↔ NH4+(aq) + OH-(aq) Kb = [NH4+] [OH-] / [NH3] EXPERIMENT : Prepare a 1 M solution of NH3 from the 6 M NH3 found in the lab. Determine the [OH-] in the 1 M NH3 by using the indicators alizarin yellow and indigo carmine. Record the colors in your data table. Using these data calculate the dissociation constant Kb for NH3. Include this in your results. REACTIONS OF NH4+ AND OHPredict what will happen when solutions of ammonium chloride (NH4Cl), and sodium hydroxide are mixed. EXPERIMENT: Prepare 12 mL of 1 M NaOH by adding 2 mL of 6 M NaOH to 10 mL of water. Mix thoroughly and then place 5 mL of the 1 M NaOH in each of two test tubes. Add several drops of indigo carmine to each test tube. Record the color in your data table. Slowly add 5 ml of 1 M NH4Cl to one test tube and record the color changes that occur. Does the OH- concentration increase or decrease as you add NH4Cl to the NaOH solution? Write an equation for the net reaction that occurs. Include this reaction and in your data section and report it in your result section. 24 DISSOCIATION CONSTANT OF ACETIC ACID In the following experiment we shall determine the dissociation constant of acetic acid by mixing solutions containing equal numbers of moles of acetic acid and sodium acetate. The resulting buffer solution will contain approximately equal concentrations of HAc and NaAc. [HAc] ≈ [Ac-] It then follows from the law of chemical equilibrium that the concentration of H+ in the solution is equal to the dissociation constant, HAc ↔ H+ + Ac- Ka = [ H+ ] [ Ac- ] = [ H+ ] [HAc] Furthermore, if we define pKa = -logKa, then in this solution -logKa = -log[H+], or pKa = pH. EXPERIMENT: Add 10 mL of 1 M HAc to 10 mL of 1 M NaAc. Determine the pH and [H+] of the resulting solution using the appropriate indicators. Record the name of each indicator and its color in this solution in your notebook. Complete the table in your worksheet for this solution (which does not contain 0.5 M HAc and 0.5 M NaAc). Using these data, calculate the dissociation constant Ka and pKa for acetic acid. Include these numbers in your results. 25 Name _______________ Date ________________ HYDROGEN ION CONCENTRATION AND pH OF AQUEOUS SOLUTIONS; DISSOCIATION CONSTANTS OF ACETIC ACID AND AMMONIA SOLUTIONS OBJECTIVE: PROCEDURE: DATA: pH Unknowns Indicator Color of Unk# Malachite Green Methyl Violet Methyl Orange Bromocresol Green Methyl Red Bromothymol Blue Cresol Red Thymol Blue Phenolphthalein Alizarin Yellow R Indigo Carmine pH of Unknown # _____ = ______ pH of Unknown # _____ = ______ 26 pH Color of Unk# pH For 1 M NH3: Complete the following table Indicator Color pH Alizarin Yellow R Indigo Carmine pH of 1 M NH3 = _____________ Substance H+ OH- NH3 Equilibrium Concentration Kb for NH3 = ________________ For 1 M NaOH: Complete the following table Indigo Carmine Color Before adding NH4Cl After adding NH4Cl The solution becomes (more less) acidic as NH4Cl is added to the NaOH (Circle one). Net reaction between NaOH and NH4Cl: For 1 M HAc and 1 M NaAc: Complete the following table Indicator Used Color pH pH of 10 mL 1 M HAc mixed with 10 mL 1 M NaAc = __________ 27 NH4+ Substance H+ OH- HAc Equilibrium Concentration Ka for HAc = ___________ pKa for HAc = __________ RESULTS: The pH of Unknown #_______ is _________ The pH of Unknown #_______ is _________ The Kb for NH3 = ___________ The net reaction between NaOH and NH4Cl is, The Ka for HAc = ___________ and the pKa for HAc is ____________ 28 Ac- Na+ PROBLEMS 1. A solution is prepared by dissolving 0.050 mole of HCl and 0.030 mole of NaOH in sufficient water to give a final volume of 2.0 liters. Calculate the concentration of all the ionic (Na+, etc.) species in the final solution. 2. Instead of determining the dissociation constant of ammonia by our experimental procedure we could mix equal volumes of 1 M NH3, and 1 M NH4Cl. Show that for such a solution Kb = [OH-], and pKb = 14 - pH. 3. Calculate the H+, and OH- of a 0.10 M NH3 solution. (Use Kb = 1.8 x 10 following problems. 29 -5 in this and the 4. A solution is prepared by mixing 0.050 mole of NH4Cl and 0.010 mole of NaOH in sufficient water to give a final volume of 200 mL. Calculate the concentration of all the molecular and ionic species in the resulting solution. 5. A solution is prepared by mixing 0.050 mole of NH3 and 0.20 mole of HCl in sufficient water to give a final volume of 500 mL. Calculate the concentration of all the molecular and ionic species in the resulting solution. 30 5 EXPERIMENT NEUTRALIZATION AND HYDROLYSIS INTRODUCTION The neutralization of an acid by a base to form a salt and water proceeds to an equilibrium position which varies depending on the strengths of the acid and base that are used. In this assignment we shall investigate the equilibrium positions reached for the reaction of; H+ + OH- ↔ H2O A. a strong acid and a strong base, B. a weak acid (HX) and a strong base, HX + OH- ↔ X- + H2O H+ + BOH ↔ B+ + H2O C. a strong acid and a weak base, or, in the special case of ammonia, H+ + NH3 ↔ NH4+ In each of the above cases the same equilibrium position may be reached by mixing equivalent amounts of the reactants (as in a titration) or by preparing an aqueous solution of the product (the salt of the acid and base) of equivalent concentration. Since it is much easier experimentally to prepare the salt solutions, we shall prepare them and measure their pH and [H+], from which we can infer the equilibrium position. We shall therefore study the above reactions in the reverse direction. The reverse of the neutralization reaction is called the "hydrolysis" reaction, because in the typical case a water molecule is split by the reaction. HYDROLYSIS OF THE SALT OF A STRONG ACID AND STRONG BASE EXPERIMENT: Test about 5mL of a 50 mL portion of distilled water with bromothymol blue indicator. It should give a green color. If it is yellow, boil the water until it gives a green color with bromothymol blue. If your solution is blue inform the instructional assistant. Prepare an approximately 1 M NaCl solution by dissolving about 1.2 grams of sodium chloride (compare with the sample in the laboratory) in 20 mL of your distilled water. Determine the pH and [H+] of your NaCl solution by using appropriate indicators, including bromothymol blue. Use the same procedure you used in the preceding assignment for determining an unknown pH. Record the names and colors of these indicators in your laboratory notebook. Because sodium ion is the ion of a strong base, there is no tendency for Na+ to combine with OH-, nor for H+ to combine with Cl-, and therefore the following reactions DO NOT occur. Na+ + H2O ↔ NaOH + H+ Cl- + H2O ↔ HCl + OHMore precisely, the equilibrium positions of these two reactions are so far to the left that [NaOH] = 0 and [HCl] = 0. 31 HYDROLYSIS OF THE SALT OF A WEAK ACID AND A STRONG BASE. The neutralization reaction for acetic acid, CH3COOH, by a strong base is CH3COOH + OH- ↔ CH3COO- + H2O The hydrolysis of acetate ion, CH3COO-, is represented by the reverse of the above reaction: CH3COO- + H2O ↔ CH3COOH + OHEXPERIMENT: Determine the pH and [H+] of 1 M CH3COONa by using appropriate indicators, including thymol blue and phenolphthalein. Record the names and color of these indicators in your laboratory notebook. From your measured value of the pH and Kw, calculate [H+] and [OH-] in 1 M CH3COONa. Using this value for [OH-], calculate [CH3COO-]. HYDROLYSIS OF THE SALT OF A STRONG ACID AND A WEAK BASE The neutralization reaction for ammonia, NH3, by a strong acid is NH3 + H+ ↔ NH4+ The reverse of the above reaction represents the “hydrolysis” reaction of ammonium ion, NH4+; NH4+ ↔ NH3 + H+ EXPERIMENT: Determine the pH and [H+] of 1 M ammonium chloride by using appropriate indicators, including bromocresol green and bromothymol blue. Record the names and colors of these indicators in your laboratory notebook. From your measured value of the pH and Kw, calculate [H+] and [OH-] in 1 M NH4Cl. Using this value for [H+], calculate [NH3] and [NH4+]. HYDROLYSIS OF SODIUM CARBONATE SOLUTIONS The neutralization reaction for bicarbonate ion, HCO3- , by a strong base is HCO3- + OH- ↔ CO32- + H2O The reverse of the above reaction represents the hydrolysis of carbonate ion, CO32CO32- + H2O ↔ HCO3- + OHEXPERIMENT: Determine the pH and [H+] of 1 M sodium carbonate by using appropriate indicators, including alizarin yellow and indigo carmine. Record the names and colors of these indicators in your laboratory notebook. From your measured value of the pH and Kw, calculate [H+] and [OH-] in 1 M Na2CO3. Using this value for [OH-], calculate [HCO3-] and [CO32-]. 32 Name ___________________ Date_________________ NEUTRALIZATION AND HYDROLYSIS WORKSHEET OBJECTIVE: PROCEDURE: Data: Strong Acid and a Strong Base: Complete the following table for 1M NaCl pH = Substance H+ OH- Na+ Cl- HCl NaOH Equilibrium Concentration Weak Acid and Strong Base: Complete the following table for 1 M CH3COONa. pH = Substance H+ OH- Ac- HAc Na+ NaOH NH3 HCl Equilibrium Concentration Weak Base and Strong Acid: Complete the following table for 1 M NH4Cl. pH = Substance H+ OH- NH4+ Equilibrium Concentration 33 Cl- Weak Diprotic Acid and Strong Base: Complete the following table for 1 M Na2CO3. pH = Substance H+ OH- Na+ CO32- HCO3- Equilibrium Concentration PROBLEMS 1) a) Is the [H+] of the NaCl solution greater than, less than or equal to that of your boiled distilled water? b) Which indicator would you choose for the most exact endpoint for the titration of an HCl solution with NaOH? 2a) Would you expect a solution of sodium acetate to be acidic, basic or neutral? b) What is the pH of sodium acetate? c) Which indicator would you choose for the titration of CH3COOH with NaOH? 3a) Would you expect a solution of ammonium chloride to be acidic, basic or neutral? b) What is the pH of 1 M NH4Cl? c) Which indicator would you choose for the titration of NH3 with HCl? 4) Would you expect a solution of sodium carbonate to be acidic, basic or neutral? 5) The Ka for acetic acid is 1.8 x 10-5, and the Kb for ammonia is also 1.8 x 10-5. Calculate the equilibrium constant for the (a) neutralization of a strong acid by a strong base, (b) neutralization of acetic acid by sodium hydroxide, (c) neutralization of ammonia by hydrochloric acid. 34 6) Ammonium acetate, NH4CH3COO, is a strong electrolyte. When a 1 M solution of ammonium acetate is tested with bromothymol blue, the solution is green. Including all ions and molecules, what substances are present in a 1 M NH4CH3COO solution? Write an equation for the net reaction for the hydrolysis of NH4CH3COO. Calculate the equilibrium constant for this reaction, using the information given in problem #5. 35 6 EXPERIMENT CARBONIC ACID AND ITS SALTS INTRODUCTION Carbon dioxide is a colorless gas at room temperature that is somewhat soluble in water. Water in equilibrium with gaseous CO2 at 1 atm pressure contains 0.034 M H2CO3. CO2(1 atm) + H2O ↔ H2CO3 K = 0.034 Solid carbon dioxide is a convenient source of carbon dioxide (dry ice), but in this lab you will use a carbon dioxide generator to make saturated solutions of H2CO3. Carbonic acid is a typical diprotic acid in that it dissociates stepwise as shown below, H2CO3 ↔ H+ + HCO3- Ka1 = 4.3 X 10-7 HCO3- ↔ H+ + CO32- Ka2 = 4.7 X 10-11 Thus we see that there are two possible salts, NaHCO3 (sodium bicarbonate) and Na2CO3 (sodium carbonate). In the following experiments we will investigate the properties of H2CO3, NaHCO3, and Na2CO3. CARBONIC ACID SOLUTIONS EXPERIMENT: Prepare about 100 mL of a saturated solution of carbon dioxide by adding a few mL of concentrated H2SO4 to some solid sodium bicarbonate inside a CO2 generator found in the lab. Allow the resulting gas to bubble through 100 mL of distilled water for about 5 minutes. Test the resulting solution with bromocresol green and methyl orange indicators. Record the pH and label the solution 0.034 M H2CO3. Note- the pH of the solution is due almost entirely to the first acid dissociation. EXPERIMENT: Bubble some CO2 through a 10 mL sample of 1M CaCl2. Does a precipitate form? If we assume that the solution is saturated with CO2, what is the concentration of H2CO3 in the solution? Calcium carbonate is very slightly soluble in water. CaCO3 ↔ Ca2+ + CO32- Ksp = 4.7 x 10-9 Calculate the maximum concentration of CO32- that could exist in a 1 M CaCl2 solution. Is the concentration of CO32- in your saturated H2CO3 solution larger or smaller than this value? Using these data complete the table in your lab report. Use the value of Ka2 to calculate [CO32-]. 36 SODIUM CARBONATE SOLUTIONS Sodium carbonate, Na2CO3 is a solid that is very soluble in water and it is a strong electrolyte which dissociates to give Na+ and CO32- in aqueous solutions. Sodium carbonate solutions can hydrolyze stepwise in the following manner, CO32- + H2O ↔ HCO3- + OHHCO3- + H2O ↔ H2CO3 + OHEXPERIMENT: Place 5 mL of 1 M Na2CO3 into a test tube and add 5 mL of 1 M CaCl2. Centrifuge the mixture and note the amount of solid CaCO3 produced. Record your results. Write an equation for the net reaction. SODIUM BICARBONATE SOLUTIONS Sodium bicarbonate, NaHCO3, is a solid that is very soluble in water and it is a strong electrolyte which dissociates to give Na+ and HCO3- in aqueous solutions. Sodium bicarbonate solutions may also be prepared by the reaction of solutions containing 1 mole of H2CO3 and 1 mole NaOH. The equation for the net reaction is, H2CO3 + OH- ↔ HCO3- + H2O Since HCO3- is a weak acid, it should dissociate to give H+ and CO32-. However the H+ produced would react with another HCO3- to form the weak acid H2CO3. As a result we predict that the principal equilibrium in a NaHCO3 solution would be, 2 HCO3- ↔ H2CO3 + CO32- * The position of this equilibrium is far to the left, the concentrations of H2CO3 and CO32- being much smaller than the concentration of HCO3-. EXPERIMENT: Place 5 mL of 1 M NaHCO3 and 5 mL of 1 M CaCl2 in a centrifuge tube. Centrifuge the mixture and note the amount of precipitate obtained. Decant the solution into a second centrifuge tube. Now, place the tube containing the decanted solution in a beaker of boiling water. Keep it in the boiling water for 10 minutes, and then cool it by placing it in a beaker of water at room temperature. After it has cooled to room temperature centrifuge it. Note the amount of precipitate obtained. Are the results in accord with what you would predict from a shift in equilibrium above?* Compare the amounts of CaCO3 obtained before and after boiling the mixture with that obtained when you used the Na2CO3 solution. Record your observations. EXPERIMENT: Determine the pH and H+ concentration of 1 M NaHCO3 by testing 5 mL portions with the following indicators: bromothymol blue, cresol red, and phenolphthalein. Record your observations. Calculate [OH-]. What would happen to the [OH-] if the 1 M NaHCO3 solution is boiled for a few minutes and then cooled? EXPERIMENT: Place 5 mL of 1 M NaHCO3 in a test tube, add 2 drops of phenolphthalein, and place the test tube in a beaker of boiling water for 5 minutes. Record the color changes. Does [OH] increase or decrease? 37 RESULTS: In the result section of your lab report please list all the ions and molecules (except for water) for the following solutions a) the carbonic acid solution (made by bubbling CO2 through water); b) the 1 M Na2CO3 and c) the 1 M NaHCO3 (without the CaCl2 and before heating). Now for each of the 3 solutions list the ions in order of decreasing concentration. It will be much simpler if you group them in pairs. 38 Name ___________________ Date_________________ CARBONIC ACID AND ITS SALTS OBJECTIVE: PROCEDURE: Data: Carbonic Acid Solutions Bromocresol Green Color __________ Methyl Orange Color ____________ pH = _______ Substance H2CO3 H+ HCO3- OH- CO32- Equilibrium Concentration Sodium Carbonate Solutions Amount of CaCO3 produced (mm) = _____________________ Net ionic reaction = Sodium Bicarbonate Solutions Amount of CaCO3 initially __________ Amount of CaCO3 after heating________________ pH of NaHCO3 Bromothymol Blue Color ___________ Cresol Red Color ______________ Phenolphthalein Color _____________ pH = ________ [OH-] = __________________ Phenolphthalein Color After Boiling _____________________ [OH-] increase decrease Ion and Molecules Present: From highest to lowest concentration: 0.034 M H2CO3 1 M Na2CO3 1 M NaHCO3 _________ _________ _________ _________ _________ _________ _________ _________ _________ _________ _________ _________ _________ _________ _________ _________ _________ _________ 39 PROBLEMS 1) In some of the following cases the two substances cannot coexist in the same solution because they will react together. For example, an acid will react with a base, and some combination of ions will form precipitates. For these cases write a net equation for the reaction that would occur. In other cases write "No Reaction". (a) OH- and HCO3- (e) Ca2+ and HCO3- (b) OH- and CO32- (f) H2CO3 and CO32- (c) OH- and H2CO3 (g) H+ and HCO3- (d) H+ and H2CO3 (h) H+ and CO32- 2a) Using the equilibrium constants given in the lab calculate the value of the equilibrium constant for 2 HCO3- ↔ H2CO3 + CO32-. 2b) Calculate the concentration of H2CO3 and CO32- in a 1 M solution of NaHCO3. 3a) Calculate the value of the equilibrium constant for H2CO3 ↔ 2 H+ + CO32- 3b) What is your experimental value for the pH of a solution of 1 M NaHCO3? 3c) Using your results from questions 2 and 3, show that for HCO3- that pH = (pK1 + pK2)/2 40 EXPERIMENT TITRATION CURVE FOR POTASSIUM ACID PHTHALATE 7 INTRODUCTION In this experiment a pH meter equipped with a combination electrode (a glass electrode and a calomel reference electrode in the same housing) and drop counter will be used to construct the titration curve for KHP titrated with NaOH. From the titration curve and the concentration of the base you can determine the concentration of the unknown KHP solution and both the first and the second dissociation constants of phthalic acid. When molar concentration units are used, the concentration quotient (1) Kc2 = [H+][P2-] [HP-] is only approximately constant over the whole accessible range of concentrations. When activities are used, however, the ionization constant defined by (2) Ka2 = (aH+)(aP2-) (aHP-) is truly constant at all concentrations of the ions. The activity may be regarded, for our purposes, as an effective concentration. In ionic solutions the nearest neighbor shell around a cation will contain a (slight) predominance of anions, and vice versa. This partial screening of the ions by their counter ions makes the effective concentration of an ionic species somewhat smaller than its stoichiometric concentration. The activity and concentration of an ion i are related by (3) ai = Cifi where fi is an empirical parameter, called the activity coefficient, which depends on the size of the ion and the total ionic strength of the solution. Electrochemical experiments, such as the measurement of pH, yield ion activities rather than ion concentrations. At the ionic-.strengths to be used in this experiment the activity coefficients of the phthalate anionic species differ appreciably from unity and a serious misestimate of Ka results if they are ignored, The appropriate equations are (4) Ka2 = (aH+)(aP2-) = (aH+)[P2-](fP2-) (aHP-) [HP-](fHP-) 41 Taking the log of both sides, (5) log Ka2 = log (aH+) + log [P2-]/[HP-] + log (fP2-)/(fHP-) Now since (6) pH = - log(aH+) and pKa2 = -log Ka2, it follows that, (7) pH = pKa2 + log [P2-]/[HP-] + log (fP2-)/(fHP-) Thus when exactly half the KHP has been neutralized (when [P2-] = [HP-]) the measured pH and pKa are roughly the same, except for the activity coefficient term. At the equivalence point on the titration curve of a weak acid with strong base, the slope of the curve is a maximum. We can use this fact to find the equivalence point very precisely. Using the utilities available in Vernier LabPro, the program will calculate the first and second derivative of our data. The first derivative produces a plot with a peak in the center. The top of this peak is your final endpoint. If using the second derivative, the final endpoint is where the plot crosses the X axis. PROCEDURE: A. Supply your instructor with a clean, dry, 125 mL Erlenmeyer flask labeled with your name and locker number. You will receive 55 mL of your unknown KHP solution in this flask. B. Set up a titration assembly, consisting of a pH meter, a magnetic stirrer, and a buret filled with your standard 0.1 M NaOH.. C. Just before you are ready to prepare a titration curve standardize the pH meter according to the directions given in a separate handout. D. Using a pipet deliver 25 mL of the unknown KHP solution into a clean, dry 100 mL beaker which already contains a clean magnetic stirring bar. Add 20 mL of distilled water and 2 drops of phenolphthalein indicator. Put the beaker onto the magnetic stirrer so that the bar rotates freely, but there is still room alongside it for the electrode to be inserted near the bottom of the beaker. Mix the solution thoroughly with the stirrer. E. Wash the standardized electrode with your wash bottle, blot it dry, and position it through the large hole of the drop counter. Read the pH meter and the buret, and record both readings. F. Turn on the magnetic stirrer so it is rotating slowly. You are now ready to begin collecting data. Click the Collect button. No data will be collected until the first drop goes through the drop counter slot. Adjust the drop rate to about 1 drop every 2 seconds. When the first drop passes through, check to see that the data has begun recording. G. Continue watching your graph to see when the large increase in pH takes place - this will be the equivalence point. When this jump in pH occurs, add about 3 more milliliters of NaOH, click STOP, and stop the titration. Remove the electrode, wash it thoroughly, and put it back into its storage bottle. Remove the stir bar from the beaker and pour the solution down the sink. Clean and dry the stir bar. 42 H. Find the equivalence point. The best method for determining the equivalence point is to take the second derivative of the pH vs. volume data. Do the following, a. Open Page 3 by clicking on the Page window of the menu bar. b. Analyze the second derivative plot and record the volume of NaOH at the equivalence point. I. From this curve determine the pH at half and three-fourths neutralization of the KHP. Go to the worksheet and use it to compute an average value for pKa2, taking into account the activity coefficients. α’s (Na+ = 0.4, P2- = 0.6, HP- = 0.6, K+ = 0.3). The Debye-Hückel equation is below. 0.51Z 2 log f a 1 3.329 J. Having Ka2 in hand you can obtain Ka1 from the initial pH of the diluted unknown. Starting with the known equilibrium expressions, (8) K a1 [H ][HP ] f H f HP [H 2 P] f H2P K a2 [H ][P 2 ] f H f P 2 [HP - ] f HP - We can solve each of these equations for [H+] and multiply them together we get; (9) [H ]2 K a1K a2 H 2 PHP f H P f HP HP P f 2 2 2 H f HP f P 2 Cancelling terms that appear on the top and bottom of this equation we obtain, (10) [H ]2 K a1K a2 H 2 P f H P P f 2 2 2 H f P2 For the hydrolysis of pure KHP, the [H2P] = [P2-] so they cancel. In the Debye-Hückel equation the charge for H2P = 0 and fH2P = 1, so canceling the [H2P] and [P2-], setting fH2P = 1, and remembering that the activity of H+ = aH+ = [H+] fH+we get; (11) [H ]2 K a1K a2 2 f H f P 2 and [aH ]2 K a1K a2 f P 2 Since we have earlier defined pH as -log(aH+), Eq. 11 reduces to the simple form (12) 2pH = pKa1 + pKa2 + log fP2- The ionic strength of your diluted starting solution is easily calculated from the concentration of your unknown. The activity coefficient of P2- at this ionic strength can be calculated from the Debye-Hückel equation. With this value, the measured pH, and average pKa2 calculated earlier, pKa1 is in hand. 43 REPORT: Fill out the attached report sheet and turn it in. You should include your titration curve, the concentration of the undiluted KHP unknown, and the calculation of pKa1 and pKa2 using equations 7 and 12. 44 45 ______________ Vol. KHP used For Initial Solution Calculation of pKa1 3/4 Neutralization 1/2 Neutralization Calculation of PKa2 ¾ Neut ½ Neut Initial Vol. NaOH μ μ Total Vol. pH pH Mol. Na+ fHP- fHP- Conc. Na+ fP2- ____ fP2- Mol. K+ pKa1 ____ pKa2 Conc. K+ Avg.pKa2 Mol. HP- Conc. HP- Mol. P2- Conc. P2- Ionic Strength Date _________________________ EXPERIMENT SEVEN : Titration Curve of KHP not needed Conc. of Unknown. ______________ ______________ Vol. NaOH used Conc. NaOH used Name EXPERIMENT DETERMINATION OF THE EQUILIBRIUM CONSTANT FOR FeSCN2+ 8 INTRODUCTION When 2 reactants are mixed, the reaction typically does not go to completion. Rather, they will react to form products until a state is reached whereby the concentrations of the reactants and products remain constant. This is a dynamic state in which the rate of formation of the products is equal to the rate of formation of the reactants. The reactants and products are in chemical equilibrium and will remain so until affected by some external force. The equilibrium constant Kc for the reaction relates the concentration of the reactants and products. In this experiment we will study the equilibrium properties of the reaction between iron (III) ion and thiocyanate ion: Fe3+ (aq) + SCN– (aq) → FeSCN2+ (aq) When solutions containing Fe3+ ion and thiocyanate ion are mixed, the deep red thiocyanatoiron (III) ion (FeSCN2+) is formed. As a result of the reaction, the starting concentrations of Fe3+ and SCN- will decrease: so for every mole of FeSCN2+ that is formed, one mole of Fe3+ and one mole of SCN- will react. The equilibrium constant expression Kc, according to the Law of Chemical Equilibrium, for this reaction is formulated as follows: [FeSCN2+] / [Fe3+][ SCN- ] = Kc Square brackets ([]) are used to indicate concentration in mols/liter, i.e., molarity (M). The value of Kc is constant at a given temperature. This means that mixtures containing Fe3+ and SCN– will react until the above equation is satisfied, so that the same value of the Kc will be obtained no matter what initial amounts of Fe3+ and SCN– were used. Our purpose in this experiment will be to find Kc for this reaction for several mixtures made up in different ways, and to show that Kc indeed has the same value in each of the mixtures. The reaction is a particularly good one to study because Kc is of a convenient magnitude and the red color of the FeSCN2+ ion makes for an easy analysis of the equilibrium mixture using a spectrophotometer. The amount of light absorbed by the red complex is measured at 447 nm, the wavelength at which the complex most strongly absorbs. The absorbance, A, of the complex is proportional to its concentration, M, and can be measured directly on the spectrophotometer using the Beer-Lambert law: A = c Where = extinction coefficient, c = concentration (Molarity), and = path length (1cm) 46 EQUIPMENT 5 test tubes 10 ml graduated cylinder 10 ml graduated measuring pipette (0.1 ml graduations) stirring rod small labels or markers 5 cuvettes for the spectrophotometers 250 ml bottle acetone for rinsing HAZARD: As always wear Safety glasses while performing this experiment CONTAIMINATION NOTES: If your flask is wet before you prepare your standard/sample solutions ensure that the flask is wet with diluant (in this case it is the 0.0200M Fe(NO3)3 in 1.0 M HNO3 ). EXPERIMENT: Beer-Lambert Data A solution of 0.0200M Fe(NO3)3 in 0.5 M HNO3 has been prepared for you. Dilute 5.0, 10.0 and 15.0 ml portions of 2.00 x10-4 M KSCN to 100 ml with the 0.0200M Fe(NO3)3 in 0.5 M HNO3. This will give you 3 solutions that can be assumed to be 1.0x10-5, 2.0x10-5 and 3.0x10-5 Min FeSCN2+. Measure the absorbance of these solutions at 447nm, using a solution of 0.0200M Fe(NO3)3 in 0.5 M HNO3 as the reference solution. Measure the absorbance of these solutions at this wavelength. Plot Absorbance vs [FeSCN2+] using Excel. Find the slope and intercept for this plot. EXPERIMENT: Determination of Kc The mixtures will be prepared by mixing solutions containing known concentrations of iron (III) nitrate, Fe(NO3)3, and potassium thiocyanate, KSCN. The color of the FeSCN2+ ion formed will allow us to determine its equilibrium concentration. Knowing the initial composition of a mixture and the equilibrium concentration of FeSCN2+, we can calculate the equilibrium concentrations of the rest of the pertinent species and then determine Kc. Label five regular test tubes 1 to 5, with labels or by noting their positions in your beakers. Pour about 30 mL 0.02 M Fe(NO3)3 in 0.5 M HNO3 into a dry 100 mL beaker. Pipet 5.00 mL of this solution into each test tube. Then add about 20 mL 2.00 x 10-4 M KSCN to another dry 100-mL beaker. Pipet 1,2,3,4, and 5 mL from the KSCN beaker into each of the corresponding test tubes labeled 1 to 5, then pipet the proper number of milliliters of 0.5 M HNO3 into each test tube to bring the total volume in each tube to 10.00 mL. 47 The volumes of reagents to be added to each tube are summarized in the table below. Reagents/mLs) Fe(NO3)3 KSCN HNO3 1 5.00 1.00 4.00 Test Tube # 2 3 4 5.00 5.00 5.00 2.00 3.00 4.00 3.00 2.00 1.00 5 5.00 5.00 0.00 Mix each solution thoroughly with a glass stirring rod. Be sure to dry the stirring rod after mixing each solution to prevent cross-contamination. Measure the absorbance of each mixture at 447 nm as demonstrated by your instructor and put the data in the worksheet provided. Determine the concentration of FeSCN2+ from your calibration curve. Record the value on your Report form. Repeat the measurement using the mixtures in each of the other test tubes. 48 Name _______________________ Date____________________ Determination of the Equilibrium Constant for FeSCN2+ OBJECTIVE: PROCEDURE: Beer-Lambert Data Data For Beer-Lambert Plot 5 mL 10 mL Data 15 mL Absorbance Data for Equilibrium Calculation Data Test Tube # 1 2 3 Absorbance Conc of FeSCN2+ Conc of Fe3+ Conc of SCN- Value of Keq RESULTS: Average Keq = Staple Beer-Lambert plot to the back of this sheet. 49 4 5 EXPERIMENT DETERMINATION OF THE HEAT OF REACTION 9 INTRODUCTION The purpose of the experiment is to determine the quantity of heat liberated in the reaction, Mg(s) + 2 HCl (aq) → H2 (g) + MgCl2 (aq) The heat liberated in this reaction can be trapped and utilized to melt ice. Ice is less dense than water and as a consequence there is a volume change when ice melts. This volume change can be measured and used to calculate how much energy was transferred to the ice. This in turn is used to measure the amount of energy released in the reaction (the enthalpy). EXPERIMENT: Students may work in pairs with one student calling out the time and the other student reading the volume and recording both the time and volume. The calculations are to be made individually. Fill a Styrofoam container with crushed ice from the ice provided in the lab and place in it a small flask containing 3 mL of 6 M HCl and 6 mL of water. While this solution is cooling, assemble the bottle and stopper as shown in the lab discussion. Fill the bottle with tap water and push the stopper in tightly. Water will spurt out the top of the pipet. Watch the water level in the pipet over a 5 minute period. If it does not drop, the apparatus is free of leaks and you may continue with the experiment. If the level drops, try to repair the leak with the aid of your instructor or instructional assistant. (Do not use grease on the stopper.) When your assembly is leak-proof, fill the bottle with crushed ice to the brim. Add ice water from the Styrofoam container to fill the bottle completely and insert the stopper. Pour the 9 mL of HCl solution that you cooled to 0°C in the reaction test tube. Immediately place the bottle in the Styrofoam container and surround the bottle with crushed ice and water to just below the rim of the test tube. Insert the cork in the test tube loosely. Allow the apparatus to stand for 15 minutes during which time the temperature should become constant at 0°C. During this time interval weigh a 0.1 g sample of magnesium ribbon to the nearest milligram. Record the mass in your notebook. After the apparatus has been allowed to stand for 15 minutes, adjust the water level in the pipet to a reading of 0.8 mL or greater by placing a small piece of latex tubing on the end of the pipet and filling it with water. By pushing on the stopper it is possible to raise the level of the water inside of the pipet until it meets the water in the tubing. Releasing the stopper will pull the water from the tubing into the pipet thus filling it. This technique may require practice. Begin reading the pipet volume once the water level has reached some convenient mark on the pipet. Read the pipet volume each minute for 5 minutes and record these readings in your notebook. If the volume change in the first 5 minutes is greater than 0.05 mL then you have a leak and must readjust the stopper and start again at the beginning of this paragraph. If the volume change during this 5minute period is 0.05 mL or less, roll the magnesium ribbon into a loose coil, and drop it into the HCl solution. Place the cork LOOSELY over the test tube and continue to record the volume each minute for a period of 15-20 minutes. 50 Plot your data using a computer-graphing program, or ask for a sheet of graph paper. The results should be similar to those shown in the figure below. CALCULATIONS: Draw parallel lines through the initial and final points. The volume change due to the heat of reaction of your sample of magnesium is the vertical distance between these parallel lines. The density of ice at 0°C is 0.9167 g/mL and for water it is 0.9998 g/mL. Inverting each of these values we see that the volume of ice is 1.09087 mL/gram and for water it is 1.0001605 mL/gram at 0°C. This means that for every gram of ice that melts the volume will change by, 1.09087 mL/gram ice - 1.00016 mL/gram water = 0.09071 mL/g of ice melted Since it require 6.01 kJ to melt one mole of ice, we can calculate how much energy is needed to melt one gram of ice, 6010 J x 1 mole ice = 333.89 J/gram ice 1 mole ice 18 grams Therefore, we can calculate how many Joules of energy must be added to our system to change the volume by one mL, 333.89 J/gram ice = 3680.9 J/mL of volume change 0.09071 mL/gram ice 51 So, to calculate the energy required to melt your volume of ice, multiply your measured volume by 3680.9 J/mL. This amount of energy was released when approximately 0.1 gram of magnesium reacted with HCl. To calculate the amount of energy released when one mole of magnesium reacts, you must calculate the number of moles of magnesium used in your experiment. Calculate the number of moles of magnesium used in your experiment (the atomic mass of magnesium is 24.3 g/mole). By dividing the number of calories calculated earlier and dividing this by the number of moles of magnesium used one obtains the enthalpy for the following reaction, Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2 (g) + Enthalpy Check your calculations and report the result of this calculation in your worksheet. 52 Name_________________ Date______________ THE DETERMINATION OF A HEAT OF REACTION OBJECTIVE: PROCEDURE: DATA: Time and Volume Measurements for the Reaction between Mg and HCl Time Volume Time Volume Time Volume CALCULATIONS: (Include graph) 53 RESULTS: SOURCES OF ERROR: PROBLEM The standard heat of formation of HCl (aq) is –167.5 kJ/mole. From this, using Hess’ law, calculate the standard heat of formation of MgCl2 (aq). (Hint: What is meant by the standard heat of formation?) 54 EXPERIMENT DETERMINING THE ENTHALPY OF A CHEMICAL REACTION USING HESS’S LAW 10 INTRODUCTION All chemical reactions involve an exchange of heat energy; therefore, it is tempting to plan to follow a reaction by measuring the enthalpy change (ΔH). However, it is often not possible to directly measure the heat energy change of the reactants and products (the system). We can measure the heat change that occurs in the surroundings by monitoring temperature changes. If we conduct a reaction between two substances in aqueous solution, then the enthalpy of the reaction can be indirectly calculated with the following equation. H = Cp × m × ΔT The term H represents the heat energy that is gained or lost. Cp is the specific heat of water, m is the mass of water, and ΔT is the temperature change of the reaction mixture. The specific heat and mass of water are used because water will either gain or lose heat energy in a reaction that occurs in aqueous solution. Furthermore, according to a principle known as Hess’s law, the enthalpy changes of a series of reactions can be combined to calculate the enthalpy change of a reaction that is the sum of the components of the series. In this experiment, you will measure the temperature change of two reactions, and use Hess’s law to determine the enthalpy change, H of a third reaction. You will use a Styrofoam cup nested in a beaker as a calorimeter, as shown in Figure 1. For purposes of this experiment, you may assume that the heat loss to the calorimeter and the surrounding air is negligible. MATERIALS Vernier computer interface 2.0 M hydrochloric acid solution 2.0 M sodium hydroxide solution 2.0 M acetic acid solution 2.0 M sodium acetate solution 250 mL beaker 50 mL or 100 mL graduated cylinders Utility clamp Glass stirring rod Temperature Probe Computer Styrofoam cup calorimeter Ring stand Figure 1: Calorimetry Setup 55 PROCEDURE Obtain and wear goggles. Connect a Temperature Probe to Channel 1 of the Vernier computer interface. Connect the interface to the computer with the proper cable. Use a utility clamp to suspend the Temperature Probe from a ring stand, as shown in Figure 1. Start the Logger Pro program on your computer. Open the file “13 Enthalpy” from the Advanced Chemistry with Vernier folder. Part I - Reaction Between Solutions of NaOH and HCl Nest a Styrofoam cup in a beaker (see Figure 1). Measure 50.0 mL of 2.0 M HCl solution into the cup. Lower the tip of the Temperature Probe into the HCl solution. CAUTION: Handle the hydrochloric acid with care. It can cause painful burns if it comes in contact with the skin. Measure out 50.0 mL of NaOH solution, but do not add it to the HCl solution yet. CAUTION: Handle the sodium hydroxide solution with care. Click to begin the data collection and obtain the initial temperature of the HCl solution. After three or four readings have been recorded at the same temperature, add the 50.0 mL of NaOH solution to the Styrofoam cup all at once. Stir the mixture throughout the reaction. Data collection will end after three minutes. If the temperature readings are no longer changing, you may terminate the trial early by clicking . Click the Statistics button, . The minimum and maximum temperatures are listed in the statistics box on the graph. If the lowest temperature is not a suitable initial temperature, examine the graph and determine the initial temperature. Record the initial and maximum temperatures in your data table. Rinse and dry the Temperature Probe, Styrofoam cup, and the stirring rod. Dispose of the solution as directed. Part II - Reaction Between Solutions of NaOH and HAc Measure out 50.0 mL of 2.0 M NaOH solution into a nested Styrofoam cup (see Figure 1). Lower the tip of the Temperature Probe into the cup of NaOH solution. Measure out 50.0 mL of 2.0 M HAc solution, but do not add it to the NaOH solution yet. Click to begin the data collection. After three or four readings have been recorded at the same temperature, add the 50.0 mL of HAc solution to the Styrofoam cup all at once. Stir the mixture throughout the reaction. Data collection will end after three minutes. If the temperature readings are no longer changing, you may terminate the trial early by clicking . Examine the graph as before to determine and record the initial and maximum temperatures of the reaction. Rinse and dry the Temperature Probe, Styrofoam cup, and the stirring rod. Dispose of the solution as directed. 56 Part III - Reaction Between Solutions of HCl and NaAc Measure out 50.0 mL of 2.0 M HCl solution into a nested Styrofoam cup (see Figure 1). Lower the tip of the Temperature Probe into the cup of HCl solution. Measure out 50.0 mL of 2.0 M NaAc solution, but do not add it to the HCl solution yet. Click to begin the data collection. After three or four readings have been recorded at the same temperature, add the 50.0 mL of NaAc solution to the Styrofoam cup all at once. Stir the mixture throughout the reaction. Data collection will end after three minutes. If the temperature readings are no longer changing, you may terminate the trial early by clicking . Examine the graph as before to determine and record the initial and maximum temperatures of the reaction. Rinse and dry the Temperature Probe, Styrofoam cup, and the stirring rod. Dispose of the solution as directed. 57 Name _____________________ Date__________________________ Determining the Enthalpy of a Chemical Reaction using Hess’s Law OBJECTIVE: PROCEDURE DATA TABLE Reaction 1 Reaction 2 Reaction 3 Maximum Temp (°C) Initial Temp (°C) Temperature Change CALCULATIONS 1. Calculate the amount of heat energy, q, produced in each reaction. Use 1.03 g/mL for the density of all solutions. Use the specific heat of water, 4.18 J/(g°C), for all solutions. 2. Calculate the enthalpy change, H, for each reaction in terms of kJ/mol of each reactant. 58 3. Using Hess’ law and your experimentally determined enthalpies for reactions 2 and 3, determine the experimental molar enthalpy for reaction 1. 4. How does your value compare to the accepted value of -56.4 kJ/mol? 5. Does this experimental process support Hess’s law? Suggest ways of improving your results. SOURCES OF ERROR 59 11 EXPERIMENT ELECTROCHEMICAL CELLS INTRODUCTION In this experiment you will investigate reactions in which electrons are transferred from one reactant to another. Since electrons are being transferred we can force them to travel through an electrical connection and thereby be measured. We call these set-ups electrochemical cells. We can observe the effect of electron transfer in other ways. In the first experiment we will observe the formation of elements by electron transfer. EXPERIMENT: For each of the following experiments, record your observations and write an net equation for each reaction. Also determine which of the two substances is the better oxidizer (oxidizers get reduced and are usually called oxidizing agents). Place a small piece of; a) Metallic lead in 5 mL of 1 M Cu(NO3)2. b) Metallic zinc in 5 mL of 1 M Pb(NO3)2. c) Metallic copper in 5 mL of 0.10 M AgNO3. d) Metallic zinc in 5 mL of 1 M HCl. e) Metallic lead in 5 mL of 1 M HCl. f) Metallic copper in HOT 6M HNO3. Heat in a water bath in the hood. (Silver also reacts with HNO3). g) 1 mL of CCl4 and 3 mL of 1 M KI and 3 mL of Bromine water. h) Metallic copper in 5 mL of Bromine water. Heat in a water bath in the hood to remove any excess bromine then add 2 mL of 6 M NH3 to see the blue color of the copper ions. i) 1 mL of CCl4 and 3 mL of 1 M KI and 3 mL of 6 M HNO3. Heat in a water bath. Some of these reactions may be slow, allow them to stand for 15 minutes and look carefully at the surface of metal of some of the slower reactions. In your result section, write all of the half reactions as reductions. Now arrange these reactions in such a way that a reaction will occur when you take a reaction and reverse any of the reactions above it (reversing a reaction makes it an oxidation). In other words, reverse the oxidation reaction and put it on top of the reduction reaction. It is sometimes difficult to see the lead reaction with H+, a reaction does occur so place the lead appropriately. What additional experiments would you need to perform to place bromine and iodine in their proper position? 60 ELECTROCHEMICAL CELLS In this portion of the experiment you will be measuring the voltage produced by three electrochemical cells. The electrodes used in electrochemical cells are usually metals and occasionally carbon rods. The oxidation reaction occurs at the anode and reduction at the cathode so that in electrochemical cells electrons flow from the anode to the cathode and the anode will be negatively charged compared to the cathode. You may work in pairs. EXPERIMENT: Put about 40 mL of 1 M CuSO4 into a 250 mL beaker and about 40 mL of 1 M ZnSO4 into a porous cup. Place the porous cup into the 250 mL beaker containing the CuSO4. Put a copper strip into the copper sulfate solution and a zinc strip into the zinc solution. Attach either a voltmeter to the zinc and copper electrodes and measure the voltage. Draw the cell and determine which electrode is acting as the anode and cathode. Write the reaction occurring at each electrode. EXPERIMENT: Repeat the experiment above using Bromine water and a carbon electrode to replace the copper sulfate solution and the copper electrode. Measure the voltage. Draw the cell and determine which electrode is acting as the anode and cathode and write the reaction occurring at each electrode. MEASUREMENT OF A SOLUBILITY PRODUCT Electrochemical cells can be used to determine the equilibrium constant of chemical reactions. In this lab you will calculate the Ksp of Cu(OH)2 by the following set of half-reactions, Cu(s) + 2 OH- ↔ Cu(OH)2(s) + 2eCu2+(1 M) + 2e- ↔ Cu(s) Cu2+(1 M) + 2 OH- ↔ Cu(OH)2 o = ? From the measured value of o one could evaluate the Ksp for Cu(OH)2(s) using, ° = (0.0592/n) log (1/Ksp) where n is the number of electrons transferred in the reaction. EXPERIMENT: Clean the electrochemical cell used above. Place 40 mL of 1 M CuSO4 in the beaker and 40 mL of 1 M NaOH in the cup. Add 4 drops of 1 M CuSO4 to the 40 mL of NaOH. Place a copper strip into each cell and measure the voltage as previously. Note the negative terminal. Which electrode is the anode? Calculate the Ksp for Cu(OH)2(s). 61 Name__________________ Date________________ Electrochemical Cell Worksheet OBJECTIVE: PROCEDURE: DATA: Ingredients Half-Reactions ox: Pb + Cu(NO3)2 red: ox: Zn + Pb(NO3)2 red: ox: Cu + AgNO3 red: ox: Zn + HCl red: ox: Pb + HCl red: ox: Cu + HNO3 (hot) red: ox: Br2 + KI red: ox: Cu + Br2 red: ox: KI + HNO3 (warm) red: 62 Observations Electrochemical Cell Data: Cu/Zn Cell Voltage = Zn/Br2 Cell Voltage = ___________ Calculation of Ksp for Cu(OH)2 ° = (0.0592/n) log (1/Ksp) Cu(OH)2 Cell Voltage = _____________ 63 RESULTS: Ordered Half-Reactions Cu/Zn Cell Voltage = Zn/Br2 Cell Voltage = ___________ Cu(OH)2 Cell Voltage = _____________ Ksp of Cu(OH)2 = _________________ CONCLUSIONS: 64 PROBLEMS 1) Use your table of oxidizing and reducing agents to predict whether the following reactions will occur as written, or will occur in the opposite direction. a) I2 + Zn(s) = Zn2+ + 2 Ib) 2 NO3- + 8 H+ + 6 Cl- = 2 NO(g) + 4 H2O + 3 Cl2 c) Pb(s) + Cu2+ = Pb2+ + Cu(s) d) Pb2+ + 2 Br- = Pb(s) + Br2 2) You are given the half-reaction for pure water, 2 H+ (10-7M) + 2e- = H2 (g, 1atm) ° = -0.41 eV On a separate piece of paper, rank the following metals according their standard electrode potentials as found in the CRC Handbook, (Na, K, Mg, Al, Zn, Fe, Sn, Pb, Cu, and Ag) by placing the most positive one at the top of the list and then determine which of the following metals should, a) React with water evolving gas. b) React with 1 M HCl but not with water to produce hydrogen gas. c) What reagent and conditions would you use to dissolve the metals that cannot be dissolved in 1 M HCl? 3) Calculate the voltage of an electrochemical cell which uses the reaction, Cu2+(1 M) = Cu2+(0.001 M) 4) Would your results have changed if you had added more drops of CuSO4 to your Ksp cell? Why or why not? 65 12 EXPERIMENT Name ____________________ Date______________ Title: Objective: Procedure: Diagram of Apparatus: Data: Calculations: Results: Conclusion: Sources of Error: 66 Exam I Rate Laws Activation Energies Mechanisms Radioactive Decay Kinetics and Activation Energy 1) Rate information was obtained for the following reaction at 25°C and 33°C; Cr(H2O)63+ + SCN- → Cr(H2O)5NCS2+ + H2O(l) [Cr(H2O)63+] [SCN-] 2.0x10-11 2.0x10-10 9.0x10-10 2.4x10-9 1.0x10-4 1.0x10-3 2.0x10-3 3.0x10-3 0.10 0.10 0.15 0.20 1.4x10-10 1.0x10-4 0.20 Initial Rate @ 33̊C a) Write a rate law consistent with the experimental data. b) What is the value of the rate constant at 25̊C? c) What is the value of the rate constant at 33̊C? d) What is the activation energy for this reaction? e) What is the reverse rate law? 2) The mechanism for the decomposition of phosgene COCl2(g) → CO(g) + Cl2(g) is thought to be, fast eq. slow fast eq. Cl2(g) ⇆ 2 Cl(g) COCl2(g) + Cl(g) → COCl(g) + Cl2(g) COCl(g) ⇆ CO(g) + Cl(g) Based on this mechanism, what is the rate law for this reaction? 3) The following experimental data were obtained for the reaction at 250 K, F2 + 2 ClO2 → 2 FClO2 [F2]/M [ClO2]/M 0.10 0.10 0.20 0.010 0.040 0.010 Rate/(M/sec) 1.2x10-3 4.8x10-3 4.8x10-3 3a) Write a rate law consistent with this data. Rate = 3b) What is the value and units of the rate constant? 4) The bromination of acetone is acid catalyzed; CH3COCH3 + Br2 → CH3COCH2Br + H+ + BrThe rate of disappearance of bromine was measure for several different concentrations of acetone, bromine and hydrogen ions; Rate/Msec-1 6x10-5 6x10-5 1.2x10-4 3.2x10-4 8x10-5 [Acetone]/M 0.30 0.30 0.30 0.40 0.40 [Br2]/M 0.050 0.100 0.050 0.050 0.050 [H+]/M 0.050 0.050 0.100 0.200 0.050 a) What is the forward rate law for this reaction? b) What is the value and units of the forward rate constant? c) What is the reverse rate law? d) If the equilibrium constant is 1.3x103, what is the value of the reverse rate constant? 5) At low temperatures, the rate law for the reaction, CO(g) + NO2(g) → CO2(g) + NO(g) can be determined by the following data; [NO]/10-3 M [CO]/10-3 M [NO2]/10-3 M Initial Rate/10-4 1.20 1.20 1.20 2.40 1.50 3.00 0.75 1.50 0.80 0.80 0.40 0.40 3.60 7.20 0.90 0.90 Write a rate law in agreement with the data. Which of the following mechanisms is consistent with this rate law? (Circle A, B, or C) A) CO + NO2 → CO2 + NO slow B) NO2 ⇆ NO + O equil. O + CO → CO2 slow C) 2 NO2 ⇆ 2 NO + O2 equil. CO + O2 → CO2 + O slow O + NO → NO2 fast D) None of the above 6) The following experimental data were obtained for the reaction; 2 NO + 2 H2 → N2 + 2 H2O Initial Rate/10-5 0.60 2.40 0.30 [NO]/10-2 M 0.50 1.00 0.25 [H2]/10-2 M 0.20 0.20 0.40 a) Write a rate law in agreement with the data. Rate = b) What is the value of the rate constant? c) What is the new rate when [NO] = 3x10-2 M and [H2] = 1.2x10-2 M? d) Which of the following mechanisms are consistent with the rate law above? i) 2 NO + H2 → N2O + H2O (slow) ii) 2 NO ⇆ N2O2 (equil) N2O2 + H2 → N2O + H2O (slow) iii) NO + H2 ⇆ H2O + N (equil) N + NO → N2O (slow) e) Platinum acts as a catalyst for this reaction. What term must be added to the rate law to account for the presence of the catalyst? f) Would platinum be a homogeneous or heterogeneous catalyst for this reaction? g) If the temperature is increased by 6.8°C the reaction rate increases 1.65 times. What is the activation energy of the reaction? 7) The following experimental data were obtained for the reaction at 25°C, 2 NO + H2 → N2O + H2O Initial [NO] /10-3 M 6.40 12.8 6.40 Initial [H2] /10-3 M Initial Rate /10-5 Msec-1 2.20 1.10 4.40 2.60 5.20 5.20 Which of the following mechanisms are consistent with the rate law for this reaction? a) 2 NO + H2 → N2O + H2O (slow) b) 2 NO ⇆ N2O2 (equil) N2O2 + H2 → N2O + H2O (slow) c) NO + H2 ⇆ H2O + N (equil) N + NO → N2O (slow) 8) At low temperatures, the rate law for the reaction, CO(g) + NO2(g) → CO2(g) + NO(g) is; Rate = k [NO2]2 Which of the following mechanisms is consistent with this rate law? (Circle A, B, or C) a) CO + NO2 → CO2 + NO slow b) 2 NO2 ⇆ N2O4 fast equil. N2O4 + 2 CO → 2 CO2 + 2 NO slow c) 2 NO2 → N2O4 N2O4 ⇆ 2 NO + O2 O2 + CO2 → 2 CO2 slow fast equil. fast 9) Given the following rate data, calculate the rate law at 25°C. Rate/10-4 [I-]/M [OCl-]/M 6.10 12.2 36.6 0.20 0.40 0.60 0.050 0.050 0.100 14.4 0.20 0.050 @ 25̊C @ 33̊C What is the rate law? Rate = What is the value of the rate constant at 25°C and at 33°C? What is the activation energy for this reaction? What is the half-life of this reaction at 45°C if [I-] = [OCl-] = 0.25 M? 10) Thiosulfate (S2O32-) can react with triiodide (I3-) according to the following reaction at 25°C, 2 S2O32- + I3- ⇆ S4O62- + 3 I- Keq = 3.75x105 Experimentally the forward rate law can be determined from the following data, Rxn Rate/Msec-1 #1 #2 #3 2.56x10-4 1.28x10-4 1.92x10-4 [S2O32-]/M [I3-]/M 0.040 0.020 0.060 0.12 0.12 0.06 a) What is the rate law? b) What is the value of the rate constant? (Include units) c) What are the minimum number of steps required in the mechanism of the forward rate law? Circle one 1 2 3 d) If reaction #1 is heated to 35°C the rate increases to 4.8x10-4 M/sec. What is the activation energy of this reaction? e) At what temperature would the reaction rate double from 25°C? 11) A cook finds that it takes 30 minutes to boil potatoes at 100°C in an open sauce pan and only 12 minutes to boil them in a pressure cooker at 110°C. Estimate the activation energy for cooking potatoes, which involves the conversion of cellulose into starch. Remember that there is an inverse relationship between time and the rate constant. 12) When N2O4 decomposes it forms NO2, N2O4 → 2 NO2 If the half-life of this reaction is 1386 seconds, how much of a 10 gram sample would be left after 1500 seconds? (Is the reaction first or second order?) 13) An archaeologist measured the amount of radioactivity of a piece of cloth used to wrap an Egyptian mummy. The cloth was found to have a decay rate of 9.1 dpm. If the decay rate is 15.3 dpm in living tissue, how old is the mummy? t1/2 = 5730 years. 14) Uranium is radioactive and decays into lead. This process can be used to date rocks. A piece of zirconium was dated using this process and it was found that this rock contained 3.2x10-3 grams of uranium and 2.2x10-5 grams of lead. If the half-life of uranium is 4.41x109 years how old is the rock? 15) Assuming that the loss of ability to recall learned material is a first-order process with a halflife of 35 days. Compute the number of days required to forget 90% of the material that you learned in preparation for this exam. 16a) Under acidic conditions sucrose (table sugar) can be broken down into its individual sugars, glucose and fructose. At 27°C it takes 54.5 minutes to convert half the sucrose to glucose and fructose and at 37°C it takes 13.7 minutes. Estimate the activation energy for the breakdown of sucrose. 16b) The above reaction is known to second order which means that the half-life is dependent on the initial concentration. Will this fact effect your calculation of the activation energy above? Why or why not. (Hint: What must you assume about the initial concentration of [A] when using the Arrhenius equation?) 17) A 10 gram sample of 131I was sent from a pharmaceutical company to a hospital for use in the treatment of hyperthyroidism. If the half life of 131I is 8.07 days, how much of the sample would be left after a 2 day mail delivery? 18) The denaturation of the virus that causes the rabbit disease Myxomatosis can be followed by heating the virus under a microscope. It is observed that the reaction is first-order and that it takes 22.35 minutes at 50°C and 0.35 minutes at 60°C for the virus to denaturate. Estimate the activation energy for the denaturation of the Myxomatosis virus. Answer Key - Kinetics and Activation Energy 1a) Rate = k[Cr(H2O)63+] [SCN-]2 1b) 2x10-11 = k [1x10-4] [0.1]2 1c) 1.4x10-10 = k [1x10-4] [0.2]2 k = 3.5x10-5 @ 33°C 1d) ln 1e) kF [Cr(H2O)63+] [SCN-] = kR [X] k = 2x10-5 @ 25°C 2 x105 Ea 298 306 5 3.5 x10 8.314 298 306 Keq kF [X] [Cr(H 2O)5 SCN 2 ] k R [Cr(H 2O)36 ][SCN ]2 [Cr(H 2O)36 ][SCN ] [X] = [Cr(H2O)5SCN2+] [SCN-] 2) Ea = 53,033 J/mol rateR = kR[Cr(H2O)5SCN2+] [SCN-] Start with the slow step rate = k [COCl2] [Cl] Get rid of [Cl] using the equilibrium, [Cl] 2 Keq [Cl] Keq[Cl 2 ] [Cl 2 ] 1 1 1 rate k Keq[Cl 2 ][COCl 2 ] kKeq 2 [Cl 2 ] 2 [COCl 2 ] k ' [Cl 2 ] 2 [COCl 2 ] 3a) rate = k[F2]2 [ClO2] 3b) 1.2x10-3 M/s = k [0.1 M]2 [0.01 M] k = 12 1/M2s 4a) rate = k [Acetone] [ H+] 4b) 6x10-5 M/s = k [0.3 M] [ 0.05 M] k = 4x10-3 1/Ms 4c) kF [Acetone] [H+] = kR [X] kF [X] [AcetoneBr][H ][Br ] k R [Acetone] [H ] [Acetone] [Br2 ] [X] [AcetoneBr][H ]2 [Br ] [Br2 ] so, rate R k R [AcetoneBr][H ]2 [Br ] [Br2 ] [CO] [NO 2 ] [NO] 5a) rate k 5b) A) rate = k [CO] [NO2] No, doesn’t match B) rate = k [O] [CO] and Keq [NO] [O] [NO 2 ] so, [O] Keq [NO 2 ] [NO] [CO] [NO 2 ] [CO] [NO 2 ] or rate k' [NO] [NO] This matches our experimentally determined rate law so it is a possible mechanism. Substituting, we get rate kKeq C) rate = k [O2] [CO] The only reason to use the equilibrium is to get rid of something that is not in the overall reaction. Since the rate law does not contain anything that is not already in the overall reaction (no intermediates) there is no reason to continue. This does not match. 6a) rate = k [NO2]2 [H2] 6b) 2.4x10-5 = k [1x10-2 M]2 [0.2x10-2 M]2 so, K = 120 1/M2s 6c) rate = (120 1/M2s) (3x10-2)2 (1.2x10-2) = 1.296x10-3 M/s 6d) Both i & ii are consistent but i) is a single step 3rd order reaction so ii is the answer. 6e) kcat [Pt] + 1 6f) Heterogeneous catalyst. Platinum is a solid. 6g) Assume k1 @ 298 K and 1.65 k1 @ 304.8 K 7) 1.65k1 Ea 304.8 298 k1 8.314 304.8 298 rate = k [NO2]2 [H2] ln Ea = 55,6130 J/mol A & B are consistent but A is a single step third order process so it doesn’t occur. Therefore the answer is B. 8) Only C matches the experimentally determined rate law. 9a) rate = k [I-] [OCl-] 9b) 6.1x10-4 = k [0.20 M] [0.05 M] so, k = 0.061 1/Ms @ 25ºC 14.4x10-4 = k [0.20 M] [0.05 M] so, k = 0.144 1/Ms @ 33ºC 9c) ln 0.144 Ea 306 298 0.061 8.314 306 298 9d) ln 81,399 318 298 0.061 8.314 318 298 K 45 Ea = 81,399 J/mol K45º = 0.4816 t½ = 1/[A]K for 2nd order reaction so, t½ = 1/[0.25](0.4816) = 8.305 seconds 10a) rate = k [S2O32-] [I3-] 10b) 2.56x10-4 M/s = k [0.04 M] [ 0.12 M] so, k = 5.33x10-2 1/Ms 10c) The reaction is 2nd order so it will take a minimum of 1 step. 10d) 2.56x10-4 M/s = k [0.04 M] [ 0.12 M] so, k1 = 5.33x10-2 1/Ms @ 25 C 4.80x10-4 M/s = k [0.04 M] [ 0.12 M] so, k2 = 0.10 1/Ms @ 35 C ln 0.10 Ea 308 298 2 5.33x10 8.314 308 298 10e) ln 2k1 47,969 1 1 k1 8.314 298 T2 11) t1 = 30 min @ 100ºC and t2 = 12 min @ 110ºC 30 min Ea 383 373 12 min 8.314 383 373 10 0.693 ln (1500 sec) X 1386 sec ln 12) 13) ln 15.3dpm 0.693 t 9.1dpm 5730yrs Ea = 47,969 J/mol T2 = 309.6 K Ea = 108,830 J/mol X = 4.72 grams t = 4296 yrs 14) moles U + moles Pb = moles U initially in the sample (Parent) 3.2x103 g U 2.2x105 g Pb 1.3445x10 5 mol U (Parent) and 1.062x10 5 mol Pb (daughter) 238g/mol 207.2 g/mol Total Parent = 1.3445x10-5 mol U + 1.062x10-7 mol Pb = 1.355x10-5 mol U ln 1.355x10-5 mol U 0.693 t -5 1.3445x10 mol U 4.41x109 yrs 15) ln 100 0.693 t 10 35 days 16a) 54.5 min @ 27ºC and 13.7 min @ 37ºC ln 16b) t = 5.01x107 yrs t = 116.3 days = summer vacation! 54.5 min Ea 310 300 13.7 min 8.314 310 300 Ea = 106, 764 J/mol No, it makes no difference. For 1st order reaction t 1 2 For 2nd order reaction t 1 2 0.693 k For both of these t½ 1/k 1 [A] k As long as [A] is the same for all runs, you can directly substitute t½ for 1/k 10 0.693 2 days X 8.07 days 17) ln 18) 22.35 min @ 50ºC and 0.35 min @ 60ºC ln 22.35 min Ea 333 323 0.35 min 8.314 333 323 X = 8.42 grams Ea = 371,706 J/mol Nuclear Chemistry - Radioactive Decay 1. What particle is emitted when a Fr-210 nucleus decays to At-206? (a) alpha (b) beta (c) neutron (d) positron (e) proton 2. What particle is emitted when a Ra-221 nucleus decays to Rn-217? (a) alpha (b) beta (c) neutron (d) positron (e) proton 3. What particle is emitted when a Th-228 nucleus decays to Ra-224? (a) alpha (b) beta (c) neutron (d) positron (e) proton 4. What particle is emitted when a F-20 nucleus decays to Ne-20? (a) alpha (b) beta (c) neutron (d) positron (e) proton 5. What particle is emitted when an Ar-39 nucleus decays to K-39? (a) alpha (b) beta (c) neutron (d) positron (e) proton 6. What particle is emitted when a Sr-90 nucleus decays radioactively to Y-90? (a) alpha (b) beta (c) neutron (d) positron (e) proton 7. What particle is emitted when a carbon-11 nucleus decays to boron-11? (a) alpha (b) beta (c) neutron (d) positron (e) proton 8. What particle is emitted when a fluorine-17 nucleus decays to oxygen-17? (a) alpha (b) beta (c) neutron (d) positron (e) proton 9. What particle is emitted when a neon-19 nucleus decays to fluorine-19? (a) alpha (b) beta (c) neutron (d) positron (e) proton 10. What nuclide is produced when Pt-175 decays by alpha emission? (a) (b) (c) (d) (e) 171 76 175 76 171 78 171 79 175 79 Os Os Pt Au Au 11. What nuclide is produced when U-235 decays by alpha emission? (a) (b) (c) (d) (e) 231 90 235 90 231 92 235 93 231 93 Th Th U Np Np 12. What nuclide is produced when Ra-223 decays by alpha and gamma emission? (a) (b) (c) (d) (e) 219 86 227 86 227 88 219 90 227 90 Rn Rn Ra Th Th 13. What radionuclide decays to Pb-210 by alpha emission? (a) (b) (c) (d) (e) 206 80 214 80 206 82 206 84 214 84 Hg Hg Pb Po Po 14. What nuclide is produced when K-43 decays by beta emission? 43 (a) 18 Ar (b) (c) (d) (e) 42 19 42 20 43 20 44 20 K Ca Ca Ca 15. What nuclide is produced when Pb-210 decays by beta emission? (a) (b) (c) (d) (e) 210 81 212 81 211 82 210 83 211 83 Tl Tl Pb Bi Bi 16. What nuclide is produced when Ar-39 decays by beta and gamma emission? (a) (b) (c) (d) (e) 39 17 40 17 40 18 39 19 40 19 Cl Cl Ar K K 17. What radionuclide decays to Fe-56 by beta emission? (a) (b) (c) (d) (e) 55 25 56 25 55 26 56 27 57 27 Mn Mn Fe Co Co 18. What nuclide is produced when N-13 decays by positron emission? 12 (a) 6 C (b) (c) (d) (e) 13 6 14 6 14 7 13 8 C C N O 19. What nuclide is produced when O-15 decays by positron emission? 14 (a) 7 N (b) (c) (d) (e) 15 7 14 8 15 9 16 9 N O F F 20. What nuclide is produced when K-40 decays by positron emission? 39 (a) 18 Ar (b) (c) (d) (e) 40 18 41 18 40 19 40 20 Ar Ar K Ca 21. What radionuclide decays to Br-73 by positron emission? (a) (b) (c) (d) (e) 72 34 74 34 72 35 74 35 73 36 Se Se Br Br Kr 22. What nuclide is produced when a Cs-129 nucleus decays by electron capture? 128 (a) 54 Xe (b) (c) (d) (e) 129 54 128 55 128 56 129 56 Xe Cs Ba Ba 23. What nuclide is produced when a W-181 nucleus decays by electron capture? 180 (a) 73 Ta (b) (c) (d) (e) 181 73 180 74 180 75 181 75 Ta W Re Re 24. What nuclide is produced when a Mn-52 nucleus decays by electron capture? 52 (a) 24 Cr (b) (c) (d) (e) 53 24 53 25 52 26 53 26 Cr Mn Fe Fe 25. What radionuclide decays to Cs-133 by electron capture? (a) (b) (c) (d) (e) 132 54 133 54 134 55 133 56 134 56 Xe Xe Cs Ba Ba Radioactive Decay Series 26. In the final step of the uranium-238 disintegration series, the parent nuclide decays to lead-206 and an alpha particle. What is the parent nuclide? 202 (a) 80 Hg (b) (c) (d) (e) 210 83 206 84 210 84 Bi Po Po none of the above 27. In the final step of the uranium-235 disintegration series, the parent nuclide decays to lead207 and a beta particle. What is the parent nuclide? 207 (a) 81 Tl (b) (c) 206 82 208 82 207 83 Pb Pb (d) Bi (e) none of the above 28. In the final step of the thorium-232 disintegration series, the parent nuclide decays to lead208 and an alpha particle. What is the parent nuclide? 208 (a) 83 Bi (b) (c) (d) (e) 212 83 208 84 212 84 Bi Po Po none of the above 29. The uranium-238 decay series begins with the emission of an alpha particle. If the daughter decays by beta emission, what is the resulting nuclide? 234 (a) 89 Ac (b) (c) (d) (e) 233 90 234 90 233 91 234 91 Th Th Pa Pa 30. The uranium-235 decay series begins with the emission of an alpha particle. If the daughter decays by beta emission, what is the resulting nuclide? 231 (a) 89 Ac (b) (c) (d) (e) 230 90 231 90 230 91 231 91 Th Th Pa Pa 31. The thorium-232 decay series begins with the emission of an alpha particle. If the daughter decays by beta emission, what is the resulting nuclide? 228 (a) 87 Fr (b) (c) (d) (e) 227 88 228 88 227 89 228 89 Ra Ra Ac Ac Answer Key 1A, 2A, 3A, 4B, 5B, 6B, 7D, 8D, 9D, 10A, 11A, 12A, 13E, 14D, 15D, 16D, 17B, , 18B, 19B, 20B, 21E, 22B, , 23B, 24A, 25D, 26D, 27A, 28D, 29E, 30E, 31E Chemistry 121 First Exam Name______________________ February 17, 2011 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit 1 (20) 2 (25) 3 (15) 4 (15) 5(25) TOTAL Important Equations ln(Ai/Af) = kt t½ = 0.693/k ln(k1/k2) = Ea/R (1/T2 - 1/T1) 1/Af - 1/Ai = kt t½ = 1/[A]k R = 8.314 J/mol-K 1) As you know, ozone (O3) protects us from ultraviolet light in our upper atmosphere. Ozone can break up to form oxygen by the following reaction, 2 O3 → 3 O2 The rate law for this reaction is known to be rate = k [O3]2/[O2]. Which of the following is a possible mechanism for this reaction? Show your work a) 2 O3 → 3 O2 slow b) O3 ⇆ O2 + O fast eq. O3 + O → 2 O2 slow c) O2 ⇆ 2 O fast eq. O3 + O → 2 O2 slow 2) The following experimental data was obtained for the bromination of acetone at 30°C. 2 I- + 2 H+ + H2O2 → I2 + 2 H2O Rate/M/sec 1.64x10-5 3.28x10-5 7.38x10-5 1.64x10-4 [I-]/M 0.15 0.30 0.45 0.20 [H+]/M 0.10 0.10 0.15 0.25 [H2O2]/M 0.10 0.10 0.10 0.30 2a) Write a rate law consistent with this data. Rate = 2b) What is the value and units of the rate constant? 2c) Copper ions (Cu2+) are known to be a catalyst for this reaction. Please write down the catalyzed rate law for this reaction. 3) Cryosurgical procedures involve lowering the body temperature of the patient prior to surgery. The activation energy for the beating of the heart muscle is about 30 kJ/mol (30,000 J/mol) A persons normal body temperature is 98.6ºF (or 37ºC) and most people have a pulse rate of about 75 beats/min. Using this information estimate a person’s pulse if their body temperature is dropped to 72ºF (22ºC). 4) Technetium is a relatively light element that has never been found on earth. As a consequence, the only technetium we have is man-made and radioactive. If a sample of Tc has a decay rate of 983 dpm and exactly 25 minutes later it has a decay rate of 937 dpm, what is halflife of Technetium? 5a) A reaction was set up using concentrations of 1.2 M and 0.30 M. When the half-life was measured one took 3.1 minutes and the other took 3.6 minutes respectively. Was the reaction first or second order? Please explain your choice. 5b) Can a 5th order reaction occur? Please explain. 5c) Please give two reasons why a reaction might be zero order. 5d) Why do reactions tend to speed up when you heat them? Chemistry 121 First Exam Name ______________________ February 17, 2011 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit 1 (20) 2 (25) 3 (15) 4 (15) 5(25) TOTAL ln(Ai/Af) = kt t½ = 0.693/k ln(k1/k2) = Ea/R (1/T2 - 1/T1) 1/Af - 1/Ai = kt R = 8.314 J/mol-K t½ = 1/[A]k 1) As you know, ozone (O3) protects us from ultraviolet light in our upper atmosphere. Ozone can break up to form oxygen by the following reaction, 2 O3 –> 3 O2 The rate law for this reaction is known to be rate = k [O3]2/[O2]. Which of the following is a possible mechanism for this reaction? Show your work a) 2 O3 → 3 O2 slow rate = k [O3]2 does not match b) O3 ⇆ O2 + O fast eq. rate = k [O3] [O] [O ] [O] [O ] Keq 2 [O] Keq 3 [O3 ] [O 2 ] O3 + O → 2 O2 slow Rate = k Keq c) O2 ⇆ 2 O fast eq. O3 + O → 2 O2 slow [O3 ]2 [O ]2 = k’ 3 matches [O2 ] [O2 ] rate = k [O3] [O] [O]2 Keq [O] Keq [O 2 ] [O 2 ] Rate = k Keq½ [O3] [O2]½ = k’ [O3] [O2]½ does not match 2) The following experimental data was obtained for the bromination of acetone at 30°C. 2 I- + 2 H+ + H2O2 → I2 + 2 H2O Rate/M/sec 1.64x10-5 3.28x10-5 7.38x10-5 1.64x10-4 [I-]/M 0.15 0.30 0.45 0.20 [H+]/M 0.10 0.10 0.15 0.25 [H2O2]/M 0.10 0.10 0.10 0.30 2a) Write a rate law consistent with this data. Rate = k [I-] [H+] [H2O2] 2b) What is the value and units of the rate constant? 1.64x10-5 M/s = k [0.15 M] [0.10 M] [0.10 M] k = 0.01093 1/M2s 2c) Copper ions (Cu2+) are known to be a catalyst for this reaction. Please write down the catalyzed rate law for this reaction. Rate = (kcat[Cu2+] + 1) k [I-] [H+] [H2O2] 3) Cryosurgical procedures involve lowering the body temperature of the patient prior to surgery. The activation energy for the beating of the heart muscle is about 30 kJ/mol (30,000 J/mol). A person’s normal body temperature is 98.6ºF (or 37ºC) and most people have a pulse rate of about 75 beats/min. Using this information estimate a persons pulse if their body temperature is dropped to 72ºF (22ºC). ln 75 bpm 30,000 310 295 X bpm 8.314 310 295 X bpm = 41.50 bpm 4) Technetium is a relatively light element that has never been found on earth. As a consequence, the only technetium we have is man-made and radioactive. If a sample of Tc has a decay rate of 983 dpm and exactly 25 minutes later it has a decay rate of 937 dpm, what is halflife of Technetium? ln 983 dpm 0.693 25 min 937 dpm t1/2 t½ = 361.5 min 5a) A reaction was set up using concentrations of 1.2 M and 0.30 M. When the half life was measured one took 3.1 minutes and the other took 3.6 minutes respectively. Was the reaction first or second order? Please explain your choice. The reaction is first order. First order reactions have half-lives that are independent of concentration. Although the concentration changed by a factor of four, the time stayed essentially constant. The small difference in time is experimental error. With a four-fold change in concentration, a 2nd order reaction would have changed by a factor four. 5b) Can a 5th order reaction occur? Please explain. Yes, but it would take at least four steps for a 5th order reaction to occur. 5c) Please give two reasons why a reaction might be zero order. 1. Though required for the reaction to occur, the compound may not be involved in the slow step of the reaction. 2. The component may be a spectator ion and not involved with the reaction at all. 3. Adding more reactant to a saturated catalyst will not cause the reaction to speed up and will appear to be zero order. 5d) Why do reactions tend to speed up when you heat them? Reactions speed up because they are hitting each other more often so more of the molecules have enough energy to exceed the activation energy. Chemistry 121 First Exam Name_______________________ February 21, 2008 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit 1 (20) 2 (35) 3 (35) 4 (10) TOTAL Important Equations ln(Ai/Af) = kΔt t1/2 = 0.693/k 1/Af - 1/Ai = kΔt t1/2 = 1/[A]k ln(k1/k2) = Ea/R (1/T2 - 1/T1) R = 8.314 J/mol-K 1a) A sample of 131I was purified and then analyzed for iodine content. It was found that the sample was 99.8% pure. 25 hours later the sample was tested again and it was found that the sample was now just 78.2% pure. What is the half-life of 131I? 1b) How long would it take for 90% of a sample of 131I to decay? 2) The following experimental data were obtained for the reaction at 295 K, H2 + Br2 → 2 HBr [H2]/M [Br2]/M 0.10 0.40 0.10 0.010 0.010 0.040 Rate/(M/sec) 1.2x10-3 4.8x10-3 2.4x10-3 2a) Write a rate law consistent with this data. Rate = 2b) What is the value and units of the rate constant? 2c) What is the rate of the reaction when [H2] = 0.30 M and [Br2] = 0.025 M? 2d) Which of the following are a possible mechanism for the above reaction? Show your work. a) H2 + Br2 → 2 HBr slow b) Br2 ⇆ 2 Br Br + H2 → HBr + H slow H + Br → HBr fast c) H2 + ½ Br2 → HBr + H slow H + ½ Br2 → HBr fast 3) If a reaction has an activation energy of 112.4 kJ, how hot must you get the reaction to increase the rate of the reaction 1.5 times? Assume an initial temperature of 25ºC. 4a) Why do reactions tend to speed up when you heat them? 4b) What term must be added to a reaction to account for the presence of a catalyst? 5) Assume that the cooking of an egg is a first order process. Assume further that it takes 3 minutes to half-cook (soft boil) an egg at 100ºC. If the activation energy of the cooking of an egg is 75 kJ/mole, how long would it take to soft boil an egg at the top of Mt. Everest where water boils at 80ºC? Chemistry 121 First Exam Name_______________________ February 21, 2008 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit 1 (20) 2 (35) 3 (35) 4 (10) TOTAL Important Equations ln(Ai/Af) = kΔt t1/2 = 0.693/k 1/Af - 1/Ai = kΔt t1/2 = 1/[A]k ln(k1/k2) = Ea/R (1/T2 - 1/T1) R = 8.314 J/mol-K 1a) A sample of 131I was purified and then analyzed for iodine content. It was found that the sample was 99.8% pure. 25 hours later the sample was tested again and it was found that the sample was now just 78.2% pure. What is the half-life of 131I? ln 99.8% 0.693 25 hours 78.2% t1/2 t½ = 71.03 hours (2.96 days) 1b) How long would it take for 90% of a sample of 131I to decay? ln 100% 0.693 t 10% 71.03 t = 236 hours 2) The following experimental data were obtained for the reaction at 295 K, H2 + Br2 → 2 HBr [H2]/M [Br2]/M 0.10 0.40 0.10 0.010 0.010 0.040 Rate/(M/sec) 1.2x10-3 4.8x10-3 2.4x10-3 2a) Write a rate law consistent with this data. Rate = k [H2] [Br2]½ 2b) What is the value and units of the rate constant? 1.2x10-3 M/s = k (0.10 M) (0.010 M) k = 1.2 1/Ms 2c) What is the rate of the reaction when [H2] = 0.30 M and [Br2] = 0.025 M? rate = 1.2 1/Ms (0.30 M) (0.025 M) = 9x10-3 M/s 2d) Which of the following are a possible mechanism for the above reaction? Show your work. a) H2 + Br2 → 2 HBr slow rate = k [H2] [Br2] doesn’t match b) Br2 ⇆ 2 Br Br + H2 → HBr + H slow H + Br → HBr fast rate = k [Br] [H2] [Br]2 Keq [Br] Keq [Br2 ] [Br2 ] rate = k Keq½ [H2] [Br2]½ rate = k’[H2] [Br2]½ possible right answer c) H2 + ½ Br2 —> HBr + H slow H + ½ Br2 —> HBr fast rate = k [H2] [Br2]½ possible right answer 3) If a reaction has an activation energy of 112.4 kJ, how hot must you get the reaction to increase the rate of the reaction 1.5 times? Assume an initial temperature of 25ºC. ln 1.5 k 2 112,400 1 1 k2 8.314 298K T2 T2 = 300.68 K 4a) What do catalysts do to speed up a reaction? They provide an alternate route for the reaction to occur. 4b) What term must be added to a reaction to account for the presence of a catalyst? [kcat [Cat] + 1] 5) Assume that the cooking of an egg is a first order process. Assume further that it takes 3 minutes to half-cook (soft boil) an egg at 100ºC. If the activation energy of the cooking of an egg is 75 kJ/mole, how long would it take to soft boil an egg at the top of Mt. Everest where water boils at 80ºC? t1 = 3 min @ 100ºC and t2 @ 80ºC ln t2 75,000 373 353 3 min 8.314 373 353 t2 = 11.81 min Exam II Chemical Equlibria LeChatlier’s Principle Acid Dissociation Base Hydrolysis Titration Curves Solubility Product pH Controlled Solubility Amphoterism Distribution Coefficients Equilibria and LeChatliers Principle 1) For each of the following sets of compounds write the equilibrium reaction that would occur when the compounds are mixed together. a) HNO3 and NaBenz b) NH4Cl and KOH c) NaCN and NaOH 2a) If H2 and Cl2 are added to a container, both at 2 atm, what will the pressure of HCl be after the system reaches equilibrium? H2 + Cl2 → 2 HCl K = 150 2b) If the equilibrium pressure of HCl is 2 atm what pressures of H2 and Cl2 must have been added to the container originally? 3) If 100 grams of NaF and 70 grams of KOH are added to 250 mL water what is the equilibrium concentration of all ionic species. Ka = 6.76x10-4 4) What is the pH of 100 mL of 1.2 M HAc after 20 mL of 2 M NaOH is added? Ka = 1.8x10-5 5) What is the pH of a solution that is made from 0.10 M HBenz, 0.25 M NaBenz, and 0.25 M KOH? Ka = 6.46x10-5 6) What is the pH of a solution where 100 mL of 0.50 M NH4OH is completely neutralized by 65 mL of HCl? Kb = 1.8x10-5 7) Ca(IO3)2 is a slightly soluble precipitate. What would the concentration of IO3- be in a solution saturated with Ca(IO3)2? Ksp = 1.5 x 10-15 8) What is the pH of a 0.10 M solution of NaAc? Ka = 1.8x10-5 Ac- + H2O ↔ HAc + OH- 9) When ammonia is heated it decomposes to N2 and H2 according to the following reaction, 2 NH3 ↔ N2 + 3 H2 Given 3 atm of NH3 and an equilibrium constant of 3x10-3, what will the final pressure (total pressure) be in the system? 10) What is the pH of a solution made by adding 0.2 mole of NaAc to 250 mL of 1 M acetic acid? Ka = 1.8x10-5 11) A 0.10 M solution of calcium ions can be precipitated with 0.10 M NaOH but not in 0.10 M NH4OH. The solubility product for Ca(OH)2 is 7.88x10-6. Please explain why 0.10 M Ca2+ will not precipitate in 0.10 M NH4OH (Kb = 1.8x10-5). 12) What is the pH of the resulting solution when 0.25 mole of NaCH3CO2 and 0.15 mole HCl are added to 200 mL water? The Ka for CH3COOH is 1.8x10-5M. 13) What is the pH of a solution made by adding 0.30 mole NH3(aq) to 0.50 mole NH4Cl and 0.25 mole KOH? Kb for NH3(aq) = 1.8x10-5M 14) What is the concentration of all ionic and molecular species when you add 30 mL of 0.5 M NaOH to 120 mL of 0.75M HCN? Ka for HCN is 4.8x10-10. 15a) What is the pH of the following solutions when the following amounts of 1.20 M NaOH are added to 160 mL of 3 M dl Aspartic acid? H2Asp ↔ H+ + HAspHAsp- ↔ H+ + Asp2- 1.38x10-4 1.51x10-10 0 mL 200 mL 400 mL 700 mL 800 mL 900 mL 15b) An Aspartic acid buffer of pH 5 was made by adding some NaHAsp to 0.5 moles of H2Asp. How much 1.2 M NaOH or 1.2M HCl must you add to this buffer in order to make a new buffer of pH 4.5? 16) What is the pH of the solution when 75 mL of 0.8 M dl-Histidine is titrated with the following volumes of 1.20 M NaOH? H3His ↔ H+ + H2HisH2His- ↔ H+ + HHis2HHis2- ↔ H+ + His3- pKa1 = 2.40 pKa2 = 6.04 pKa3 = 9.33 0 mL 30 mL 50 mL 75 mL 100 mL 125 mL 135 mL 150 mL 175 mL 17) How much NH4Cl and NH4OH must you add to 250 mL of water to make a buffer of pH = 8.0? Kb = 1.8x10-5 18) How many moles of sodium acetate must be added to 100 mL of 0.25 M acetic acid to make a buffer of pH = 4.0? pKa = 4.74 for acetic acid. 19) Another way of making the buffer from 2 above would be to titrate the acid with a base. What volume of 0.40 M NaOH must be added to 100 mL of 0.25 M acetic acid to make a buffer of pH = 4.0? pKa = 4.74 for acetic acid. 20) How much 0.5 M NaOH must be added to 0.75 M H3PO4 to make a buffer of pH = 8? Ka1 = 2.12, Ka2 = 7.21, Ka3 = 12.67 Solubility Products 21) Are the following molecules acidic, basic or neutral in aqueous solution? NaF Cr(NO3)3 KCl NH4CN acid acid acid acid basic basic basic basic neutral neutral neutral neutral can't tell can't tell can't tell can't tell 22) Are the following compounds soluble or insoluble in water? NaIO3 Cr(OH)3 PbSO4 CaCl2 Soluble Soluble Soluble Soluble Insoluble Insoluble Insoluble Insoluble FeSO4 ScCl3 Na2O PbSO4 Soluble Soluble Soluble Soluble Insoluble Insoluble Insoluble Insoluble 23) What is the solubility of PbBr2 in pure water? Ksp = 4x10-5 24) What is the solubility of Ca(OH)2 in 0.05 M NaOH? Ksp = 5.5x10-6 25) What is the solubility of Ca3(PO4)2 in water? Ksp = 5.87x10-8 26) What is the solubility of Ca3(PO4)2 in 0.5 M Na3PO4? Ksp = 5.87x10-8 27) What is the solubility of AgCl in a solution of 1 M HCl? Ksp = 1.8x10-10 28) At what pH will the concentration of Cu2+ exceed 0.02 M given the following equilibrium? Cu(OH)2 ↔ Cu2+ + 2 OH- Ksp = 2.2x10-20 29) How much NH3(aq) must you add to 100 grams of AgCl in order to dissolve all of the AgCl. Assume a liter of solution and calculate the concentration of NH3(aq). AgCl ↔ Ag+ + Cl- Ksp = 1.8x10-10 Ag(NH3)2+ ↔ Ag+ + 2 NH3(aq) Keq = 1.6x10-9 30) Copper(I) ions in aqueous solution react with NH3 according to, Cu+ + 2 NH3 ↔ Cu(NH3)2+ Keq = 6.3x1010 Calculate the solubility of CuBr (Ksp = 5.3x10-9) in a solution in which the equilibrium concentration of NH3 is 0.185 M. 31) What is the solubility of PbCl2 in a solution of 0.10 M H2S at a pH=0? Ksp =1.6x10-5 for PbCl2, Ksp = 8x10-28 for PbS, Keq = 1x10-20 for H2S (to 2 H+ + S2-). PbS ↔ Pb2+ + S2- Ksp = 8x10-28 H2S ↔ 2 H+ + S2- Keq = 1x10-20 32) Calculate the maximum concentration of Fe3+ in an 0.10 M H2S solution buffered at pH=6. Ksp = 1.6x10-21 for Fe2S3 and Keq = 1x10-20 for H2S (to 2 H+ + S2-). 33) What is the solubility cadmium sulfide (CdS) in a saturated solution of H2S = 0.10 M at pH = 3? CdS ↔ Cd2+ + S2- Ksp = 8x10-27 H2S ↔ 2H+ + S2Keq = 1x10-20 34) What is the solubility (silver ion concentration) of silver sulfide (Ag2S) in a saturated solution of H2S = 0.10 M at pH = 3? Ag2S ↔ 2 Ag+ + S2H2S ↔ 2H+ + S2- Ksp = 1.6x10-49 Keq = 1x10-20 35) How much of each precipitate will form and what will the concentration of lead be if 0.75 mole sodium oxalate (Na2C2O4 = Na2Ox) is added to 100 mL of solution containing 0.30 M Mg(NO3)2 and 0.5 M Pb(NO3)2? MgOx(s) ↔ Mg2+ + Ox2- Ksp = 8.6x10-5 PbOx(s) ↔ Pb2+ + Ox2- Ksp = 4.8x10-10 36) What is the solubility of silver carbonate (Ag2CO3) in a solution saturated with carbonic acid (0.034M) and which is buffered at pH = 9? Ag2CO3 ↔ 2 Ag+ + CO32H2CO3 ↔ 2 H+ + CO32- Ksp = 8.1x10-12 Ka = 2.02x10-17 Equilibria and LeChatliers Principle 1) For each of the following sets of compounds write the equilibrium reaction that would occur when the compounds are mixed together. a) HNO3 and NaBenz b) NH4Cl and KOH c) NaCN and NaOH HBenz ↔ H+ + BenzNH4OH ↔ NH4+ + OHCN- + H2O ↔ HCN + OH- 2a) If H2 and Cl2 are added to a container, both at 2 atm, what will the pressure of HCl be after the system reaches equilibrium? H2 + Cl2 ↔ 2 HCl K = 150 x x 4 atm – 2x Since the equilibrium constant is large (150) we must shift the ingredients to the right. This means that 2 atm H2 and 2 atm Cl2 will produce 4 atm HCl. Using this value we can now solve the equilibrium [4 2 x ]2 150 x2 You do not have to assume that 2x is small. You can solve this exactly. Take the square root of both sides and then solve for x. So, x = 0.2808 atm so the pressure of HCl = 4 atm – 2(0.2808 atm) = 3.44 atm 2b) If the equilibrium pressure of HCl is 2 atm what pressures of H2 and Cl2 must have been added to the container originally? [2atm]2 150 x2 x = 0.1633 atm H2 and Cl2 left after equilibrium. To make 2 atm of HCl, 1 atm H2 and 1 atm Cl2 must have been consumed. Therefore there must have been 1 atm + 0.1633 = 1.1633 atm H2 and Cl2 to start. 3) If 100 grams of NaF and 70 grams of KOH are added to 250 mL water what is the equilibrium concentration of all ionic species. Ka = 6.76x10-4 Note: HF is not ionic but was included. 100g NaF/42 g/mol = 2.381 mol NaF and 70g KOH/56.1 g/mol = 1.248 mol KOH [ HF ][OH ] [ HF ][4.99 x] 1x10 14 F- + H2O ↔ HF + OH [ HF ] 2.822 x10 11 4 [F ] [9.52 x] 6.76 x10 [H+] = 1x10-14/[OH-] = 1x10-14/[4.99 M] = 2x10-15 M H+ Ion Moles Before Rxn Moles After Rxn Conc (250 mL total) Na+ FK+ OHHF H+ 2.381 2.381 1.248 1.248 0 0 2.381 2.381 1.248 1.248 9.52 M 9.52 M 4.99 M 4.99 M 2.822x10-11 2x10-15 From equilibrium From equilibrium 4) What is the pH of 100 mL of 1.2 M HAc after 20 mL of 2 M NaOH is added? Ka = 1.8x10-5 (1.2M HAc)(0.100 L) = 0.12 mol HAc and (2M)(0.020L) = 0.040 mol NaOH HAc + NaOH ↔ H2O + NaAc 0.12 0.040 0 -0.04 -0.040 +0.040 0.08 0 0.040 HAc ↔ H+ + Ac0.08-x x 0.04+x [ H ][ Ac ] [ H ][0.040 x] 1.8 x105 [ H ] 3.6 x105 pH = 4.44 [ HAc ] [0.080 x] 5) What is the pH of a solution that is made from 0.10 M HBenz, 0.25 M NaBenz, and 0.25 M KOH? Ka = 6.46x10-5 Benz- + H2O ↔ OH- + 0.25 0.25 +0.10 -0.10 0.35-x 0.15+x HBenz 0.10 -0.10 0+x If you ignore x, [OH-] = 0.15 M and [H+] = 1x10-14/0.15 = 6.66x10-14 so pH = 13.18 6) What is the pH of a solution where 100 mL of 0.50 M NH4OH is completely neutralized by 65 mL of HCl? Kb = 1.8x10-5 If the NH4OH is completely neutralized, then the only thing left in solution is NH4Cl (an acid). The only question is, what is the concentration of the NH4Cl? Total volume = 100 mL + 65 mL (0.50 M)(0.100 L) = 0.050 mol NH4OH = 0.050 mol NH4Cl 0.050 mol NH 4 Cl 0.3030 M NH 4Cl 0.165 L NH4+ ↔ H+ + NH3 0.3030-x x x [H ][NH 3 ] x2 5.55x1010 x 1.297 x105 M H pH = 4.89 [NH 4 ] 0.3030 x 7) Ca(IO3)2 is a slightly soluble precipitate. What would the concentration of IO3- be in a solution saturated with Ca(IO3)2? Ksp = 1.5 x 10-15 [Ca2+] [IO3-]2 = x(2x)2 = 4x3 = 1.5x10-15 x = 7.21x10-6 8) What is the pH of a 0.10 M solution of NaAc? Ka = 1.8x10-5 Ac- + H2O ↔ HAc + OH[OH ][HAc] x2 5.55x1010 x 7.45x106 M OH pH = 8.87 [Ac ] 0.1 x 9) When ammonia is heated it decomposes to N2 and H2 according to the following reaction, 2 NH3 ↔ N2 + 3 H2 Given 3 atm of NH3 and an equilibrium constant of 3x10-3, what will the final pressure (total pressure) be in the system? 2 NH3 ↔ N2 + 3 H2 3 – 2x x 3x x(3 x)3 27 x 4 3x103 (3 2 x ) 2 (3 2 x ) 2 ignore the 2x on the bottom and solve. x = 0.1778 atm Total pressure = 3-2x+x+3x = 3+2x = 3+2(0.1778) = 3.3556 atm 10) What is the pH of a solution made by adding 0.2 mole of NaAc to 250 mL of 1 M acetic acid? Ka = 1.8x10-5 (1 M)(0.250 L) = 0.250 mol HAc Use the HH equation pH = 4.74 + log 0.20/0.25 pH = 4.64 11) A 0.10 M solution of calcium ions can be precipitated with 0.10 M NaOH but not in 0.10 M NH4OH. The solubility product for Ca(OH)2 is 7.88x10-6. Please explain why 0.10 M Ca2+ will not precipitate in 0.10 M NH4OH (Kb = 1.8x10-5). A 0.10 M Ca2+ solution will require. [Ca2+] [OH-]2 = [0.10 M] [OH-]2 = 7.88x10-6 [OH-] = 0.00888 M OH- to precipitate 0.10 M NH4OH can only produce, [OH ][NH 4 ] x2 1.8x105 x 0.001342 M OH this is not enough to ppt. Ca2+ [NH 4OH] 0.1 x 12) What is the pH of the resulting solution when 0.25 mole of NaCH3CO2 and 0.15 mole HCl are added to 200 mL water? The Ka for CH3COOH is 1.8x10-5M. Ac- + H+ → HAc [H ][Ac ] [H ][0.10 x] 0.25 0.15 1.8x105 [H ] 2.7 x105 pH = 4.57 [HAc] [0.15 x] -0.15 -0.15 +0.15 0.10 0 0.15 13) What is the pH of a solution made by adding 0.30 mole NH3(aq) to 0.50 mole NH4Cl and 0.25 mole KOH? Kb for NH3(aq) = 1.8x10-5M NH4OH ↔ NH4+ + OH0.30 0.50 0.25 Shift +0.25 -0.25 -0.25 0.55 0.25 0 → trickle 0.55-x 0.25+x 0+x [OH ][NH 4 ] x(0.25 x) 1.8x105 x 3.96 x105 M OH [NH 4OH] 0.55 x pH = 9.60 14) What is the concentration of all ionic and molecular species when you add 30 mL of 0.5 M NaOH to 120 mL of 0.75M HCN? Ka for HCN is 4.8x10-10. (0.5 M)(0.030 L) = 0.015 mol NaOH and (0.75 M)(0.120 L) = 0.090 mol HCN HCN + OH- ↔ H2O + CN0.09 0.015 -0.015 -0.015 +0.015 0.075 0 0.015 [H+] = 2.4x10-9 Ion + Na OHHCN H+ CN- [ H ][CN ] [ H ][0.015 x ] 4.8 x10 10 [ HCN ] [0.075 x] [OH-] = 1x10-14/[2.4x10-9] = 4.17x10-6 M OHMoles Before Rxn 0.015 0.015 0.090 0 0 Moles After Rxn Conc (150 mL total) 0.015 0.10 M 4.17x10-6 0.05 M 2.4x10-9 0.05 M From equilibrium 0.075 From equilibrium 0.015 15a) What is the pH of the following solutions when the following amounts of 1.20 M NaOH are added to 160 mL of 3 M dl Aspartic acid? H2Asp ↔ H+ + HAspHAsp- ↔ H+ + Asp2- 0 mL 1.38x10-4 1.51x10-10 pKA1 = 3.86 pKA2 = 9.82 x2 1.38x104 x 0.002035 M H pH = 1.69 3 x M1V1 = M2V2 (3 M)(0.160 L) = (1.20 M)(V2) V2 = 400 mL Also, (3 M)(0.160 L) = 0.48 mol H 2Asp 200 mL 400 mL 700 mL pH = 3.86 pH pK A1 pK A2 3.86 9.82 2 2 300 mL into 2nd Region (1.2 M)(0.300 L) = 0.36 mol OHpH 9.82 log 800 mL [Asp2- ] 0.36 0.48 0.36 pH = 10.30 0.48 mol 1x10-14 0.50 M and Kb 6.62 x10 5 0.960 L 1.51x10-10 x2 6.62 x105 0.50 x 900 mL pH = 6.84 x = 5.75x10-3 M OH- (0.100L)(1.20M) = 0.12 mol OH0.12 mol OH 0.1132 M OH 1.060 L pH = 11.76 pH = 13.05 15b) An Aspartic acid buffer of pH 5 was made by adding some NaHAsp to 0.5 moles of H2Asp. How much 1.2 M NaOH or 1.2M HCl must you add to this buffer in order to make a new buffer of pH 4.5? First, calculate the composition of the existing buffer, pH pKa log B A [ HAsp ] [HAsp-] = 6.90 mol 0.5 Now find out how much acid must be added to make new buffer (pH is going down so add acid). 5 3.86 log 4.5 3.86 log 6.90 x 0. 5 x x = 0.8792 mol H+ = (1.2 M)(V) V = 732.7 mL of 1.2 M HCl 16) What is the pH of the solution when 75 mL of 0.8 M dl-Histidine is titrated with the following volumes of 1.20 M NaOH? H3His ↔ H+ + H2HisH2His- ↔ H+ + HHis2HHis2- ↔ H+ + His3- M1V1 = M2V2 (0.8 M)(0.075 L) = (1.20 M)(V2) V2 = 50 mL pKa1 = 2.40 pKa2 = 6.04 pKa3 = 9.33 Also, (0.8 M)(0.075 L) = 0.060 mol H 3His 0 mL x2 3.98x103 x 0.0564 M H 0. 8 x 30 mL (1.2 M)(0.030 L) = 0.036 mol OHpH 2.40 log 0.036 0.060 0.036 pK A1 pK A2 2.40 6.04 2 2 pH = 1.248 pH = 2.58 50 mL pH 75 mL pH = 6.04 100 mL pH 125 mL pH = 9.33 135 mL 35 mL into 3rd Region (1.2 M)(0.035 L) = 0.042 mol OH- pK A2 pK A3 6.04 9.33 2 2 pH 9.33 log 150 mL 0.042 0.060 0.042 pH = 7.69 pH = 9.70 0.060 mol 1x10-14 [His ] 0.2667 M and Kb 2.14 x10 5 -10 0.225 L 4.68x10 3- x2 2.14 x105 0.2667 x 175 mL pH = 4.22 x = 2,39x10-3 M OH- (0.025 L)(1.20M) = 0.030 mol OH0.030 mol OH 0.120 M OH pH = 13.08 0.250 L pH = 11.38 17) How much NH4Cl and NH4OH must you add to 250 mL of water to make a buffer of pH = 8.0? Kb = 1.8x10-5 Using the base version of the HH equation; 6 4.74 log [NH 4Cl] [NH 4Cl] 18.19 0.1819 [NH 4OH] [NH 4 OH] 1 0.01 Adding 18.19 moles to 250 mL of water will make a solution that is 72.76M, so divide top and bottom by 100. Add 0.1819 mol NH4Cl and 0.010 mole NH4OH to 250 mL of water to make a buffer of pH = 8 18) How many moles of sodium acetate must be added to 100 mL of 0.25 M acetic acid to make a buffer of pH = 4.0? pKa = 4.74 for acetic acid. (0.25 M HAc)(0.100 L) = 0.025 mol HAc (x) 4 4.74 log x 0.00455 mol NaAc (0.025 mol HAc) 19) Another way of making the buffer from 18 above would be to titrate the acid with a base. What volume of 0.40 M NaOH must be added to 100 mL of 0.25 M acetic acid to make a buffer of pH = 4.0? pKa = 4.74 for acetic acid. 4 4.74 log (x) x 0.003849 mol NaOH (0.025 - x) (0.40 M)(V) = 0.003849 mol NaOH V = 9.62 mL 20) How much 0.5 M NaOH must be added to 80 mL of 0.75 M H3PO4 to make a buffer of pH = 8? Ka1 = 2.12, Ka2 = 7.21, Ka3 = 12.67 The pH is nearest the 2nd pKa so the buffer is in Region II. You must add, (0.75 M)(0.080 L) = (0.50 M)(V) V = 120 mL so add 120 mL to get to the starting point. 8 7.21 log (x) x 0.05163 mol NaOH = (0.50 M)(V) (0.060 - x) Total volume = 103.25 mL + 120 mL = 223.35 mL NaOH V = 103.25 mL Solubility Products 21) Are the following molecules acidic, basic or neutral in aqueous solution? NaF Cr(NO3)3 KCl NH4CN acid acid acid acid basic basic basic basic neutral neutral neutral neutral can't tell can't tell can't tell can't tell 22) Are the following compounds soluble or insoluble in water? NaIO3 Cr(OH)3 PbSO4 CaCl2 Soluble Soluble Soluble Soluble Insoluble Insoluble Insoluble Insoluble FeSO4 ScCl3 Na2O Hg2SO4 Soluble Soluble Soluble Soluble Insoluble Insoluble Insoluble Insoluble 23) What is the solubility of PbBr2 in pure water? Ksp = 4x10-5 [Pb2+] [Br-]2 = x(2x)2 = 4x3 = 4x10-5 x = 2.15x10-2 M 24) What is the solubility of Ca(OH)2 in 0.05 M NaOH? Ksp = 5.5x10-6 [Ca2+] [0.50 M]2 = 5.5x10-6 x = 2.20x10-5 25) What is the solubility of Ca3(PO4)2 in water? Ksp = 5.87x10-8 [Ca2+]3 [PO43-]2 = (3x)3(2x)2 = 108x5 = 5.87x10-8 x = 1.40x10-2 M 26) What is the solubility of Ca3(PO4)2 in 0.5 M Na3PO4? Ksp = 5.87x10-8 [Ca2+]3 [PO43-]2 = (3x)3(0.50)2 = 0.75x3 = 5.87x10-8 x = 4.278x10-3 M 27) What is the solubility of AgCl in a solution of 1 M HCl? Ksp = 1.8x10-10 [Ag+] [1 M] = 1.8x10-10 [Ag+] = 1.8x10-10 28) At what pH will the concentration of Cu2+ exceed 0.02 M given the following equilibrium? Cu(OH)2 ↔ Cu2+ + 2 OH- Ksp = 2.2x10-20 [Cu2+] [OH-]2 = [0.02 M] [OH-]2 = 2.2x10-20 [OH-] = 1.05x10-9 pH = 6.02 29) How much NH3(aq) must you add to 100 grams of AgCl in order to dissolve all of the AgCl. Assume a liter of solution and calculate the concentration of NH3(aq). AgCl ↔ Ag+ + Cl- Ksp = 1.8x10-10 Ag(NH3)2+ ↔ Ag+ + 2 NH3(aq) Keq = 1.6x10-9 If you reverse the bottom reaction and add it to the top reaction you get the following (see below). AgCl ↔ Ag+ + ClKsp = 1.8x10-10 + + Ag + 2 NH3(aq) ↔ Ag(NH3)2 Keq = 6.25x10-8 AgCl + 2 NH3(aq) ↔ Ag(NH3)2+ + ClK = 0.1125 This means that all the silver in solution will end up being Ag(NH3)2+. If you dissolve all of the AgCl in 1 liter of solution, the concentration of Ag(NH3)2+ must be, 100 g AgCl/143.35 g/mol = 0.6976 mole AgCl => 0.6976 mole Ag+ becomes, 0.6976 mole Ag(NH3)2+ and 0.6976 mole Cl- in 1 liter of solution. So, [Ag(NH3)2+] = 0.6976 M and [Cl-] = 0.6976 M Solving for the concentration of NH3(aq), [Ag(NH 3 ) 2 ] [Cl ] [0.6976 M] [0.6976 M] 0.1125 [NH3] = 2.08 M [NH 3 ]2 [NH3 ]2 30) Copper(I) ions in aqueous solution react with NH3 according to, Cu+ + 2 NH3 ↔ Cu(NH3)2+ Keq = 6.3x1010 Calculate the solubility of CuBr (Ksp = 5.3x10-9) in a solution in which the equilibrium concentration of NH3 is 0.185 M. CuBr ↔ Cu+ + BrCu+ + 2 NH3 ↔ Cu(NH3)2+ CuBr + 2 NH3 ↔ Cu(NH3)2+ + Br- Ksp = 5.3x10-9 Keq = 6.3x1010 K = 333.9 For every Cu(NH3)2+ there must be one Br-, so [Cu(NH3)2+] = [Br-] = x x2 x2 333.9 x = 3.38 M CuBr [NH 3 ]2 [0.185]2 31) What is the solubility of PbCl2 in a solution of 0.10 M H2S at a pH=0? Ksp =1.6x10-5 for PbCl2, Ksp = 8x10-28 for PbS, Keq = 1x10-20 for H2S (to 2 H+ + S2-). PbS ↔ Pb2+ + S2H2S ↔ 2 H+ + S2- Ksp = 8x10-28 Keq = 1x10-20 Note: the Pb2+ concentration can only be as high as the least soluble compound so the solubility of PbS will determine the concentration of Pb2+ [H+]2 [S2-]/[H2S] = 1x10-20 [S2-] = 1x10-20 [H2S]/[H+]2 = 1x10-20 [0.1]/[1]2 = 1x10-21 M S2Substitute this result into the following equation, [Pb2+] [S2-] = 8x10-28 [Pb2+] = 8x10-28/[S2-] = 8x10-28/[1x10-21] = 8x10-7 M Now, all equilibriums must be satisfied simultaneously. Since this is the maximum concentration of Pb2+ that can be in solution, we can use this value to calculate the Clconcentration from the PbCl2. PbCl2(s) ↔ Pb2+ + 2 Cl[Pb2+] [Cl-]2 = 1.6x10-5 => [8x10-7 M] [Cl-]2 = 1.6x10-5 [Cl-] = 4.47 M This concentration of Cl- is fairly high. It can only be this high if all of the PbCl2 dissolved. So, if you put H2S in a solution of PbCl2, all of the PbCl2 would dissolve and become PbS. 32) Calculate the maximum concentration of Fe3+ in an 0.10 M H2S solution buffered at pH=6. Ksp = 1.6x10-21 for Fe2S3 and Keq = 1x10-20 for H2S (to 2 H+ + S2-). [H+]2 [S2-]/[H2S] = 1x10-20 [S2-] = 1x10-20 [H2S]/[H+]2 = 1x10-20 [0.1]/[1x10-6]2 = 1x10-9 M S2Substitute this result into the following equation, Fe2S3(s) ↔ 2 Fe3+ + 3 S2[Fe3+]2 [S2-]3 = 1.6x10-21 [Fe3+] = (1.6x10-21/[S2-]3)½ = (1.6x10-21/[1x10-9]3)½ = 1264.9 M Fe3+ => 632.45 M Fe2S3 33) What is the solubility cadmium sulfide (CdS) in a saturated solution of H2S = 0.10 M at pH = 3? CdS ↔ Cd2+ + S2- Ksp = 8x10-27 H2S ↔ 2H+ + S2Keq = 1x10-20 [H+]2 [S2-]/[H2S] = 1x10-20 [S2-] = 1x10-20 [H2S]/[H+]2 = 1x10-20 [0.1]/[1x10-3]2 = 1x10-15 M S2Substitute this result into the following equation, [Cd2+] [S2-] = 8x10-27 [Cd2+] = 8x10-27/[S2-] = 8x10-27/[1x10-17] = 8x10-12 M 34) What is the solubility (silver ion concentration) of silver sulfide (Ag2S) in a saturated solution of H2S = 0.10 M at pH = 3? Ag2S ↔ 2 Ag+ + S2- Ksp = 1.6x10-49 H2S ↔ 2H+ + S2Keq = 1.1x10-20 [H+]2 [S2-]/[H2S] = 1x10-20 [S2-] = 1x10-20 [H2S]/[H+]2 = 1x10-20 [0.1]/[1x10-3]2 = 1x10-15 M S2Substitute this result into the following equation, [Ag+]2 [S2-] = 1.6x10-49 [Ag+] = 1.6x10-49/[S2-] = 1.6x10-49/[1x10-17] = 1.6x10-32 M Ag+ 35) How much of each precipitate will form and what will the concentration of lead be if 0.075 mole sodium oxalate (Na2C2O4 = Na2Ox) is added to 100 mL of solution containing 0.30 M Mg(NO3)2 and 0.5 M Pb(NO3)2? MgOx(s) ↔ Mg2+ + Ox2PbOx(s) ↔ Pb2+ + Ox2- Ksp = 8.6x10-5 Ksp = 4.8x10-10 Neither compound is very soluble so you can begin by assuming that 100% of the Pb2+ and Mg2+ will react to form solids. (0.30 M)(0.100 L) = 0.03 mole Mg(NO3)2 (0.50 M)(0.100 L) = 0.05 mole Pb(NO3)2 Start with the least soluble compound (PbOx), You have 0.075 mole Ox2- and 0.50 mole of Pb2+, so all of the Pb2+ will react to make 0,50 mole of PbOx and leave 0.025 mole of Ox2- in solution. The remaining 0.025 mole of Ox2- will react with the 0.03 mole Mg(NO3)2 leaving 0.005 mole of Mg2+ in solution and no Ox2-. The [Mg2+] = (0.005 mol)(0.100 L) = 0.05 M Mg2+. [Mg2+] [Ox2-] = [0.05 M] [Ox2-] = 8.6x10-5 [Ox2-] = 1.72x10-3 M [Pb2+] [Ox2-] = [Pb2+] [1.72x10-3 M] = 4.8x10-10 [Pb2+] = 2.79x10-7 M 36) What is the solubility of silver carbonate (Ag2CO3) in a solution saturated with carbonic acid (0.034M) and which is buffered at pH = 9? Ag2CO3 ↔ 2 Ag+ + CO32H2CO3 ↔ 2 H+ + CO32- Ksp = 8.1x10-12 Ka = 2.02x10-17 [H ]2 [CO32 ] (1x109 M) 2 (CO32 ) 2.02x1017 [CO32 ] 0.6868 M [H 2CO3 ] 0.034M [Ag+]2 [CO32-] = [Ag+]2 [0.6868 M] = 8.1x10-12 [Ag+] = 3.43x10-6 M Solubility of Ag2CO3 = (3.43x10-6 M)/2 = 1.72x10-6 M Ag2CO3 Distribution Coefficients 1. If 0.500 grams of caffeine is dissolved in 100 mL of water, what percentage of caffeine can be separated from the water using a single 40 mL sample of methylene chloride. P = 4.6 2. Benzoic acid can be separated from water using octanol as the organic solvent. The distribution coefficient for this water/octanol system is P = 1.87. Assuming that 1 gram of benzoic acid has been dissolved in 100 mL of water, how many 20 mL extractions must be done to extract 60+ percent of the benzoic acid from the water? 3. A 0.100 gram sample of phthalic acid was dissolved in 100 mL of water. When 25 mL of diethyl ether was used to extract the phthalic acid, 0.042 grams of phthalic acid were recovered. What is the distribution coefficient for this extraction? 4. Draw a graph (using Excel) that shows the percentage of solute (extractable compound) extracted from water using increasingly large volumes of organic solvent. Express the amount extracted as a percent of the total. Make the spreadsheet very general. Allow for the input of various distribution coefficients, amount of dissolved compound, and volume of aqueous solution. ݈2ܥ2ܪܥܮ ݉ ⁄40ݔ ܱ2ܪܮ ݉ )⁄100ݔ (0.5 − ݔ ݔ 0.5 − = 1.84 1. = 4.6 0.92 – 1.84x = x 0.92 = 2.84x 0.92 ݂݂ܽܥ݃= 0.3239 2.84 =ݔ ݃0.3239 ݀݁ݐܿܽݎݐݔ݂݂݁݁݊݅݁ܽܥ× 100 = 64.78 % ݃0.500 ݈݊ܽݐܱܿܮ ݉ ⁄20ݔ ܱ2ܪܮ ݉ )⁄100ݔ (1 − ݔ ݔ1− 2. First extraction = 1.87 = 0.374 0.374 – 0.374x = x 0.374 = 1.374x 0.374 ݏ݊݅ܽ ݉݁ݎ݃0.7278ݏ ݅ܿܽܿ݅݀,ݖ݊݁ܤ݃= 0.2722 1.374 ݃0.2722 ݀݁ݐܿܽݎݐݔ݁ܿ݅ݖ݊݁ܤ× 100 = 27.22 % ݃1 ݈݊ܽݐܱܿܮ ݉ ⁄20ݔ ܱ2ܪܮ ݉ )⁄100ݔ (0.7278 − ݔ ݔ 0.7278 − =ݔ Second extraction = 1.87 = 0.374 0.2722 – 0.374x = x 0.2722 = 1.374x 0.2722 ݏ݊݅ܽ ݉݁ݎ݃0.5291ݏ ݅ܿܽܿ݅݀,ݖ݊݁ܤ݃= 0.1987 1.374 =ݔ Third extraction 0.2722 + 0.1987݃ × 100 = 47.09 %݀݁ݐܿܽݎݐݔ݁ܿ݅ݖ݊݁ܤ 1݃ 1.87 = ݔ⁄20 ݉ ݈݊ܽݐܱܿܮ (0.5291 − )ݔ⁄100 ݉ ܪܮ2ܱ 0.374 = ݔ 0.5291 − ݔ 0.1979 – 0.374x = x 0.1979 = 1.374x =ݔ 3. 0.1979 = 0.1440݃݀݅ܿܽܿ݅ݖ݊݁ܤ, ݏ0.3851݃ݏ݊݅ܽ ݉݁ݎ 1.374 0.2722 + 0.1987 + 0.1440݃ × 100 = 61.49 %݀݁ݐܿܽݎݐݔ݁ܿ݅ݖ݊݁ܤ 1݃ ܲ= 0.042݃⁄25 ݉ ݐ݁ܮℎ݁ݎ = 2.897 (0.100 − 0.042݃)⁄100 ݉ ܪܮ2ܱ 4. You should begin by solving the extraction equation so that the grams of extracted solute is in terms of the volume of extracting solvent. ܲ= ݃݁݁ݐݑ݈ݏ݀݁ݐܿܽݎݐݔ⁄݈ݒ. ݁ݐ݊݁ݒ݈ݏ݃݊݅ݐܿܽݎݐݔ (݃݁ݐݑ݈ݏ− ݃݁)݁ݐݑ݈ݏ݀݁ݐܿܽݎݐݔ⁄݈ݒ. ݎ݁ݐܽݓ It will be difficult to solve this equation using the long words to describe each part so let’s simplify the variables, g extracted solute = gx g solute = gs vol. extracting solvent = vx vol. water = vw Our equation is now, ܲ= ݃ݔ⁄ݔݒ (݃ݏ− ݃)ݔ⁄ݓݒ We want to rearrange this equation so that we have gx in terms of vx. So, rearranging, ܲ(݃ݏ− ݃ݔ݃ )ݔ = ݓݒ ݔݒ ܲ ∙ ݏ݃(ݔݒ− ݃ݓݒ ∙ ݔ݃ = )ݔ ܲ ∙ ݏ݃ ∙ ݔݒ− ܲ ∙ ݓݒ ∙ ݔ݃ = ݔ݃ ∙ ݔݒ ܲ ∙ ݓݒ ∙ ݔ݃ = ݏ݃ ∙ ݔݒ+ ܲ ∙ ݔ݃ ∙ ݔݒ ܲ ∙ ݓݒ( = ݏ݃ ∙ ݔݒ+ ܲ ∙ ݔ݃ ∙ )ݔݒ ܲ ∙ ݏ݃ ∙ ݔݒ = ݃ݔ ( ݓݒ+ ܲ ∙ )ݔݒ This is our equation. We can test is by using the number given in problem #1. P = 4.6 vx = 40 mL gs = 0.500 gram sample vw = 100 mL 4.6 ∙ 40 ∙ 0.500 = 0.3239݂݃݁ݐݑ݀݁ݐܿܽݎݐݔ0.5000 (64.78%) (100 + 4.6 ∙ 40) Finally, we can calculate the percentage directly by dividing our answer (gx) by the grams of sample (gs) and multiplying by 100. ܲ ∙ ݏ݃ ∙ ݔݒ ( ݓݒ+ ܲ ∙ )ݔݒ × 100 = %݁݀݁ݐܿܽݎݐݔ ݃ݏ So now, we need to make an Excel spreadsheet to do this calculation for us for various volumes of extracting solvent. The Excel file that does this for you is online with this problem set. You will notice that the amount extracted increases with the volume of extracting solvent but it also increases when you lower the volume of the water. So, it is easier to extract stuff out of a more concentrated water solution. You will also get more extracted if the distribution coefficient increases. Chemistry 121 Second Exam Name_____________ April 18, 2007 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit 1(10) 2(15) 3(45) 4(20) 5(10) Total 1) Which of the following compounds is the most soluble in water? Circle one. Show your work. BaSO4 Ksp = 1.6x10-10 PbI2 Ksp = 1.2x10-13 Ag3PO4 Ksp = 1x10-22 2) I found a bottle marked “Acetic Acid” in our back room, but there was no indication of concentration. I wanted to know the concentration so I measured the pH and found it to be pH = 2.87. What was the concentration of the acetic acid in the bottle? Ka = 1.8x10-5 3) Calculate the pH of 100 mL of 1.2 M Arsenous acid upon addition of the following amounts of 0.8 M NaOH H3AsO4 ↔ H2AsO4- + H+ H2AsO4- ↔ HAsO42- + H+ HAsO42- ↔ AsO43- + H+ Ka1= 5.62x10-5 Ka2= 1.7x10-8 Ka3= 3.95x10-12 a) 0 mL b) 150 mL c) 225 mL d) 400 mL e) 450 mL f) 600 mL f) If you wanted to titrate to the first endpoint which indicator would you use? Circle one. Indicator Methyl Red Bromothymol Blue Cresol Red Phenolphthalein Range 4.8 - 6.0 6.2 - 7.6 7.0 - 8.8 8.2 - 10 g) How much 0.8 M NaOH must you add to 100 mL of 1.2 M Arsenous acid to make a buffer of pH = 11.6? 4) What is the solubility of magnesium carbonate (MgCO3) in a solution saturated with carbonic acid (0.034M) and which is buffered at pH = 8? MgCO3 ↔ Mg2+ + CO32H2CO3 ↔ 2 H+ + CO32- Ksp = 8.1x10-12 Ka = 2.02x10-17 5) Define amphoterism. Draw a graph of pH vs. concentration as part of your example. Chemistry 121 Second Exam Name_____________ April 18, 2007 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit 1(10) 2(15) 3(45) 4(20) 5(10) Total 1) Which of the following compounds is the most soluble in water? Circle one. Show your work. BaSO4 Ksp = 1.6x10-10 (x)(x) = x2 = 1.6x10-10 x = 1.26x10-5 PbI2 Ksp = 1.2x10-13 (x)(2x)2 = 4x3 = 1.2x10-13 x = 3.11x10-5 Ag3PO4 Ksp = 1x10-22 (3x)3(x) = 27x4 = 1x10-22 x = 1.39x10-6 2) I found a bottle marked “Acetic Acid” in our back room, but there was no indication of concentration. I wanted to know the concentration so I measured the pH and found it to be pH = 2.87. What was the concentration of the acetic acid in the bottle? Ka = 1.8x10-5 -log[H+] = 2.87 [H+] = 1.35x10-3 M also, when HAc dissociates, [H+] = [Ac-] HAc ↔ H+ + Ac-3 x 1.35x10 1.35x10-3 1.35x10 -3 2 x 1.6 x10 5 x = 0.1013 M HAc 3) Calculate the pH of 100 mL of 1.2 M Arsenous acid upon addition of the following amounts of 0.8 M NaOH H3AsO4 ↔ H2AsO4- + H+ H2AsO4- ↔ HAsO42- + H+ HAsO42- ↔ AsO43- + H+ a) 0 mL x2 5.62 x10 5 1.2 x Ka1= 5.62x10-5 pKA1 = 4.25 Ka2= 1.7x10-8 pKA2 = 7.74 Ka3= 3.95x10-12 pKA3 = 11.40 x = 8.21x10-3 pH = 2.09 M1V1 = M2V2 (1.2M)((0.100L) = (0.08 M)(V2) V2 = 150 mL for each region Also (1.2M)(0.100L) = 0.12 mol H3AsO4 pK A1 pK A2 4.25 7.74 2 2 b) 150 mL pH c) 225 mL pH = 7.74 d) 400 mL 100 mL into 3rd Region (0.8 M)(0.100 L) = 0.08 mol OHpH 11.40 log e) 450 mL [AsO34 ] 0.08 0.12 0.08 pH = 11.70 0.12mol 1x10-14 0.2182M and Kb 2.53 x10 3 0.550L 3.95x10-12 x2 2.54 x10 3 0.2182 x f) 600 mL pH = 5.995 x = 7.43x10-2 M OH- (0.150L)((0.80M) = 0.12 mol OH0.12 mol OH 0.1714 M OH 0.700 L pH = 12.87 pH = 13.23 g) If you wanted to titrate to the first endpoint which indicator would you use? Circle one. Indicator Methyl Red Bromothymol Blue Cresol Red Phenolphthalein Range 4.8 - 6.0 6.2 - 7.6 7.0 - 8.8 8.2 – 10 since pH = 5.995 @ first endpoint choose an indicator that changes just after 5.995 g) How much 0.8 M NaOH must you add to 100 mL of 1.2 M Arsenous acid to make a buffer of pH = 11.6? The pH is in the 3rd region so you will have to add 300 mL of NaOH to your answer, 11.6 11.4 log x 0.12 x x = 0.7357 mol OH- 0.7357 mol OH- = (0.8 M)(V) V = 91.97 mL + 300 mL = 391.97 mL total 4) What is the solubility of magnesium carbonate (MgCO3) in a solution saturated with carbonic acid (0.034M) and which is buffered at pH = 8? MgCO3 ↔ Mg2+ + CO32H2CO3 ↔ 2 H+ + CO32- Ksp = 8.1x10-12 Ka = 2.02x10-17 [H ][CO32 ] (1x108 M)(CO32 ) 2.02x1017 [CO32 ] 6.868x103 M [H 2 CO3 ] 0.034M [Mg2+] [CO32-] = [Mg2+] [6.868x10-3 M] = 8.1x10-12 [Mg2+] = 1.18x10-9 M 5) Define amphoterism. Draw a graph of pH vs. concentration as part of your example. Concentration of Metal Ion Amphoterism = the ability of a metal ion to be soluble in acidic and basic solutions 0 7 pH 14 Chemistry 121 Second Exam Name ________________ April 10, 2008 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit 1(10) 2(15) 3(45) 4(20) 5(10) Total 1) Which of the following compounds is the most soluble in water? Show your work. BaF2 Ksp = 1.84x10-9 Mg3(PO4)2 Ksp = 1.04×10-15 Ag3PO4 Ksp = 1.2x10-15 2) Please write the equilibrium reaction that will occur when the following compounds are mixed. Write all reactions as dissociations where appropriate. NaOH + HAc Na3PO4 + HCl HCl + KOH 3) Calculate the pH of 80 mL of 1.5 M Carbonic acid upon addition of the following amounts of 0.6 M NaOH H2CO3 ↔ H+ + HCO3HCO3- ↔ H+ + CO32- Ka1= 4.3x10-7 Ka2= 4.7x10-11 a) 0 mL b) 100 mL c) 150 mL d) 200 mL e) 300 mL f) 365 mL g) 400 mL h) 450 mL i) How much 0.8 M NaOH must be added to 80 mL of 1.5 M H2CO3 to make a buffer of pH = 10? 4) What is the solubility of Ag2S in a saturated solution of H2S = 0.01 M at pH = 3? Ag2S ↔ 2 Ag+ + S2- Ksp = 6.0x10-50 H2S ↔ 2 H+ + S2- Keq = 6.84x10-23 5a) Iron can be separated from copper using pH controlled solubility in H2S. CuS is not soluble in acid solutions and FeS is soluble. Draw and approximate ion separation curve for CuS and FeS. 5b) Define amphoterism and give an example. Chemistry 121 Second Exam Name April 10, 2008 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit 1(10) 2(15) 3(45) 4(20) 5(10) Total 1) Which of the following compounds is the most soluble in water? Show your work. BaF2 Ksp = 1.84x10-9 (x)(2x)2 = 4x3 = 1.84x10-9 x = 7.72x10-4 most soluble Mg3(PO4)2 Ksp = 1.04×10-15 (3x)3(2x)2 = 108x5 = 1.04x10-15 x = 3.95x10-4 Ag3PO4 Ksp = 1.2x10-15 (3x)3 (x) = 27x4 = 1.2x10-15 x = 8.16x10-5 2) Please write the equilibrium reaction that will occur when the following compounds are mixed. Write all reactions as dissociations where appropriate. NaOH + HAc H2O + Ac- ↔ HAc + OH- Na3PO4 + HCl HPO42- ↔ H+ + PO43HCl + KOH H2O ↔ H+ + OH- 3) Calculate the pH of 80 mL of 1.5 M Carbonic acid upon addition of the following amounts of 0.6 M NaOH H2CO3 ↔ H+ + HCO3HCO3- ↔ H+ + CO32- Ka1= 4.3x10-7 Ka2= 4.7x10-11 a) 0 mL x2/(1.5-x) = 4.3x10-7 x = 8.03x10-4M H+ pH = 3.095 b) 100 mL ½ way point so pH = 6.37 c) 150 mL - use H-H equation (0.6M)(0.150L) = .09 mole NaOH and (1.5M)(0.080L) = 0.12 mole H2CO3 pH = 6.37 + log(0.09/(0.12-0.09)) pH= 6.847 d) 200 mL first end point so pH = (pka1 + pka2)/2 → pH = (6.37 + 10.33)/2 = 8.35 e) 300 mL 2nd midpoint so pH = pKa2 = 10.33 f) 365 mL - you are now 165 mL into the second region - use H-H equation (0.06M)(0.165L) = 0.099 mole NaOH and you still have 0.12 mole of acid (HCO3-) pH = 10.33 + log(0.099/(0.12-0.099)) pH = 11.00 g) 400 mL - at final endpoint pKb = Kw/Ka2 = 1x10-14/4.7x10-11 = 2.13x10-4 also [CO32-] = 0.12 mole/ (0.480L total) = 0.25M CO32x2/(0.25 - x) = 2.13x10-4 → x = 0.007297 M OH- pH = 11.86 h) 450 mL - this is 50 mL past the final endpoint (0.050L)(0.60M) = 0.030 mole excess OH- in a total of 530 mL of solution. Therefore, [OH-] = 0.030 mole / 0.530 L = 0.0566 M OH- → pH = 12.75 i) How much 0.8 M NaOH must be added to 80 mL of 1.5 M H2CO3 to make a buffer of pH = 10? The pH is in the 2nd region so you must add 200 mL to start this titration. Use H-H, 10 = 10.33 + log(x/(0.12-x)) → x = 0.0382 mole NaOH 0.0382 mole NaOH = MV = (0.80M)(V) → V = 0.0478 L or 47.8 mL, therefore, 200 mL NaOH to get to first endpoint + 47.8 mL = 247.8 mL of 0.80 M NaOH 4) What is the solubility of Ag2S in a saturated solution of H2S = 0.01 M at pH = 3? Ag2S ↔ 2 Ag+ + S2- Ksp = 6.0x10-50 H2S ↔ 2 H+ + S2- Keq = 6.84x10-23 [H+]2 [S2-]/[H2S] = 6.84x10-23 [S2-] = 6.84x10-23 [H2S]/[H+]2 = 6.84x10-23 [0.01]/[1x10-3]2 = 6.84x10-19 M S2Substitute this result into the following equation, [Ag+]2 [S2-] = 6x10-50 [Ag+]2 = 6x10-50/[S2-] = 6x10-50/[6.84x10-19] = 8.77x10-32 taking the square root of this answer to find [Ag+], [Ag+] = 2.96x10-16 M Ag+ 5a) Iron can be separated from copper using pH controlled solubility in H2S. CuS is not soluble in acid solutions and FeS is soluble. Draw and approximate ion separation curve for CuS and FeS. [Ions] CuS FeS 0 At this pH FeS is highly soluble and CuS is not very soluble. 7 pH 14 5b) Define amphoterism and give an example. Amphoterism is the ability of an ion to be soluble in either an acid or a base but not in between. This property is commonly exhibited with transition metals like Zinc, Zn(OH)2(s) ↔ Zn2+ + 2 OHZn(OH)2(s) + 2 OH- ↔ Zn(OH)42- soluble in acid soluble in base Exam III Thermodynamics 1 , 2nd, and 3rd Laws Enthalpy, Entropy, and Gibbs Energy Heats of Formation Hess’ Law Spontaniety Gibbs Equation Electrochemical Cells Balancing Redox Reactions Nernst Equation Batteries Electroplating and Hydrolysis st Thermodynamics 1) A 100 gram block of copper was heated with a bunsen burner and then thrown into a cup of water (250 mL) initially at 25°C. When the system reached thermal equilibrium the final temperature of the water and the block was 61°C. What was the initial temperature of the block of copper? Cp(Cu)=24.5 J/mol-K, Cp(H2O)=75.2 J/mol-K, FWT(Cu) = 63.55 g/mol, FWT(H2O) = 18 g/mol, density of water = 1 g/ml 2) Given the following information calculate the heat of formation of C2H4. C2H4 + 3 O2 → 2 CO2 + 2 H2O ΔH° = -414 kJ/mol C + O2 → CO2 ΔH° = -393.5 kJ/mol H2 + ½ O2 → H2O ΔH° = -241.8 kJ/mol 3) When ethanol (C2H5OH) burns in oxygen it forms CO2 and water. If the heat of combustion is -1237.7 kJ/mol, the heat of formation of water is -241.8 kJ/mol, and the heat of formation of carbon dioxide is -393.5 kJ/mol, what is the heat of formation of C2H5OH? 4) Given a block of ice at 0°C how much of the ice will melt if you put a 100 gram block of aluminum on it that is initially at 100°C? Cp(Al)=24.1 J/mol-K, Cp(H2O)=75.2 J/mol-K, Heat of Fusion = 6.01 kJ/mol for ice, FWT(Al) = 26.98 g/mol, FWT (water) = 18 g/mol 5) 10 grams of diethyl ether was placed into a bomb calorimeter and then combusted in excess oxygen. The balanced combustion reaction was as follows; C4H10O + 6 O2 -----> 4 CO2 + 5 H2O The heat released by this reaction heated 2500 mL of water 8.6°C. What is the heat of formation of the diethyl ether? Cp(H2O) = 75.2 J/mol-K, ΔH°f(CO2) = -393.5 kJ/mol, ΔH°f(H2O) = -241.8 kJ/mol, FWT (Diethyl ether) = 74 g/mol, FWT (water) = 18 g/mol. 6) If a 25 gram block of copper at 45°C is added to 50 grams of liquid ammonia at -60°C, calculate the final temperature of the system. For NH3: FWT(NH3) = 17 g/mol, ΔHvap = 23.4 kJ/mol, Cp(liq)= 35.1 J/mol.K and for Cu : Cp = 24.4 J/mol.K, FWT(Cu) = 63.55 g/mol 7) You have 50 grams of ice at -2.5°C sitting in a dixie cup. If you heat 5 nickels and throw them into the cup and 8 grams of the ice melts, how hot were the nickels? Each nickel weighs 5 grams and has a heat capacity of 24.3 J/mol-K. Assume that nickels are pure nickel with an atomic mass of 57.81 g/mol. 8) What is the ΔG and ΔS for the dilution of 4.5 M H2SO4 to 0.03 M H2SO4 at 25°C? The enthalpy for this reaction is - 210 kJ/mol. 9) Calculate the ΔG for the making of 6M NaOH from solid NaOH. Saturated solutions of NaOH have a concentration of 19.25 M. 10) Mathematically we write the first law of thermodynamics as U = q + w. The work, w, is pressure-volume work. What is the sign on this work (+ or -) and why does it have this sign? 11) Please give me a definition of a heat capacity. 12) What is the ΔG° for the following reaction at 25°C? BaBr2(aq) + CuSO4(aq) ----> BaSO4(s) + CuBr2(s) ΔH° BaBr2 CuSO4 BaSO4 CuBr2 S° -186.47 kJ/mol -201.51 kJ/mol -345.57 kJ/mol -25.1 kJ/mol 42.0 J/mol K 19.5 J/mol K 7.0 J/mol K 21.9 J/mol K Circle all that apply about this reaction, spontaneous exothermic entropy increases the reaction occurs quickly not spontaneous endothermic entropy decreases the reaction occurs slowly What is the equilibrium constant for this reaction? 13) Given the following set of thermodynamic data calculate the ΔG°, ΔH°, ΔS°, and Keq for the following reaction at 25°C (Enthalpies are in kJ/mol and entropies are in J/mol-k) CoCl2(s) ⇆ Co2+ + 2 ClCoCl2 Co2+ Cl- ΔH° -325.5 -67.36 -167.4 S° 106.3 -155.2 55.1 Electrochemical Cells 14) Please balance each of the following redox reactions; MnO4- + Fe2+ → Fe3+ + Mn2+ CrO42- + I- → I2 + Cr3+ Co2+ + NO3- → Co2O3 + NO I3- + S2O32- → I- + S4O62H2S + Cr2O72- → Cr3+ + S Fe2+ + NiOOH → Fe3+ + Ni(OH)2 15) Consider the following electrochemical reaction, PbI2 + 2 e- → Pb + 2 IPb → Pb2+ + 2 ePbI2 → Pb2+ + 2 IHow would the voltage change if you, a) Add KI(s) to the oxidation cell b) Add AgNO3 to the reduction cell c) Add heat d) Add water to the reduction cell? e) Add NaNO3(s) to the oxidation cell f) Enlarge the oxidation electrode? Inc. Inc. Inc. Inc. Inc. Inc. N.C. N.C. N.C. N.C. N.C. N.C. 16) Consider the following electrochemical reaction, Au → Au3+ + 3eAuCl4- + 3e- → Au + 4 ClAuCl4- → Au3+ + 4 ClHow would the voltage change if you, a) Enlarge the Au electrodes? b) Add KCl(s)? c) Add AgNO3? d) Add water to the reduction cell? e) Add NaNO3(s) f) Cool the reaction vessel? Inc. Inc. Inc. Inc. Inc. Inc. N.C. N.C. N.C. N.C. N.C. N.C. Dec. Dec. Dec. Dec. Dec. Dec. Dec. Dec. Dec. Dec. Dec. Dec. 17) Given the following half-reactions draw an electrochemical cell, calculate the voltage of the cell and lable the anode and cathode. You may use a carbon electrode for the I- solution. Fe3+ + e- → Fe2+ I2 → 2 I- +0.771 V +0.535 V 18) What is the value of the half reaction, Cu2+ + e- → Cu+ given that, Cu2+ + 2e- → Cu° ° = +0.3402 Cu+ + e- → Cu° ° = +0.522 19) Given the following half-reactions draw an electrochemical cell that would work. Calculate the voltage of the cell and label the anode and cathode, tell which electrode is positive and which is negative, and where the oxidation and reduction reactions are occurring. In addition indicate the direction of electron flow, and the concentration of any ionic species in solution. Co3+ + e- → Co2+ +1.842 V Pb2+ + 2e- → Pb -0.126 V 20) Given the following half-reactions, properly set up a WORKING electrochemical cell. Ag + I- → AgI + ePbI2 + 2e-→ Pb + 2 I- ° = 0.1519 V ° = 0.3580 V Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all ionic species. Indicate what the electrodes are made of. 20b) Draw the cell diagram for the above cell. 20c) What is the Keq for the above cell? 20d) Look very carefully at the overall reaction, and explain whether your answer for 20c) is reasonable or not. 21) Balance the following half-reactions in a BASE and then use these reactions to set up a WORKING electrochemical cell. Bi(s) → Bi2O3(s) Hg2O(s) → Hg(l) 0.460 Volts -0.123 Volts Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all ionic species. Indicate the composition of the electrodes. 21b) Draw the cell diagram for the above cell. 21c) What is the Keq for the above cell? 22) How long will it take to deposit 1 ounce of gold (31.1 g) at a constant current flow of 10 amps? MW (Au) = 197.8 23) How long will it take to deposit 1 lb. of Palladium (453.6 g) from a solution of PdCl4 at a constant current of 25 amps? MW (Pd) = 106.4 Thermodynamics 1) A 100 gram block of copper was heated with a bunsen burner and then thrown into a cup of water (250 mL) initially at 25°C. When the system reached thermal equilibrium the final temperature of the water and the block was 61°C. What was the initial temperature of the block of copper? Cp(Cu)=24.5 J/mol-K, Cp(H2O)=75.2 J/mol-K, FWT(Cu) = 63.55 g/mol, FWT(H2O) = 18 g/mol, density of water = 1 g/ml -nCpTCu = nCpTH2O 100g J 250g J 24.5 61C - Ti 75.2 61C - 25C 63.55 g/mol molK 18 g/mol molK Ti = 1036.3°C 2) Given the following information calculate the heat of formation of C2H4. C2H4 + 3 O2 → 2 CO2 + 2 H2O Δ H° = -414 kJ/mol C + O2 → CO2 Δ H° = -393.5 kJ/mol H2 + ½ O2 → H2O Δ H° = -241.8 kJ/mol 2 CO2 + 2 H2O → C2H4 + 3 O2 ΔH° = 414 kJ/mol 2 (C + O2 → CO2) ΔH° = -393.5 kJ/mol x2 2 (H2 + ½ O2 → H2O) ΔH° = -241.8 kJ/mol x2 2 C + 2 H2 → C2H4 ΔH° = 414 kJ/mol + 2 (-393.5 kJ/mol )+ 2(-241.8 kJ/mol) ΔH° = -856.6 kJ/mol 3) When ethanol (C2H5OH) burns in oxygen it forms CO2 and water. If the heat of combustion is -1237.7 kJ/mol, the heat of formation of water is -241.8 kJ/mol, and the heat of formation of carbon dioxide is -393.5 kJ/mol, what is the heat of formation of C2H5OH? C2H5OH + 3 O2 → 2 CO2 + 3 H2O ΔH° = ΔH°prod - ΔH°react -1237.7 kJ/mol= [2(-393.5kJ/mol) + 3(-241.8kJ/mol)] – [ΔH°C2H5OH + 3(0kJ/mol)] ΔH°C2H5OH = -274.7 kJ/mol 4) Given a block of ice at 0°C how much of the ice will melt if you put a 100 gram block of aluminum on it that is initially at 100°C? Cp(Al)=24.1 J/mol-K, Cp(H2O)=75.2 J/mol-K, Heat of Fusion = 6.01 kJ/mol for ice, FWT(Al) = 26.98 g/mol, FWT (water) = 18 g/mol -nCpTAl = nfus 100 g (24.1 J/molK)(100C) n (6010 J/mol) 26.98 g/mol n = 1.4863 mol ice x 18g/mol = 26.75 g ice will melt 5) 10 grams of diethyl ether was placed into a bomb calorimeter and then combusted in excess oxygen. The balanced combustion reaction was as follows; C4H10O + 6 O2 → 4 CO2 + 5 H2O The heat released by this reaction heated 2500 mL of water 8.6°C. What is the heat of formation of the diethyl ether? Cp(H2O) = 75.2 J/mol-K, Δ H°f(CO2) = -393.5 kJ/mol, Δ H°f(H2O) = 241.8 kJ/mol, FWT (Diethyl ether) = 74 g/mol, FWT (water) = 18 g/mol. -nCpTH2O = rxn note: a negative sign was added because energy was given off. 2500g J (75.2 )(8.6C) 89,822 J 18 molK 89,822 J 664,683 J/mol 10 g 74 g/mol ΔH° = ΔH°prod - ΔH°react -664,683 kJ/mol= [4(-393.5kJ/mol) + 5(-241.8kJ/mol)] – [ΔH°Ether + 6(0kJ/mol)] ΔH°Ether = -2118.3 kJ/mol 6) If a 25 gram block of copper at 45°C is added to 50 grams of liquid ammonia at -60°C, calculate the final temperature of the system. For NH3: FWT(NH3) = 17 g/mol, ΔHvap = 23.4 kJ/mol, Cp(liq)= 35.1 J/mol.K and for Cu : Cp = 24.4 J/mol.K, FWT(Cu) = 63.55 g/mol -nCpTCu = nCpTNH3 25g J 50g J 24.4 Tf 45C 35.1 Tf - (-60C) 63.55 molK 17 molK -9.9587(Tf – 45°C) = 103.235(Tf + 60°C) -Tf + 45 = 10.3663(Tf + 60) = 10.3663Tf + 621.98 -11.3663Tf = 666.98 Tf = -50.76°C 7) You have 50 grams of ice at -2.5 C sitting in a dixie cup. If you heat 5 nickels and throw them into the cup and 8 grams of the ice melts, how hot were the nickels? Each nickel weighs 5 grams and has a heat capacity of 24.3 J/mol-K. Cpice = 37.7 J/molK. Assume that nickels are pure nickel with an atomic mass of 57.81 g/mol. Note: You must heat all of the ice but melt only 8 grams of it. nCpTNi = nCpTice + nfus 25g J 8g J 50g J 24.3 Tf 0C 6010 37.7 2.5C 57.81 molK 18 mol 18 molK 10.508 Tf = 2932.92 Tf = 279.1°C so Tf = 279.1°C 8) What is the ΔG and ΔS for the dilution of 4.5 M H2SO4 to 0.03 M H2SO4 at 25°C? The enthalpy for this reaction is - 210 kJ/mol. G = G° + RT ln Q = 0 + (8.314 J/molK)(298K) ln (0.03 M/ 4.5M) G = -12,414 J/mol J J - 12,414 210,000 mol mol 663 J G = H - TS => ΔG ΔH ΔS T 298 molK 9) Calculate the ΔG for the making of 6M NaOH from solid NaOH. Saturated solutions of NaOH have a concentration of 19.25 M. There are two processes going on, 1. Solid NaOH ↔ Sat’d NaOH 2. Sat’d NaOH → 6 M NaOH The first process is an equilibrium so G = 0 for the first process. For the second process, since all solutions have a standard state of 1M, the standard state for Sat’d NaOH and 6 M NaOH are both 1 M and the G° = 0 since there is no difference between the G°’s for these two solutions. So, G1 = 0 G2 = G° + RT ln Q G2 = 0 + (8.314 J/molK)(298K) ln (6 M/ 19.25M) = -2888.2 J/mol G1+ G2 = 0 + (- 2888.2 J/mol) = - 2888.2 J/mol 10) Mathematically we write the first law of thermodynamics as U = q + w. The work, w, is pressure-volume work. What is the sign on this work (+ or -) and why does it have this sign? Work = PV A chemist thinks of himself as being inside the system pushing out (like expanding a balloon). To a chemist, pushing out (expanding the balloon) is negative work. Negative work is work that you must do as opposed positive work that is done for you. In mathematics, a change in volume is always the Vfinal – Vinitial. In the case of an expanding balloon, the final volume will be larger than the initial volume so Vfinal – Vinitial will produce a positive value. Since PV = negative for an expanding balloon, and both P and V are positive, we must add a negative sign to the work expression to make this work overall negative. So, chemists define pressure-volume work as, Work = - PV (Chemist) It should be noted that engineers see the world opposite from chemists. Engineers view themselves as being outside of the system pushing in. Work for an engineer is compressing the balloon, making it smaller. Since V = Vfinal – V initial, and the Vfinal is smaller than Vinitial, the V is negative. So, for an engineer, Work = PV (Engineer) This value is still negative, like the chemist, but we see the world differently from an engineer. 11) Please give me a definition of a heat capacity. It is the amount of energy needed to raise the temperature of one mole of substance 1°C. 12) What is the ΔG° for the following reaction at 25°C? BaBr2(aq) + CuSO4(aq) ----> BaSO4(s) + CuBr2(s) ΔH° BaBr2 CuSO4 BaSO4 CuBr2 S° -186.47 kJ/mol -201.51 kJ/mol -345.57 kJ/mol -25.1 kJ/mol 42.0 J/mol K 19.5 J/mol K 7.0 J/mol K 21.9 J/mol K ΔH° = ΔH°prod - ΔH°react and ΔS° = S°prod - S°react ΔH° = ΔH°prod - ΔH°react ΔH° = [(-345.57 kJ/mol) + (-25.1 kJ/mol)] – [(-186.47 kJ/mol) + (-201.51 kj/mol)] ΔH° = 17.31 kJ/mol ΔS° = S°prod - S°react ΔS° = [(7.0 J/molK) + (21.9 J/molK)] – [(42.0 J/molK) + (21.9 J/molK)] ΔS° = -32.6 kJ/mol G° = H° - TS° G° = 17,310 J/mol + (298 K) (-32.6 J/molK) G° = 27,024.8 J/mol Circle all that apply about this reaction, spontaneous exothermic entropy increases the reaction occurs quickly not spontaneous endothermic entropy decreases the reaction occurs slowly What is the equilibrium constant for this reaction? G° = - RT ln Keq Keq e ΔG RT e ( 27 , 024 .8 ) (8.314)(29 8) 1.832 x10 -5 13) Given the following set of thermodynamic data calculate the ΔG°, ΔH°, ΔS°, and Keq for the following reaction at 25C (Enthalpies are in kJ/mol and entropies are in J/mol-k) CoCl2(s) ↔ Co2+ + 2 ClCoCl2 Co2+ Cl- ΔH° -325.5 -67.36 -167.4 S° 106.3 -155.2 55.1 ΔH° = ΔH°prod - ΔH°react and ΔS° = S°prod - S°react ΔH° = ΔH°prod - ΔH°react ΔH° = [(-67.36 kJ/mol) + 2(-167.4 kJ/mol)] – [(-325.5 kj/mol)] ΔH° = -76.66 kJ/mol (exothermic) ΔS° = S°prod - S°react ΔS° = [(-155.2 J/molK) + 2(55.1 J/molK)] – [(106.3 J/molK)] ΔS° = -151.3 J/molK (entropy decreases) G° = H° - TS° G° = -76,660 J/mol - (298 K) (-151.3 J/molK) G° = -31,572.6 J/mol (reaction is spontaneous as written) G° = - RT ln Keq Keq e ΔG RT e ( 31,572.6) (8.314)(29 8) 3.423x10 5 Electrochemical Cells 14) Please balance each of the following redox reactions; MnO4- + Fe2+ → Fe3+ + Mn2+ Fe2+ → Fe3+ + e5e- + 8 H+ + MnO4- → Mn2+ + 4 H2O CrO42- + I- → I2 + Cr3+ 3 e- + 8 H+ + CrO42- → Cr3+ + 4 H2O 2 I- → I2 + 2 eCo2+ + NO3- → Co2O3 + NO 3 H2O + 2 Co2+ → Co2O3 + 6 H+ + 2e3 e- + 4 H+ + NO3- → NO + 2 H2O I3- + S2O32- → I- + S4O622 e- + I3- → 3 IS2O32- → S4O62- + 2 eH2S + Cr2O72- → Cr3+ + S H2S → S + 2 H+ + 2 e6 e- + 14 H+ + Cr2O72- → 2 Cr3+ + 7 H2O Fe2+ + NiOOH → Fe3+ + Ni(OH)2 Fe2+ → Fe3+ + ee- + H+ + NiOOH → Ni(OH)2 15) Consider the following electrochemical reaction, PbI2 + 2 e- → Pb + 2 IPb → Pb2+ + 2 ePbI2 → Pb2+ + 2 IHow would the voltage change if you, a) Add KI(s) to the oxidation cell b) Add AgNO3 to the reduction cell c) Add heat d) Add water to the reduction cell? e) Add NaNO3(s) to the oxidation cell f) Enlarge the oxidation electrode? Inc. Inc. Inc. Inc. Inc. Inc. N.C. N.C. N.C. N.C. N.C. N.C. Dec. Dec. Dec. Dec. Dec. Dec. In electrochemical cells all concentrations are 1 M. If all concentrations are 1 M, the reverse reaction is going to occur. Therefore, the real reaction going on is, Pb + 2 I- → PbI2 + 2 e- (oxidation) Pb2+ + 2 e- → Pb (reduction) 2+ Pb + 2 I → PbI2 To understand what would happen, you need to write down the Nernst Equation, = ° - RT/n ln 1/[Pb2+][I-]2 a) When you put KI in the oxidation cell you are increasing the I- concentration. This causes 1/[Pb2+][I-]2 to become smaller. The natural log of a smaller number is a negative number. The negative of a negative is a positive so the voltage will increase. b) Nothing happens. There is nothing for the silver to react with. c) Adding heat increases T in the equation. If everything else remains the same, the - RT/n ln 1/ [Pb2+][I-]2 becomes more negative and the voltage decreases. d) Adding water is a dilution. Diluting the reduction cell lowers the concentration of I-. If I- gets smaller, the 1/[Pb2+][I-]2 gets larger. The natural log will become larger but there is a negative so it becomes more negative, so the voltage goes down. e) NaNO3 does not participate in this reaction. No change. f) Changing the size of the electrodes does not change the voltage. The electrodes are not a part of the reaction. They are not in the Nernst Equation. 16) Consider the following electrochemical reaction, Au → Au3+ + 3eAuCl4- + 3e- → Au + 4 ClAuCl4- → Au3+ + 4 ClHow would the voltage change if you, a) Enlarge the Au electrodes? b) Add KCl(s)? c) Add AgNO3? d) Add water to the reduction cell? e) Add NaNO3(s) f) Cool the reaction vessel? Inc. Inc. Inc. Inc. Inc. Inc. N.C. N.C. N.C. N.C. N.C. N.C. Dec. Dec. Dec. Dec. Dec. Dec. Note: This is NOT a Ksp problem since AuCl4- is not a solid (solids are not charged) so we will leave the reaction as is. = ° - RT/n ln [Au3+][Cl-]4/[ AuCl4-] a) No change. Changing the size of the electrode makes no difference. b) Adding Cl- makes the natural log large which decreases the voltage c) The Ag+ will react with the Cl- to make AgCl. This removes Cl-, making the natural log smaller (more negative). So you will be subtracting a smaller number so the voltage will increase. d) Diluting the reduction cell lowers the concentration of both AuCl4- and Cl- but the Clis raised to the 4th power so it reduces faster. Reducing Cl- makes the natural log smaller (more negative so you are subtracting a smaller (more negative) number so voltage increases. e) No change. f) If T is smaller, then you are subtracting a smaller number so voltage increases. 17) Given the following half-reactions draw an electrochemical cell, calculate the voltage of the cell and label the anode and cathode. You may use a carbon electrode for the I- solution. Fe3+ + e- → Fe2+ +0.771 eV 2 I- → I2 + 2 e-0.535 eV 2 Fe3+ + 2 I- → 2 Fe2+ + I2 0.236 eV e- Anode Negative Oxidation C Pt I2 1M FeCl 3 1M FeCl 2 1M KI 18) What is the value of the half reaction, Cu2+ + e- ----> Cu+ given that, Cu2+ + 2e- → Cu ° = +0.3402 Cu+ + e- → Cu ° = +0.522 Cu2+ + 2e- → Cu ° = +0.3402 Cu → Cu+ + e- ° = -0.522 Cu2+ + e- → Cu+ ° = 2(0.3402) + 1(-0.522) = 0.1584 volts 1 Cathode Positive Reduction 19) Given the following half-reactions draw an electrochemical cell that would work. Calculate the voltage of the cell and label the anode and cathode, tell which electrode is positive and which is negative, and where the oxidation and reduction reactions are occurring. In addition indicate the direction of electron flow, and the concentration of any ionic species in solution. Co3+ + e- → Co2+ Δ° = + 1.842 eV Pb2+ + 2e- → Pb Δ° = - 0.126 eV e- Anode Negative Oxidation Pb Pt 1M Pb(NO3)2 1M CoCl3 1M CoCl2 Cathode Positive Reduction Co3+ + e- → Co2+ Δ° = + 1.842 eV Pb → Pb2+ + 2e- Δ° = + 0.126 eV 3 Pb + 2 Co3+ → 3 Pb2+ + 2 Co Δ° = 1.968 V 20) Given the following half-reactions, properly set up a WORKING electrochemical cell. AgI + e----> Ag + IΔ° = -0.1519 V PbI2 + 2e ---> Pb + 2 I Δ° = 0.3580 V Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all ionic species. Indicate the composition of the electrodes. - e Ag Anode Negative Oxidation AgI(s) Pb 1M KI Cathode Positive Reduction PbI2(s) 20b) Draw the cell diagram for the above cell. Ag/AgI(s)/1 M KI/PbI2(s)/Pb 20c) What is the Keq for the above cell? = 0.5099 volts = 0.0592/2 log Keq => Keq = 1.684x1017 20d) Look very carefully at the overall reaction, and explain whether your answer for 20c) is reasonable or not. The problem is that everything is a solid except for the 1 M KI. The equilibrium constant must be, [I ] Keq Pb [ I ] Ag So the voltage is completely dependent on the difference between the amount of I- in equilibrium with the Pb and the I- in equilibrium with the Ag. This makes what is known as a concentration cell, that is, a cell whose voltage is due only to differences in concentration on the two sides of the cell. 21) Balance the following half-reactions in a BASE and then use these reactions to set up a WORKING electrochemical cell. 6 OH- + 2 Bi(s) → Bi2O3(s) + 3 H2O + 6 e2 e- + H2O + Hg2O(s) → 2 Hg(l) + 2 OH- 0.460 Volts -0.123 Volts Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all ionic species. Indicate the composition of the electrodes. e Anode Negative Oxidation - Bi Cathode Positive Reduction 1M NaOH Bi2O3(s) Hg2O Pt 21b) Draw the cell diagram for the above cell. Bi / Bi2O3(s) / 1 M NaOH / Hg2O(s) / Hg / Pt 21c) What is the Keq for the above cell? = 0.337 volts = 0.0592/6 log Keq Keq = 1.43x1034 22) How long will it take to deposit 1 ounce of gold (31.1 g) at a constant current flow of 10 amps? MW (Au) = 197.8 31.1 g/197.8 g/mol = 0.1572 mol Au has a 3+ charge, so multiply by 3. 3 mole e-/mol Au x 0.15723 mol Au = 0.4717 mol e0.4717 mol e- x 96,487 C/mol e- = 45,511.8 C Amp x sec = Coulombs (C) (10 amps)(sec) = 45,511.9 C sec = 4,551.18 sec => 75.85 minutes 23) How long will it take to deposit 1 lb. of Palladium (453.6 g) from a solution of PdCl4 at a constant current of 25 amps? MW (Pd) = 106.4 453.6 g/106.4 g/mol = 4.263 mol Pd, but Pd has a 4+ charge, so multiply by 4. 4 mole e-/mol Pd x 4.263 mol Pd = 17.053 mol e17.053 mol e- x 96,487 C/mol e- = 1,645,357 C Amp x sec = Coulombs (C) (25 amps)(sec) = 1,645,357 C sec = 65,814 sec => 18.28 hours Chemistry 121 Third Exam Name____________________ May 15, 2007 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit 1 ( 18) 2 (40) 3(12) 4(15) 5(15) Total R = 8.314 J/mol-K = - 0.0592/n log Q ΔG = -n = 96486 C/mol eΔG = ΔG + RT ln Q PV = nRT C = amp x sec ΔG= -RTlnKeq R = 0.08205 L-atm/mol-K 1) Balance each of the following oxidation-reduction half-reactions; IO3- → I3- (in acid) Mn(OH)2 → Mn2O3 (in base) 2) Given the following set of half-reactions, set up a WORKING electrochemical cell. Hg2Cl2(s) + 2 e- → 2Hg + 2Cl2 H+ + 2 e- → H2 = 0.27 V = 0.000 V Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all ionic species. Indicate the composition of the electrodes. 2b) Draw the cell diagram for the above cell. 2c) What is the voltage of the cell when all ionic species have a concentration of 0.10 M? 3) How long will it take to galvanize (coat with Zinc) a nail with 2 grams of zinc using a current (amperage) of 0.8 amps? FWT of Zn2+ = 65.41 g/mol 4) You are given 100 mL of water at 25°C. You add a 20 gram ice cube at 0°C and 50 mL of boiling water at 100°C. What is the final temperature of the system? 5) What is the ΔG and ΔS for the dilution of 4.5 M H2SO4 to 0.03 M H2SO4 at 25°C? The ΔH = - 210 kJ/mol.for this reaction. Chemistry 121 Third Exam Name__Answer Key____ May 15, 2007 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit 1 ( 18) 2 (40) 3(12) 4(15) 5(15) Total R = 8.314 J/mol-K = - 0.0592/n log Q ΔG = -n = 96486 C/mol eΔG = ΔG + RT ln Q PV = nRT C = amp x sec ΔG= -RTlnKeq R = 0.08205 L-atm/mol-K 1) Balance each of the following oxidation-reduction half-reactions; IO3-→ I3- (in acid) 16 e- + 18 H+ + 3 IO3- → I3- + 9 H2O Mn(OH)2 → Mn2O3 (in base) 2 Mn(OH)2 → Mn2O3 + H2O + 2 H+ + 2e2 OH- → 2 OH2 Mn(OH)2 + 2 OH- → Mn2O3 + 3 H2O + 2e- 2) Given the following set of half-reactions, set up a WORKING electrochemical cell. Hg2Cl2(s) + 2 e- → 2Hg + 2Cl2 H+ + 2 e- → H2 e- H2 1atm Anode Negative Oxidation = 0.27 V = 0.000 V Pt 1M HCl Cathode Positive Reduction 1M HCl Hg2Cl2 Pt Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all ionic species. Indicate the composition of the electrodes. 2b) Draw the cell diagram for the above cell. (Note: The solution in both cells is HCl so no salt bridge is required). Pt/H2 (1atm) / 1M HCl / Hg2Cl2(s) / Hg(l)/ Pt 2c) What is the voltage of the cell when all ionic species have a concentration of 0.10 M? The overall reaction is, Hg2Cl2(s) + 2 e- → 2Hg + 2ClH2 → 2 H+ + 2 eHg2Cl2(s) + H2 –> 2 Hg + 2 H+ + 2 Cl- = 0.27 V = 0.000 V = 0.27 V So the voltage is, = 0.27 - 0.0592/2 log [H+]2 [Cl-]2 = 0.27 - 0.0591/2 log [0.10]2 [0.10]2 = 0.3882 Volts 3) How long will it take to galvanize (coat with Zinc) a nail with 2 grams of zinc using a current (amperage) of 0.8 amps? FWT of Zn2+ = 65.41 g/mol 2 grams/65.41g/mol = 0.03058 mole of Zn 0.03058 mole of Zn x 2 mole e-/mol Zn = 0.06115 mole e0.06115 mole e- x 96,486 C/mol e- = 5900.4 Coulombs 1 C = 1 amp x sec 5900.4 C = 0.8 amps x sec 5900.4 C/0.8 amps = 7375.47 seconds 4) You are given 100 mL of water at 25°C. You add a 20 gram ice cube at 0°C and 50 mL of boiling water at 100°C. What is the final temperature of the system? ΔHfus = 6.01 kJ/mol and Cp(H2O) = 75.2 J/mol K, FWT H2O = 18g/mol (These were not given in the original problem). The easiest way of doing this is by mixing the hot and cold water and then adding the ice cube. First add 100 mL of 25°C water to 50 mL of 100°C water, nCpΔT25°C = - nCpΔT100°C 100g/18g/mol (75.2 J/molK)(Tf - 25) = -50g/18g/mol(75.2 J/molK)(Tf-100) The 18g/mol and the 75.2 J/molK cancel on both sides so, 100(Tf-25) = -50(Tf-100) Tf = 50°C So now we have 150 mL of water at 50̊C and we add the 20 gram ice cube. I assume that the ice cube will melt and then waters will mix to get to some final temperature so, nΔHfus + nCpΔT0°C = - nCpΔT50°C 20g/18g/mol(6010 J/mol) + 20g/18g/mol (75.2)(Tf - 0) = - 150g/18g/mol(75.2)(Tf - 50) All the 18 g/mol cancel so, 120,200 + 1504Tf = - 11280Tf + 564000 12784Tf = 443,800 Tf = 34.71°C 5) What is the ΔG and ΔS for the dilution of 4.5 M H2SO4 to 0.03 M H2SO4 at 25°C? The ΔH = - 210 kJ/mol.for this reaction. ΔG = ΔG + RT ln Q But ΔG = 0 since both concentrations have the same standard state of 1M. So, ΔG = 0 + RT ln 0.03M/4.5M ΔG = -5391.42 J/mol Now, since ΔG = ΔH - TΔS, we can solve for ΔS, -5391.42 J/mol = -210,000 J/mol - (298)ΔS ΔS = (-5391.42 J/mol + 210,000 J/mol )/298 = 686.61 J/mol K Chemistry 121 Third Exam Name____________________ May 14, 2009 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit 1 (18) 2 (28) 3(10) 4(20) Total R = 8.314 J/mol-K = - 0.0592/n log Q ΔG = -n = 96486 C/mol eΔG = ΔG + RT ln Q PV = nRT C = amp x sec ΔG= -RTlnKeq R = 0.08205 L-atm/mol-K 1a) Please balance the following half-reaction in a base. BH3 → B2O3 1b) What is the voltage of the following half-cell reactions when added together? Cu2+ + e- → Cu+ Cu+ + e- → Cu Cu2+ + 2 e- → Cu ̊ = 1.29 = 1.68 = ? 2) In lab, you were asked to experimentally determine the Ksp of PbI2. Given the information below (found in the CRC), what voltage should you have obtained for your cell, and what should the Ksp have been? PbI2(s) ↔ Pb2+ + 2 IΔH S PbI2 -174.1 kJ/mol 176.98 J/mol-K Pb2+ 1.63 kJ/mol 21.34 J/mol-K I-55.94 kJ/mol 109.37 J/mol-K The reaction is (Circle all that apply). Spontaneous Exothermic Entropy increases Occur Fast Not Spontaneous Endothermic Entropy Decreases Occurs Slow 2c) Like you did in lab, draw the electrochemical cell that was used to determine the Ksp of PbI2. Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all ionic species. Indicate the composition of the electrodes. 2c) Draw the cell diagram for the above cell. 3) It took 35 minutes and 10 amps to electroplate 10.615 grams of platinum (FWT = 195.08 g/mol). What was the charge on the platinum? 4) A 20 gram block of ice initally at -10.0°C was dropped into 250 mL of water at 25°C. What is the final temperature of the system? ΔHfus = 6.01 kJ/mol, Cp(H2O) = 75.2 J/mol-K, Cp(ice) = 35.2 J/mol-K, FWT (H2O)= 18 g/mol Chemistry 121 Third Exam Name__Answer Key___ May 14, 2009 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit 1 (18) 2 (28) 3(10) 4(20) Total R = 8.314 J/mol-K = - 0.0592/n log Q ΔG = -n = 96486 C/mol eΔG = ΔG + RT ln Q PV = nRT C = amp x sec ΔG= -RTlnKeq R = 0.08205 L-atm/mol-K 1a) Please balance the following half-reaction in a base. BH3 → B2O3 3 H2O + 2 BH3 → B2O3 + 12 H+ + 12 e12 OH- → 12 OH12 OH- + 2 BH3 → B2O3 + 9 H2O + 12 e- 1b) What is the voltage of the following half-cell reactions when added together? Cu2+ + e- —> Cu+ Cu+ + e- —> Cu Cu2+ + 2 e- —> Cu = 1.29 = 1.68 = (1.29(1) + 1.68 (1))/2 = 1.485 Volts 2) In lab, you were asked to experimentally determine the Ksp of PbI2. Given the information below (found in the CRC), what voltage should you have obtained for your cell, and what should the Ksp have been? PbI2(s) ⇆ Pb2+ + 2 IPbI2 Pb2+ I- ΔH -174.1 kJ/mol 1.63 kJ/mol -55.94 kJ/mol S 176.98 J/mol-K 21.34 J/mol-K 109.37 J/mol-K ΔH̊ = [1.63 + 2(-55.94)] - [-174.1] = 63.85 kJ/mol Δ S = [21.34 + 2(109.37)] - [176.98] = 63.1 J/molK ΔG = 63,850 - 298(63.1) = 45,046.2 J/mol 45,046.2 J/mol = -8.314(298)ln(Ksp) Ksp = 1.27x10-8 The reaction is (Circle all that apply). Spontaneous Exothermic Entropy increases Occurs Fast Not Spontaneous Endothermic Entropy Decreases Occurs Slow 2c) Like you did in lab, draw the electrochemical cell that was used to determine the Ksp of PbI2. Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all ionic species. Indicate the composition of the electrodes. e- Cathode Positive Reduction Pb Pb Anode Negative Oxidation PbI2(s) 1M Pb(NO3)2 1M KI 2c) Draw the cell diagram for the above cell. Pb / PbI2(s) / 1 M KI // 1 M Pb(NO3)2 / Pb 3) It took 35 minutes and 10 amps to electroplate 10.615 grams of platinum (FWT = 195.08 g/mol). What was the charge on the platinum? 10.615 g/195.08 g/mol = 0.05441 mole of Pt 0.05441 mole of Pt x Charge on Pt = moles of eAlso, amp x sec/ 96,486 C/mol e- = mole of eSo, 0.05441 mole of Pt x Charge on Pt = amp x sec/ 96,486 C/mol e0.05441 mole of Pt x Charge on Pt = 10 amps x 35 min x 60 sec/min / 96,486 C/mol eCharge on Pt = [10 amps x 35 min x 60 sec/min / 96,486 C/mol e-]/0.05441 mole of Pt Charge on Pt = +4 4) A 20 gram block of ice initally at -10.0°C was dropped into 250 mL of water at 25°C. What is the final temperature of the system? ΔHfus = 6.01 kJ/mol, Cp(H2O) = 75.2 J/mol-K, Cp(ice) = 35.2 J/mol-K, FWT (H2O)= 18 g/mol nCpΔTice + nΔHfus + nCpΔT0°C H2O = -nCpΔT25°C H2O (20/18)35.2(10) + (20/18)6010 + (20/18)75.2(Tf-0) = -(250/18)75.2(Tf-25) All the 18's cancel so, 7040 + 120200 + 1504Tf = -18800Tf + 470000 20304Tf = 342760 Tf = 16.88°C Final Exam Comprehensive Exam ~ 30% Cut and Paste from Old Exams ~30% Organic Chemistry ~40% New Questions on Old Topics Mechanism of Free Radical Halogenation using Bromine Chain Initiation light Br2 2 Br Chain Propagation Br H H H H H H C C C H H H H H H C C C H H H Br2 H + HBr H H H C C C H H H H Br H C C C H H H H Br H C C C H H H + Br H Chain Termination Br Br H Br H Br2 H C C C H H H H H H H H C H H C H C H H H H C H H C H C H H H H H Relative Reactivity 1 2 3 Chlorination 1 3.5 5 Bromination 1 97 C H C H H H H C H C H C H H C H H H 1) Please draw the rotational energy diagram for the rotation of 2,3-dimethyl butane around the C2-C3 bond. Energy 0 60 120 180 240 300 360 Draw the molecule below, convert it to a Newman projection and then rotate it about the C2-C3 bond. Based on your rotations, draw the energy diagram above. 2) Please draw the rotational energy diagram for the rotation of 3-methyl pentane around the C2C3 bond. Energy 0 60 120 180 240 300 360 Draw the molecule below, convert it to a Newman projection and then rotate it about the C2-C3 bond. Based on your rotations, draw the energy diagram above. 1) Please draw the rotational energy diagram for the rotation of 2,3-dimethyl butane around the C2-C3 bond. Energy 0 60 120 180 240 300 360 Draw the molecule below, convert it to a Newman projection and then rotate it about the C2-C3 bond. Based on your rotations, draw the energy diagram above. 0 CH3 CH3 H3C C C H H 120 CH3 H3C H3C CH3 CH3 CH3 4 60 H3C CH3 H HH CH3 H3C CH3 CH3 CH3 H 2 CH3 H3C H H3C H CH3 3 H H3C H3C CH3 H H3C CH3 H3C H CH3 H H CH3 H 300 240 180 2 3 1 2) Please draw the rotational energy diagram for the rotation of 3-methyl pentane around the C2C3 bond. 6 5 Energy 4 3 2 1 0 60 120 180 240 300 360 Draw the molecule below, convert it to a Newman projection and then rotate it about the C2-C3 bond. Based on your rotations, draw the energy diagram above. 0 H H3C CH3 C C H H 120 CH3 H3C H CH3 C2H5 C2H5 6 60 H CH3 H HH CH3 H3C CH3 CH3 C2H5 H 3 C2H5 H H H H C2H5 5 H C2H5 H CH3 H H CH3 C2H5 H CH3 H H CH3 H 300 240 180 2 4 1 3) What is the ratio of products in the free radical chlorination of bicyclo [2,2,0] hexane? 2° ݔ 8 3.5 ቆ °ቇ = ∙ = ݔ26.32% 3 100 − ݔ2 5 4) What is the ratio of products in the free radical chlorination of 2-methylpropane? CH3 H3C C CH3 H 1° ݔ 9 1 ቆ °ቇ = ∙ = ݔ64.29% 3 100 − ݔ1 5 5) What is the ratio of products in the free radical chlorination of 1,1,3,3tetramethylcyclopentane? H3C CH3 H3C CH3 1° ݔ 12 1 ቆ °ቇ = ∙ = ݔ36.36% 2 100 − ݔ 6 3.5 6) Please name the following compounds, Acids O O C C C C HO OH Propanoic acid O O C C H C C OH OH butadioic acid (malic acid) methanoic acid (formic acid) COOH OH C C Br Benzoic acid O C C OH 3-bromo-3-hydroxybutanoic acid Aldehydes and Ketones O O C C C C C O C C C H C H H Propanal 2-pentanone OH O O C C methanal (formaldehyde) C Propanone (acetone) C C C C 3-hydroxybutanone Alcohols H OH C C C OH C C C C C H C OH H Propanol 2-pentanol OH OH OH C C methanol C 2-propanol C C C 2,3-butadiol C Aromatics NH2 NO2 CH3 HO Cl Toluene CH3 2-chloro-4-nitrophenol Br para methylaniline Br H2N COOH Br 1,3,5-tribromobenzene para aminobenzoic acid Chemistry 121 Final Exam Name______________________ May 27, 1998 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit Question 1(15) 5(20) 2(20) 6(20) 3(20) 7(25) 4(40) 8(40) Total Total Credit TOTAL POINTS = R = 8.314 J/mol-K Δ = Δ° - 0.0592/n log Q ΔGº = -nΔ° ln(Ai/Af) = kΔt = 96486 C/mol eΔG = ΔGº + RT ln Q PV = nRT 1/Af - 1/Ai = kΔt C = amp x sec ΔGº= -RTlnKeq R = 0.08205 L-atm/mol-K ln(k1/k2) = Ea/R (1/T2 - 1/T1) 1)Write down the equilibrium that occurs when the following compounds are mixed. a) NaOH + NaBenz b) NH4OH + NH4Cl c) NaCN + HCl 2) Last night I poured myself a glass of water from the tap and took a drink. The water was not cold and didn't taste good so I put a couple of ice cubes in it. The ice cubes melted and the water tasted better. If my cup had 250 mL of water at 22ºC, and I put two ice cubes into the water that weighed 20 grams each, what was the final temperature of the water if the ice cubes were initially at 0°C? ΔHfus = 6.01 kJ/mol, Cp(H2O) = 75.2 J/molK 3) Assume that the cooking of an egg is a first order process. Assume further that it takes 3 minutes to half-cook (soft boil) an egg at 100ºC. If the activation energy of the cooking of an egg is 75 kJ/mole, how long would it take to soft boil an egg at the top of Mt. Everest where water boils at 80ºC? 4a) Calculate the pH at each of the following points for the titration of 100 mL of 2.00 M Phthalic acid with 5 M NaOH H2Phthal → H+ + HPhthalHPhthal- → H+ + Phthal2- pKA1 = 2.89 pKA2 = 5.41 INITIAL pH 20 mL NaOH 40 mL NaOH 50 mL NaOH 80 mL NaOH 100 mL NaOH 4b) How many milliliters of 5 M NaOH must you add to 100 mL of 2 M Phthalic acid to make a buffer of pH = 3.5? 5) At low temperatures, the rate law for the reaction, CO(g) + NO2(g) → CO2(g) + NO(g) can be determined by the following data; Initial Rate/10-4 [NO]/10-3 M 3.60 7.20 0.90 0.90 1.20 1.20 1.20 2.40 [CO]/10-3 M 1.50 3.00 0.75 1.50 [NO2]/10-3 M 0.80 0.80 0.40 0.40 Write a rate law in agreement with the data. Work out the rate law for each of the following mechanisms (show your work) and then circle the one that is consistent with the rate law determined above. (Circle A, B, or C) A) CO + NO2 → CO2 + NO slow B) NO2 ↔ NO + O equil. O + CO → CO2 slow C) 2 NO2 ↔ 2 NO + O2 equil. CO + O2 → CO2 + O slow O + NO → NO2 fast 6) What is the solubility of Barium Carbonate in a solution saturated with H2CO3 at a pH of 10? [H2CO3]sat'd = 0.034M. BaCO3 ↔ Ba2+ + CO32- Ksp = 8.1x10-9 H2CO3 ↔ 2 H+ + CO32- Keq = 2.02x10-17 7) A cell has the following cell diagram; Ag / AgCl / 1M HCl // 1M AgNO3 / Ag 7a) What is the reduction half reaction? 7b) What is the overall reaction? 7c) Based on the cell diagram please draw the electrochemical cell. 7d) If the Ksp for this cell is 1.8x10-10 then what is the cell voltage when all the ionic species are 0.01 M? 8) Please name or draw the following compounds. OH CH3 OH OH C H2N C C Cl O C C C C C C C O C C C C C Cl C C CH3 butanone 3,3-dichloro butanoic acid trans 1,3-cyclopentadiol o-methylaniline Chemistry 121 Final Exam Name Answer Key May 27, 1998 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit Question 1(15) 5(20) 2(20) 6(20) 3(20) 7(25) 4(40) 8(40) Total Total Credit TOTAL POINTS = R = 8.314 J/mol-K Δ = Δ° - 0.0592/n log Q ΔGº = -nΔ ° ln(Ai/Af) = kΔt = 96486 C/mol eΔG = ΔGº + RT ln Q PV = nRT 1/Af - 1/Ai = kΔt C = amp x sec ΔGº= -RTlnKeq R = 0.08205 L-atm/mol-K ln(k1/k2) = Ea/R (1/T2 - 1/T1) 1)Write down the equilibrium that occurs when the following compounds are mixed. a) NaOH + NaBenz Benz- + H2O ↔ HBenz + OHb) NH4OH + NH4Cl NH4OH ↔ NH4+ + OHc) NaCN + HCl HCN ↔ H+ + CN- 2) Last night I poured myself a glass of water from the tap and took a drink. The water was not cold and didn't taste good so I put a couple of ice cubes in it. The ice cubes melted and the water tasted better. If my cup had 250 mL of water at 22ºC, and I put two ice cubes into the water that weighed 20 grams each, what was the final temperature of the water if the ice cubes were initially at 0°C? ΔHfus = 6.01 kJ/mol, Cp(H2O) = 75.2 J/molK First, use the 22C water to melt the ice. The result will be colder water and melted ice. nCpT = nHFus 250 g/18 g/mol (75.2 J/molK)(TF - 22ºC) = 40g/18 (6010 J/mol) TF = 9.212ºC Now, mix this colder water (9.212ºC) with the melted ice (0ºC), -nCpT = nCpT -[250 g/18 g/mol (75.2 J/molK)(TF - 22ºC)] = 40 g/18 g/mol (75.2 J/molK)(TF - 0ºC) TF = 7.94ºC 3) Assume that the cooking of an egg is a first order process. Assume further that it takes 3 minutes to half-cook (soft boil) an egg at 100ºC. If the activation energy of the cooking of an egg is 75 kJ/mole, how long would it take to soft boil an egg at the top of Mt. Everest where water boils at 80ºC? t1 = 3 min @ 100ºC and t2 @ 80ºC ln t2 75,000 373 353 3 min 8.314 373 353 t2 = 11.81 min 4a) Calculate the pH at each of the following points for the titration of 100 mL of 2.00 M Phthalic acid with 5 M NaOH H2Phthal → H+ + HPhthalHPhthal- → H+ + Phthal2INITIAL pH x2 2.88 x10 3 2x pKA1 = 2.89 pKA2 = 5.41 pH = 1.29 KA1 = 2.88x10-3 KA2 = 3.89x10-6 M1V1 = M2V2 (2M)((0.100L) = (5 M)(V2) V2 = 40 mL for each region Also (2M)(0.100L) = 0.2 mol H2Phthal 20 mL NaOH pH = 2.89 40 mL NaOH pH 50 mL NaOH 10 mL into 2nd Region (5 M)(0.010 L) = 0.05 mol OH- pK A1 pK A2 2.89 5.41 2 2 pH 5.41 log 80 mL NaOH [Phthal 2- ] 0.05 0.2 0.05 pH = 4.933 0.2 mol 1x10 -14 1.111M and Kb 2.57 x10 9 0.180 L 3.89x10 -6 x2 2.57 x10 9 1.111 x 100 mL NaOH pH = 4.15 x = 5.3x10-5 M OH- (0.020 L)(5 M) = 0.10 mol OH0.10 mol OH 0.5 M OH 0.200 L pH = 9.72 pH = 13.70 4b) How many milliliters of 5 M NaOH must you add to 100 mL of 2 M Phthalic acid to make a buffer of pH = 3.5? The pH is nearest the 1st pKa so the buffer is in Region I. You must add, 3.5 2.89 log (x) x 0.1607 mol NaOH = (5 M)(V) (0.20 - x) V = 32.14 mL NaOH 5) At low temperatures, the rate law for the reaction, CO(g) + NO2(g) → CO2(g) + NO(g) can be determined by the following data; Initial Rate/10-4 [NO]/10-3 M 3.60 7.20 0.90 0.90 [CO]/10-3 M 1.20 1.20 1.20 2.40 1.50 3.00 0.75 1.50 [NO2]/10-3 M 0.80 0.80 0.40 0.40 Write a rate law in agreement with the data. rate k [CO] [NO 2 ] [NO] Work out the rate law for each of the following mechanisms (show your work) and then circle the one that is consistent with the rate law determined above. (Circle A, B, or C) A) CO + NO2 → CO2 + NO slow rate = k [CO] [NO2] B) NO2 ↔ NO + O O + CO → CO2 C) 2 NO2 ↔ 2 NO + O2 CO + O2 → CO2 + O O + NO → NO2 No, doesn’t match equil. slow [NO] [O] [NO 2 ] rate = k [O] [CO] and Keq so, [O] Keq [NO2 ] [NO] [CO] [NO 2 ] [CO] [NO 2 ] Substituting, we get rate kKeq or rate k' [NO] [NO] This matches our experimentally determined rate law so it is a possible mechanism. equil. slow fast rate = k [O2] [CO] The only reason to use the equilibrium is to get rid of something that is not in the overall reaction. Since the rate law does not contain anything that is not already in the overall reaction (no intermediates) there is no reason to continue. This does not match. 6) What is the solubility of Barium Carbonate in a solution saturated with H2CO3 at a pH of 10? [H2CO3]sat'd = 0.034M. BaCO3 ↔ Ba2+ + CO32- Ksp = 8.1x10-9 H2CO3 ↔ 2 H+ + CO32- Keq = 2.02x10-17 [H ]2 [CO32 ] (1x1010 M) 2 (CO32 ) 2.02x1017 [CO32 ] 68.68 M [H 2CO 3 ] 0.034M [Ba2+] [CO32-] = [Ba2+] [68.68 M] = 8.1x10-9 [Ba2+] = 1.18x10-10 M 7) A cell has the following cell diagram; Ag / AgCl / 1M HCl // 1M AgNO3 / Ag 7a) What is the reduction half reaction? Ag+ + e- → Ag 7b) What is the overall reaction? Ag+ + Cl- ↔ AgCl 7c) Based on the cell diagram please draw the electrochemical cell. e- Cathode Positive Reduction Ag Ag AgCl(s) 1M AgNO3 1M HCl Anode Negative Oxidation 7d) If the Ksp for this cell is 1.8x10-10 then what is the cell voltage when all the ionic species are 0.01 M? Δ° = 0.0592/n log Ksp = 0.0592/(1) log (1/1.8x10-10) = 0.5769 volts Δ = Δ° - 0.0592/n log Q = Δ° - 0.0592/n log [1/([Ag+][Cl-])] Δ = 0.5769 V- 0.0592/(1) log 1/[(0.01)(0.01)] Δ = 0.3401 volts 8) Please name or draw the following compounds. OH CH3 OH OH C C C H2N para methylaniline phenol 1,2-propadiol Cl O C C C C C C O C H3C C C C 3-methyl-2-pentanone butanone C C Cl C H CH3 cis,trans-2,4-dichloro-3-methyl--2,4-hexadiene 3,3-dichloro butanoic acid O C cyclopropyl ethyl ether CH3 Cl O C C C C HO trans 1,3-cyclopentadiol C C Cl o-methylaniline NH2 OH CH3 OH Chemistry 121 Final Exam Name May 26, 2000 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit Question 1(15) 6(20) 2(15) 7(15) 3(15) 8(30) 4(15) 9(20) 5(25) 10(30 ) Total Total Credit TOTAL POINTS = R = 8.314 J/mol-K = - 0.0592/n log Q ΔG = -n ln(Ai/Af) = kΔt = 96486 C/mol eΔG = ΔG + RT ln Q PV = nRT 1/Af - 1/Ai = kΔt C = amp x sec ΔG= -RTlnKeq R = 0.08205 L-atm/mol-K 1) Which of the following compounds is the most soluble in water? Show your work. BaSO4 Fe(OH)2 Ag3PO4 Ksp = 1.98x10-10 Ksp = 1.64x10-14 Ksp = 1.2x10-15 2) Please write the equilibrium reaction that will occur when the following compounds are mixed. Write all reactions as dissociations where appropriate. NaOH + HF Na3Asp + HCl HCl + KOH 3) The bromination of acetone is acid catalyzed; CH3COCH3 + Br2 → CH3COCH2Br + H+ + BrThe rate of disappearance of bromine was measured for several different concentrations of acetone, bromine and hydrogen ions; Rate 6x10-5 2.4x10-5 1.2x10-4 3.2x10-4 8x10-5 [Acetone] 0.30 0.30 0.30 0.40 0.40 [Br2] 0.050 0.100 0.050 0.050 0.050 [H+] 0.050 0.050 0.100 0.200 0.050 a) What is the forward rate law for this reaction? b) What is the value and units of the forward rate constant? 4) What is the solubility of Cu2S in a saturated solution of 0.10M H2S that is buffered at pH=10? Cu2S ↔ 2 Cu+ + S2H2S ↔ 2H+ + S2- Ksp = 2x10-47 Keq = 1.1x10-20 5a) Set up a WORKING electrochemical cell that will determine the solubility product for HgO. Note that this reaction is spontaneous in the reverse direction. Your overall reaction should be; HgO + H2O → Hg2+ + 2 OH- Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all ionic species. Indicate the composition of the electrodes. 5b) Draw the cell diagram for the above cell. 5c) If the = 1.056 volts, what is the voltage of the cell when concentration of the ionic species in the reduction cell are 0.5 M and the ionic species in the oxidation cell are 0.2 M? 6a) Assume that the cooking of an egg is a first order process. Assume further that it takes 3 minutes to half-cook (soft boil) an egg at 100°C. If the activation energy of the cooking of an egg is 75 kJ/mole, how long would it take to soft boil an egg at the top of Mt. Everest where water boils at 80°C? 6b) If the rate of a reaction doubles with a 15°C change in temperature, what is it's activation energy? State your assumptions. 7) A block of copper intially at 500°C was placed on top of a 100 gram block of ice at 0°C. All of the ice melted and turned into water. The water and the block of copper ended up at 20°C. How much did the block of copper weigh? ΔHfus = 6.01 kJ/mol, Cp(Cu) = 24.5 J/mol-K, Cp(H2O) = 75.2 J/mol-K, FWT (Cu) = 63.54 g/mol, FWT (H2O)= 18 g/mol 8) Calculate the pH at each of the following points for the titration of 100 mL of 3 M Phthalic acid with 5 M NaOH H2Phthal → H+ + HPhthal- pKA1 = 2.89 HPhthal- → H+ + Phthal2- pKA2 = 5.41 INITIAL pH 20 mL NaOH 60 mL NaOH 75 mL NaOH 90 mL NaOH 8b) How much 5 M NaOH would you add to 100 mL of 3 M H2Phthal to make a buffer of pH = 5? 9) How much heat must be added to 100 g of ice at -5°C to turn it into 60 grams of water and 40 grams of steam, both at 100°C? Cp H2O = 75.7 J/mol-K, Cp Ice = 37.7 J/mol-K, Cp Steam = 35.1 J/mol-K, ΔHfus = 6.01 kJ/mol, ΔHvap = 40.7 kJ/mol Chemistry 121 Final Exam Name Answer Key May 26, 2000 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit Question 1(15) 6(20) 2(15) 7(15) 3(15) 8(30) 4(15) 9(20) 5(25) 10(30 ) Total Total Credit TOTAL POINTS = R = 8.314 J/mol-K = - 0.0592/n log Q ΔG = -n ln(Ai/Af) = kΔt = 96486 C/mol eC = amp x sec ΔG = ΔG + RT ln Q ΔG= -RTlnKeq PV = nRT R = 0.08205 L-atm/mol-K 1/Af - 1/Ai = kΔt 1) Which of the following compounds is the most soluble in water? Show your work. BaSO4 Fe(OH)2 Ag3PO4 Ksp = 1.98x10-10 Ksp = 1.64x10-14 Ksp = 1.2x10-15 (x)(x) = x2 = 1.98x10-10 (x)(2x)2 = 4x3 = 1.64x10-14 (3x)3(x) = 27x4 = 1.2x10-15 x = 1.41x10-5 x = 1.6x10-5 x = 8.16x10-5 2) Please write the equilibrium reaction that will occur when the following compounds are mixed. Write all reactions as dissociations where appropriate. NaOH + HF F- + H2O ↔ HF + OH- Na3Asp + HCl HAsp2- ↔ H+ + Asp3- HCl + KOH H2O ↔ H+ + OH- 3) The bromination of acetone is acid catalyzed; CH3COCH3 + Br2 → CH3COCH2Br + H+ + BrThe rate of disappearance of bromine was measured for several different concentrations of acetone, bromine and hydrogen ions; Rate 6x10-5 2.4x10-5 1.2x10-4 3.2x10-4 8x10-5 [Acetone] 0.30 0.30 0.30 0.40 0.40 [Br2] 0.050 0.100 0.050 0.050 0.050 [H+] 0.050 0.050 0.100 0.200 0.050 a) What is the forward rate law for this reaction? rate = k [Acetone] [Br2]2 [H+] b) What is the value and units of the forward rate constant? 6x10-5 M/s = k [0.30 M] [0.050 M]2 [0.05 M] k = 1.6 1/M3s 4) What is the solubility of Cu2S in a saturated solution of 0.10M H2S that is buffered at pH=10? Cu2S ↔ 2 Cu+ + S2H2S ↔ 2H+ + S2- Ksp = 2x10-47 Keq = 1.1x10-20 [H+]2 [S2-]/[H2S] = 1x10-20 [S2-] = 1x10-20 [H2S]/[H+]2 = 1x10-20 [0.10]/[1x10-10]2 = 0.10 M S2Substitute this result into the following equation, [Cu+]2 [S2-] = 2x10-47 [Cu ] 2 x7 /[S2- ] 2 x7 /[0.10] 1.41x1023 M Cu [Cu 2S] 1.41x10-23 7.07 x10 24 M Cu 2S 2 5a) Set up a WORKING electrochemical cell that will determine the solubility product for HgO. Note that this reaction is spontaneous in the reverse direction. Your overall reaction should be; HgO + H2O ↔ Hg2+ + 2 OHThe half-reactions; Hg2+ + 2 e- → Hg Hg + 2 OH- ↔ HgO + H2O + 2 e- Cathode Positive Reduction 1 M NaOH 1 M Hg(NO3) 2 Anode Negative Oxidation Hg2O Pt e- Pt Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all ionic species. Indicate the composition of the electrodes. 5b) Draw the cell diagram for the above cell. Pt / Hg(l) / Hg2O(s) / 1M NaOH // 1M Hg(NO3)2 / Hg / Pt 5c) If the = 1.056 volts, what is the voltage of the cell when concentration of the ionic species in the reduction cell are 0.5 M and the ionic species in the oxidation cell are 0.2 M? = - 0.0592/n log 1/[Hg2+][OH-]2 = 1.056 V - 0.0592/2 log 1/[0.5M][0.2M]2 = 1.0057 Volts 6a) Assume that the cooking of an egg is a first order process. Assume further that it takes 3 minutes to half-cook (soft boil) an egg at 100°C. If the activation energy of the cooking of an egg is 75 kJ/mole, how long would it take to soft boil an egg at the top of Mt. Everest where water boils at 80°C? t1 = 3 min @ 100ºC and t2 @ 80ºC ln t2 75,000 373 353 3 min 8.314 373 353 t2 = 11.81 min 6b) If the rate of a reaction doubles with a 15°C change in temperature, what is its activation energy? State your assumptions. Assume k1 @ 298 K and 2 k1 @ 313 K ln 2k1 Ea 313 298 k1 8.314 313 298 Ea = 35,834 J/mol 7) A block of copper intially at 500°C was placed on top of a 100 gram block of ice at 0°C. All of the ice melted and turned into water. The water and the block of copper ended up at 20°C. How much did the block of copper weigh? ΔHfus = 6.01 kJ/mol, Cp(Cu) = 24.5 J/mol-K, Cp(H2O) = 75.2 J/mol-K, FWT (Cu) = 63.54 g/mol, FWT (H2O)= 18 g/mol nCpTCu = nfus + nCpTH2O xg J 100 g J 100 g J 24.5 20C 500C 6010 75.2 20C 63.54 molK 18 mol 18 molK x g = 225.55 g Cu 8) Calculate the pH at each of the following points for the titration of 100 mL of 3 M Phthalic acid with 5 M NaOH + - H2Phthal ↔ H + HPhthal pKA1 = 2.89 HPhthal- ↔ H+ + Phthal2- pKA2 = 5.41 M1V1 = M2V2 (3 M)(0.100 L) = (5 M)(V2) V2 = 60 mL Also, (3 M)(0.1-0 L) = 0.300 mol H2Phthal INITIAL pH x2 1.288x103 x 0.06217 M H 3 x 20 mL NaOH pH 2.89 log 60 mL NaOH pH 75 mL NaOH pH 5.41 log 90 mL NaOH 0.10 0.30 0.10 pK A1 pK A2 2.89 5.41 2 2 0.075 0.30 0.075 pH = 1.206 pH = 2.59 pH = 4.15 pH = 4.93 pH = 5.41 8b) How much 5 M NaOH would you add to 100 mL of 3 M H2Phthal to make a buffer of pH = 5? The pH is nearest the 2nd pKa so the buffer is in Region II. You must add, (3 M)(0.100 L) = (5 M)(V) V = 60 mL to get to the starting point. 5 5.41 log (x) x 0.0840 mol NaOH = (5 M)(V) (0.30 - x) Total volume = 16.8 mL + 60 mL = 76.8 mL NaOH V = 16.8 mL 9) How much heat must be added to 100 g of ice at -5°C to turn it into 60 grams of water and 40 grams of steam, both at 100°C? Cp H2O = 75.7 J/mol-K, Cp Ice = 37.7 J/mol-K, Cp Steam = 35.1 J/mol-K, ΔHfus = 6.01 kJ/mol, ΔHvap = 40.7 kJ/mol 100g J 100g J 100g J 40g J 37.7 5C 6010 75.2 100C 40,700 166,658 J 18 molK 18 mol 18 molK 18 mol Chem 121 Practice Problems Final Exam Nomenclature Isomers Free Radical Halogenation Hybridization Sigma and Pi bonds Intermolecular Forces Mechanism of Free Radical Halogenation using Bromine Chain Initiation light Br2 2 Br Chain Propagation Br H H H H H H C C C H H H H H H C C C H H H Br2 H + HBr H H H C C C H H H H Br H C C C H H H H Br H C C C H H H + Br H Chain Termination Br Br H Br H Br2 H C C C H H H H H H H H H C C C H H H H H H H H C C C H H H H Relative Reactivity 1 2 3 Chlorination 1 3.5 5 Bromination 1 97 H C H H H C H H C H H H C C H H H C H H Section I 3) Please draw all of the isomers of C4H7Cl. There are more than a dozen. 4) Please draw all of the isomers of C3H5Cl. 5) Please draw all of the isomers of C3H4Cl2. 6) Please draw all of the isomers of C3H8O 7) Please answer the following questions concerning the molecule given below. a) How many pi and sigma bonds are in this compound? O8 C 7 9 10 O CH3 1 6 b) What is the hybridization on the following atoms; 1 2 5 7 8 9 3 5 4 8) Please answer the following questions concerning the molecule given below. 8 2 a) How many pi and sigma bonds are in this compound? O 1 H b) What is the hybridization on the following atoms; N3 C C 4 Cl C 6 5 Cl 7 2 3 4 5 7 8 9) Please answer the following questions concerning the molecule given below. 7 O 6C N 8 3 9 C a) How many pi and sigma bonds are in this compound? 10 b) What is the hybridization on the following atoms; 5 4 C 1 2 1 2 5 7 8 14) Please describe what happens to the melting point of a large alkene (C20 or larger) if the double bond changes from cis to trans. Also explain why it does this. 15) Please explain what happens to the melting point of an alkane as it becomes more branched. 16) What kinds of compounds can hydrogen bond? What is hydrogen bonding? 17) What is the major intermolecular bonding type for each of the following compounds? CH3CH2NH2 CH2Cl2 CH3COOH 18) Which compound in each pair has the highest boiling point? CH3OCH2CH3 or CH3CH(OH)CH3 CH3CH2CH2CH2CH3 or CH3CH2CH2CH3 (CH3)2CHCH3 or CH3CH2CH2CH3 19) Please predict the primary type of bonding that will occur in each of the following liquids. CCl4 C16H34 Propanone H-bond H-bond H-bond Dipole-Dipole Dipole-Dipole Dipole-Dipole Dispersion Dispersion Dispersion Section I – Answer Key 3) Please draw all of the isomers of C4H7Cl. There are more than a dozen. Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl 4) Please draw all of the isomers of C3H5Cl. Cl Cl Cl Cl 5) Please draw all of the isomers of C3H4Cl2. Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl 6) Please draw all of the isomers of C3H8O HO OH O Cl 7) Please answer the following questions concerning the molecule given below. a) How many pi and sigma bonds are in this compound? 5 pi and 16 sigma O8 C 7 9 10 O CH3 1 6 b) What is the hybridization on the following atoms; 1 sp2 2 5 sp 7 sp2 8 sp2 9 sp3 3 5 4 8) Please answer the following questions concerning the molecule given below. 8 2 a) How many pi and sigma bonds are in this compound? 4 pi and 16 sigma O 1 N3 H C 4 Cl C 5 C 6 Cl 7 b) What is the hybridization on the following atoms; 2 sp2 3 sp3 4 sp2 5 sp 7 sp3 8 sp3 9) Please answer the following questions concerning the molecule given below. 7 O 6C N C C 8 9 10 a) How many pi and sigma bonds are in this compound? 3 pi and 21 sigma 1 b) What is the hybridization on the following atoms; 5 4 3 2 1 sp2 2 sp2 5 sp3 7 sp2 8 sp3 14) Please describe what happens to the melting point of a large alkene (C20 or larger) if the double bond changes from cis to trans. Also explain why it does this. The melting point increases. A cis bond puts a kink in the chain that makes it hard to stack. By changing the cis to a trans the chain straightens out and makes it easier to stack. This allows for more dispersion forces and increases the melting point. 15) Please explain what happens to the melting point of an alkane as it becomes more branched. The more branched an alkane becomes the lower the melting point. Branching makes it more difficult to stack the molecules and this reduces the molecules ability to use dispersion forces for bonding. 16) What kinds of compounds can hydrogen bond? What is hydrogen bonding? Hydrogen bonds are formed between molecules that have OH or NH bonds (also HF but HF is not an organic compound). A hydrogen bond is a sharing of protons between the oxygen or nitrogen molecules. The hydrogen atoms freely move between the oxygens or nitrogens by sharing grabbing onto the lone pair electrons found on these atoms. Each atom of oxygen or nitrogen can have up to 4 atoms of hydrogen surrounding them, each of them sharing their electrons. 17) What is the major intermolecular bonding type for each of the following compounds? CH3CH2NH2 = H bond CH2Cl2 = Dipole-dipole CH3COOH = H bond 18) Which compound in each pair has the highest boiling point? CH3OCH2CH3 or CH3CH(OH)CH3(because of H bonds) CH3CH2CH2CH2CH3 (because it is bigger) or CH3CH2CH2CH3 (CH3)2CHCH3 or CH3CH2CH2CH3(because it is less branched) 19) Please predict the primary type of bonding that will occur in each of the following liquids. CCl4 C16H34 Propanone H-bond H-bond H-bond Dipole-Dipole Dipole-Dipole Dipole-Dipole Dispersion Dispersion Dispersion Section II 1) Please name or draw the structure of the following compounds. H CH3 Cl CH3 H3C H H3C CH3 Cl Br H3C Br Cl 3) Please draw all of the isomers of C3H3Cl C CH3 CH3 4) Please answer the following questions concerning the molecule given below. O 3 C O 4 C CH3 a) How many pi and sigma bonds are in this compound? N2 1 N C b) What is the hybridization on the following atoms; 1 2 3 4 6a) Please write down all the steps in the mechanism of the free radical chlorination of 2methylbutane.. 6b) What is the ratio of products formed in the free radical chlorination of bicyclo [3,2,0] heptane? 7b) What happens to the boiling point of an alkane when it becomes branched? 7c) How do the intermolecular bonding forces change when acids become larger? 7d) What is the primary intermolecular bonding force in each of the following compounds? Isopropyl Alcohol H-Bond Dipole-Dipole Dispersion 1,3-dichloropropane H-Bond Dipole-Dipole Dispersion C12H25NH2 H-Bond Dipole-Dipole Dispersion methylcyclohexane H-Bond Dipole-Dipole Dispersion Section II – Answer Key 1) Please name or draw the structure of the following compounds. H CH3 Cl CH3 H3C Bicyclo [4, 2, 1] nonane H S-3-chloro-2-methylbutane H3C CH3 cis-1,4-dimethylcyclohexane 2,2,4-trimethylpentane Cl Br H3C C Br CH3 CH3 Cl Z-bromo-1,2-dichlorocyclobutane 2-bromo-2-methylpropane 3) Please draw all of the isomers of C3H3Cl H Cl C C H C H H C C H H C C Cl H Cl C C H Cl H H H H Cl H H H 4) Please answer the following questions concerning the molecule given below. O 3 C N2 1 N C O 4 C CH3 a) How many pi and sigma bonds are in this compound? 20 sigma and 6 pi b) What is the hybridization on the following atoms; 1 2 3 4 sp sp3 sp3 sp3 6a) Please write down all the steps in the mechanism of the free radical chlorination of 2methylbutane. Chain Initiation light Cl2 2 Cl Chain Propagation H Cl H3C C C2H5 HCl + H3C C C2H5 CH3 CH3 Cl H3C C Cl2 C2H5 H3C C CH3 C2H5 + Cl CH3 Chain Termination Cl Cl Cl2 Cl Cl H3C C H3C C2H5 C C2H5 CH3 CH3 CH3 H3C C C2H5 H3C C CH3 C2H5 CH3 H3C C C2H5 H3C C C2H5 CH3 6b) What is the ratio of products formed in the free radical chlorination of bicyclo [3,2,0] heptane? secondary tertiary X 100-X = 10 x 3.5 2 5 X = 77.78% secondary 22.22% tertiary 7b) What happens to the boiling point of an alkane when it becomes branched? The boiling point goes down 7c) How do the intermolecular bonding forces change when acids become larger? As acids get larger they get less soluble in water so the intermolecular bonding force changes from being mostly H-bond to mostly dispersion forces 7d) What is the primary intermolecular bonding force in each of the following compounds? Isopropyl Alcohol H-Bond Dipole-Dipole Dispersion 1,3-dichloropropane H-Bond Dipole-Dipole Dispersion C12H25NH2 H-Bond Dipole-Dipole Dispersion methylcyclohexane H-Bond Dipole-Dipole Dispersion Section III 1) Please name or draw the structure of the following compounds. Cl Cl C C CH3 Cl H3C CH C C CH3 C C CH3 3) Please draw all of the isomers of C3H4Cl2 H3C Cl H H C2H5 4) Please answer the following questions concerning the molecule given below. O 2 a) How many pi and sigma bonds are in this compound? 3 N O C OH 4 1 b) What is the hybridization on the following atoms; 5 C 1 N 2 3 4 5 7a) Hexane and cyclohexane both have six carbons but one has a higher boiling point than the other. Which one has the higher boiling point and why? 7b) Rank the following compounds from highest to lowest boiling point (highest =4, lowest = 1) C C C C C C C C C C C C C C C C C C OH C 7d) What is the primary intermolecular bonding force in each of the following compounds (Hbond, Dipole, Dispersion)? C2H5NH2 cis 2,3-dichlorobutene CH3(CH2)4CH2Br C12H25COH Section III – Answer Key 1) Please name or draw the structure of the following compounds. Cl Cl Cis-1,2-dichlorocyclpentane C C Bicyclo [4,1,0] heptane CH3 Cl H3C CH C C CH3 C C CH3 H3C Cl H H 4-ethyl-5-methyloctane C2H5 2R,3R-2,3-dichloropentane 3) Please draw all of the isomers of C3H4Cl2 C Cl Cl C C C C C C Cl C C Cl C Cl Cl Cl C C C Cl Cl Cl C C C C Cl Cl C Cl C C C Cl Cl Cl Cl Cl Cl Cl Cl 4) Please answer the following questions concerning the molecule given below. O a) How many pi and sigma bonds are in this compound? 3 2 N 18 sigma and 4 pi O C OH 4 1 5 C N b) What is the hybridization on the following atoms; 1 2 3 4 5 sp2 sp3 sp3 sp2 sp Cl 6) Please predict the percentage of products made by the free radical chlorination of bicycle [4,2,0] octane. secondary tertiary X 100-X = 12 x 3.5 2 5 X = 80.77% secondary 19.23% tertiary 7a) Hexane and cyclohexane both have six carbons but one has a higher boiling point than the other. Which one has the higher boiling point and why? Hexane has the higher boiling point because it is longer and less compact than cyclohexane so it has more opportunity to exert dispersion forces on its neighbors. 7b) Rank the following compounds from highest to lowest boiling point (highest =4, lowest = 1) C C C C C C C C C C C C C C C C C C OH C 1 3 2 4 7d) What is the primary intermolecular bonding force in each of the following compounds (Hbond, Dipole, Dispersion)? C2H5NH2 H-bond CH3(CH2)4CH2Br Dipole-Dipole (borderline) cis 2,3-dichlorobutene Dispersion C12H25COH Dispersion (too big)
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