IONIC COMPOUNDS Chapter 5 Many reactions involve ionic compounds, especially reactions in water — aqueous solutions. KMnO4 in water K+(aq) + MnO4-(aq) 1 2 An Ionic Compound, CuCl2, in Water Aqueous Solutions How do we know ions are present in aqueous solutions? The solutions conduct electricity! They are called ELECTROLYTES HCl, MgCl2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions. 3 Aqueous Solutions HCl, MgCl2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions. 4 Aqueous Solutions Acetic acid ionizes only to a small extent, so it is a weak electrolyte. CH3CO2H(aq) ---> CH3CO2-(aq) + H+(aq) 5 Aqueous Solutions Acetic acid ionizes only to a small extent, so it is a weak electrolyte. CH3CO2H(aq) ---> CH3CO2-(aq) + H+(aq) 6 Aqueous Solutions Some compounds dissolve in water but do not conduct electricity. They are called nonelectrolytes. Examples include: sugar ethanol ethylene glycol 7 Water Solubility of Ionic Compounds 8 If one ion from the “Soluble Compd.” list is present in a compound, the compound is water soluble. Screenof5.4 & Figure 5.1 Guidelines to predict the solubility ionic compounds 9 Water Solubility of Ionic Compounds Common minerals are often formed with anions that lead to insolubility: sulfide fluoride carbonate oxide Iron pyrite, a sulfide Azurite, a copper carbonate Orpiment, arsenic sulfide ACIDS 10 An acid -------> H+ in water Some strong acids are HCl H2SO4 HClO4 HNO3 hydrochloric sulfuric perchloric nitric HNO3 ACIDS An acid -------> H+ in water HCl(aq) ---> H+(aq) + Cl-(aq) 11 12 The Nature of Acids HCl Cl- H 2O hydronium ion H 3O+ Weak Acids WEAK ACIDS = weak electrolytes CH3CO2H acetic acid H2CO3 carbonic acid H3PO4 phosphoric acid HF hydrofluoric acid Acetic acid 13 ACIDS Nonmetal oxides can be acids CO2(aq) + H2O(liq) ---> H2CO3(aq) SO3(aq) + H2O(liq) ---> H2SO4(aq) and can come from burning coal and oil. 14 BASES see Screen 5.9 and Table 5.2 Base ---> OH- in water NaOH(aq) NaOH is a strong base ---> Na+(aq) + OH-(aq) 15 Ammonia, NH3 An Important Base 16 BASES Metal oxides are bases CaO(s) + H2O(liq) --> Ca(OH)2(aq) CaO in water. Indicator shows solution is basic. 17 18 Know the strong acids & bases! 19 Net Ionic Equations Mg(s) + 2 HCl(aq) --> H2(g) + MgCl2(aq) We really should write Mg(s) + 2 H+(aq) + 2 Cl-(aq) ---> H2(g) + Mg2+(aq) + 2 Cl-(aq) The two Cl- ions are SPECTATOR IONS — they do not participate. Could have used NO3-. Net Ionic Equations Mg(s) + 2 HCl(aq) --> H2(g) + MgCl2(aq) Mg(s) + 2 H+(aq) + 2 Cl-(aq) ---> H2(g) + Mg2+(aq) + 2 Cl-(aq) We leave the spectator ions out — Mg(s) + 2 H+(aq) ---> H2(g) + Mg2+(aq) to give the NET IONIC EQUATION 20 Chemical Reactions in Water 21 Sections 5.2 & 5.4-5.6—CD-ROM Ch. 5 We will look at EXCHANGE REACTIONS AX + BY AY + BX The anions exchange places between cations. Pb(NO3) 2(aq) + 2 KI(aq) ----> PbI2(s) + 2 KNO3 (aq) Precipitation Reactions The “driving force” is the formation of an insoluble compound — a precipitate. Pb(NO3)2(aq) + 2 KI(aq) -----> 2 KNO3(aq) + PbI2(s) Net ionic equation Pb2+(aq) + 2 I-(aq) ---> PbI2(s) 22 Acid-Base Reactions • The “driving force” is the formation of water. NaOH(aq) + HCl(aq) ---> NaCl(aq) + H2O(liq) • Net ionic equation OH-(aq) + H+(aq) ---> H2O(liq) • This applies to ALL reactions of STRONG acids and bases. 23 Acid-Base Reactions CCR, page 162 24 Acid-Base Reactions • A-B reactions are sometimes called NEUTRALIZATIONS because the solution is neither acidic nor basic at the end. • The other product of the A-B reaction is a SALT, MX. HX + MOH ---> MX + H2O Mn+ comes from base & Xn- comes from acid This is one way to make ionic compounds! 25 Gas-Forming Reactions This is primarily the chemistry of metal carbonates. CO2 and water ---> H2CO3 H2CO3(aq) + Ca2+ ---> 2 H+(aq) + CaCO3(s) (limestone) Adding acid reverses this reaction. MCO3 + acid ---> CO2 + salt 26 27 Gas-Forming Reactions CaCO3(s) + H2SO4(aq) ---> 2 CaSO4(s) + H2CO3(aq) Carbonic acid is unstable and forms CO2 & H2O H2CO3(aq) ---> CO2 (g) + water (Antacid tablet has citric acid + NaHCO3) 28 See also: Gas Forming Reactions in Biological Systems Three of the pioneers in working out the roles of NO forming reactions shared a Nobel Prize in 1988 for their discoveries. 29 Quantitative Aspects of Reactions in Solution Sections 5.8-5.10 Terminology In solution we need to define the • SOLVENT the component whose physical state is preserved when solution forms • SOLUTE the other solution component 30 31 Concentration of Solute The amount of solute in a solution is given by its concentration. Molarity (M) = moles solute liters of solution Concentration (M) = [ …] 32 1.0 L of water was used to make 1.0 L of solution. Notice the water left over. CCR, page 177 33 PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough water to make 250 mL of solution. Calculate molarity. Step 1: Calculate moles of NiCl2•6H2O 1 mol 5.00 g • = 0.0210 mol 237.7 g Step 2: Calculate molarity 0.0210 mol = 0.0841 M 0.250 L [NiCl2•6 H2O ] = 0.0841 M The Nature of a CuCl2 Solution Ion Concentrations CuCl2(aq) --> Cu2+(aq) + 2 Cl-(aq) If [CuCl2] = 0.30 M, then [Cu2+] = 0.30 M [Cl-] = 2 x 0.30 M 34 35 USING MOLARITY What mass of oxalic acid, H2C2O4, is required to make 250. mL of a 0.0500 M solution? Because Conc (M) = moles/volume = mol/V this means that moles = M•V USING MOLARITY What mass of oxalic acid, H2C2O4, is required to make 250. mL of a 0.0500 M solution? moles = M•V Step 1: Calculate moles of acid required. (0.0500 mol/L)(0.250 L) = 0.0125 mol Step 2: Calculate mass of acid required. (0.0125 mol )(90.00 g/mol) = 1.13 g 36 37 Preparing Solutions • Weigh out a solid solute and dissolve in a given quantity of solvent. • Dilute a concentrated solution to give one that is less concentrated. 38 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Add water to the 3.0 M solution to lower its concentration to 0.50 M Dilute the solution! 39 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? But how much water do we add? 40 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? How much water is added? The important point is that ---> moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Amount of NaOH in original solution = M•V = (3.0 mol/L)(0.050 L) = 0.15 mol NaOH Amount of NaOH in final solution must also = 0.15 mol NaOH 0.15/Volume of final solution = 0.5 M/ 1 L Volume of final solution = (0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L or 300 mL 41 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Conclusion: add 250 mL of water to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH. 42 Preparing Solutions by Dilution A shortcut Cinitial • Vinitial = Cfinal • Vfinal 43 The pH Scale 44 pH = log (1/ [H+]) = - log [H+] Remember : log a = b if 10b=a In a neutral solution, [H+] = [OH-] = 1.00 x 10-7 M at 25 oC pH = - log [H+] = -log (1.00 x 10-7) - (-7) = 7 See CD Screen 5.17 for a tutorial = pH, a Concentration Scale pH: a way to express acidity -- the concentration of H+ in solution. Low pH: high [H+] Acidic solution Neutral Basic solution High pH: low [H+] pH < 7 pH = 7 pH > 7 45 [H+] and pH If the [H+] of soda is 1.6 x 10-3 M, the pH is ____? Because pH = - log [H+] then pH= - log (1.6 x 10-3) pH = - (-2.80) pH = 2.80 What’s the origin of the name of the soda 7up ? 46 ACID-BASE REACTIONS Titrations H2C2O4(aq) + 2 NaOH(aq) ---> acid base Na2C2O4(aq) + 2 H2O(liq) Carry out this reaction using a TITRATION. Oxalic acid, H2C2O4 47 Setup for titrating an acid with a base CCR, page 186 48 Titration 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. 3. Indicator shows when exact stoichiometric reaction has occurred. 4. Net ionic equation H+ + OH- --> H2O 5. At equivalence point moles H+ = moles OH- 49 LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately determine its concentration. 1.065 g of H2C2O4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH? 50 51 1.065 g of H2C2O4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH? Step 1: Calculate amount of H2C2O4 1 mol 1.065 g • = 0.0118 mol 90.04 g Step 2: Calculate amount of NaOH req’d 2 mol NaOH 0.0118 mol acid • = 0.0236 mol NaOH 1 mol acid 52 1.065 g of H2C2O4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH? Step 1: Calculate amount of H2C2O4 = 0.0118 mol acid Step 2: Calculate amount of NaOH req’d = 0.0236 mol NaOH Step 3: Calculate concentration of NaOH 0.0236 mol NaOH 0.663 M 0.03562 L [NaOH] = 0.663 M 53 LAB PROBLEM #2: Use standardized NaOH to determine the amount of an acid in an unknown. Apples contain malic acid, C4H6O5. C4H6O5(aq) + 2 NaOH(aq) ---> Na2C4H4O5(aq) + 2 H2O(liq) 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid? 54 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid? Step 1: C•V = = Step 2: Calculate amount of NaOH used. (0.663 M)(0.03456 L) 0.0229 mol NaOH Calculate amount of acid titrated. 1 mol acid 0.0229 mol NaOH • 2 mol NaOH = 0.0115 mol acid 55 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid? Step 1: = Step 2: = Calculate amount of NaOH used. 0.0229 mol NaOH Calculate amount of acid titrated 0.0115 mol acid Step 3: Calculate mass of acid titrated. 134 g 0.0115 mol acid • = 1.54 g mol 56 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid? Step 1: = Step 2: = Step 3: = Calculate amount of NaOH used. 0.0229 mol NaOH Calculate amount of acid titrated 0.0115 mol acid Calculate mass of acid titrated. 1.54 g acid Step 4: Calculate % malic acid. 1.54 g • 100% = 2.01% 76.80 g pH and [H+] If the pH of Coke is 3.12, it is ____________. Because pH = - log [H+] then log [H+] = - pH Take antilog and get [H+] = 10-pH [H+] = 10-3.12 = 7.6 x 10-4 M 57 SOLUTION STOICHIOMETRY Gas-forming reactions • Zinc reacts with acids to produce H2 gas. • Have 10.0 g of Zn • What volume of 2.50 M HCl is needed to convert the Zn completely? 58 GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS Mass HCl Mass zinc Moles zinc Stoichiometric factor Moles HCl Volume HCl 59 60 Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? Step 1: Write the balanced equation Zn(s) + 2 HCl(aq) --> ZnCl2(aq) + H2(g) Step 2: Calculate amount of Zn 1.00 mol Zn 10.0 g Zn • = 0.153 mol Zn 65.39 g Zn Step 3: Use the stoichiometric factor 61 Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? Step 3: Use the stoichiometric factor 2 mol HCl 0.153 mol Zn • = 0.306 mol HCl 1 mol Zn Step 4: Calculate volume of HCl req’d 1.00 L 0.306 mol HCl • = 0.122 L HCl 2.50 mol 62 EXCHANGE: Precipitation Reactions EXCHANGE Gas-Forming Reactions REACTIONS REDOX REACTIONS EXCHANGE Acid-Base Reactions 63 REDOX REACTIONS Redox reactions are characterized by ELECTRON TRANSFER between an electron donor and electron acceptor. Transfer leads to— 1. increase in oxidation number of some element = OXIDATION 2. decrease in oxidation number of some element = REDUCTION 64 REDOX REACTIONS Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s) In all reactions if something has been oxidized then something has also been reduced 65 Why Study Redox Reactions Batteries Corrosion Manufacturing metals Fuels 66 OXIDATION NUMBERS The electric charge an element APPEARS to have when electrons are counted by some arbitrary rules: 1. Each atom in free element has ox. no. = 0. Zn O2 I2 S8 2. In simple ions, ox. no. = charge on ion. -1 for Cl+2 for Mg2+ OXIDATION NUMBERS 3. F always has an oxidation number of -1 when forming compounds with other elements. 4. Cl, Br and I have oxidation numbers of -1 when forming compounds with other elements, except when combined with oxygen and fluorine. 5a. O has ox. no. = -2 (except in peroxides: in H2O2, O = -1) 67 OXIDATION NUMBERS 5b. Ox. no. of H = +1 (except when H is associated with a metal as in NaH where it is -1) 6. Algebraic sum of oxidation numbers = 0 for a compound = overall charge for an ion 68 OXIDATION NUMBERS NH3 N = ClO- Cl = H3PO4 P = MnO4- Mn = Cr2O72C3H8 Cr = C = Oxidation number of F in HF? 69 Recognizing a Redox Reaction Corrosion of aluminum 2 Al(s) + 3 Cu2+(aq) --> 2 Al3+(aq) + 3 Cu(s) Al(s) --> Al3+(aq) + 3 e• Ox. no. of Al increases as e- are donated by the metal. • Therefore, Al is OXIDIZED • Al is the REDUCING AGENT in this balanced halfreaction. 70 Recognizing a Redox Reaction 71 Corrosion of aluminum 2 Al(s) + 3 Cu2+(aq) --> 2 Al3+(aq) + 3 Cu(s) Cu2+(aq) + 2 e- --> Cu(s) • Ox. no. of Cu decreases as e- are accepted by the ion. • Therefore, Cu is REDUCED • Cu is the OXIDIZING AGENT in this balanced halfreaction. Recognizing a Redox Reaction Notice that the 2 half-reactions add up to give the overall reaction —if we use 2 moles of Al and 3 moles of Cu2+. 2 Al(s) --> 2 Al3+(aq) + 6 e3 Cu2+(aq) + 6 e- --> 3 Cu(s) ----------------------------------------------------------2 Al(s) + 3 Cu2+(aq) ---> 2 Al3+(aq) + 3 Cu(s) Final eqn. is balanced for mass and charge. 72 Common Oxidizing and Reducing Agents 73 See Table 5.4 Metals (Cu) are reducing agents Metals (Na, K, Mg, Fe) are reducing agents HNO3 is an oxidizing agent Cu + HNO3 --> 2 K + 2 H2O --> Cu2+ + NO2 2 KOH + H2 74 Recognizing a Redox Reaction See Table 5.4 Reaction Type Oxidation Reduction In terms of oxygen gain loss In terms of halogen gain loss In terms of electrons loss gain Examples of Redox Reactions Metal + halogen 2 Al + 3 Br2 ---> Al2Br6 75 Examples of Redox Reactions Nonmetal (P) + Oxygen Metal (Mg) + Oxygen 76 Examples of Redox Reactions 77 Metal + acid Mg + HCl Mg = reducing agent H+ = oxidizing agent Metal + acid Cu + HNO3 Cu = reducing agent HNO3 = oxidizing agent
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