Solution
Chapter 15 Section 2 Concentrations of Solutions Concentration of Solution The ratio of solute to solvent Units – molarity (M), molality (m), mole fraction, % by mass, parts per million (ppm), parts per billion (ppb) % by Mass (m/m) grams of solute in 100 g of solution g of solute % by mass 100 g of solution Different from solubility. How? mass = density × volume m/v = mass of solute in volume of solution v/v = volume of solute in volume of solution Masses are additive mass of solute + mass of solvent = mass of solution Volumes are not additive, unless told to do so volume of solute + volume of solvent ≠ volume of solution Example 1 12.5 g of NaCl is dissolved in 100. mL of water. Find the % of NaCl in the solution. Density of water = 1 g/mL Example 2 How much water would you need to dissolve 35.6 g KNO3 to make a 25.5% m/m KNO3 solution? Parts Per Million (ppm) Useful for very diluted solution ppm=mg of solute in 1000 000 mg of solution = mg of solute in 1000 g of solution = mg of solute in 1 kg of solution mg of solute g of solute ppm 1, 000, 000 kg of solution g of solution *For a very dilute solution, g of solution = g of solvent % vs. ppm vs. ppb g of solute % (parts out of 100) 100 g of solution g of solute ppm (parts out of million) 1, 000, 000 g of solution ppb (parts out of billion) g of solute 1,000,000,000 g of solution For ppm and ppb, g of solution ≈ g of solvent Example A chemical analysis shows that there are 2.2 mg of lead in exactly 500 g of a water sample. Convert this measurement to ppm. Solution Convert 500 g to kg: 1 kg 500 g 0.5 kg 1000 g mg of solute ppm kg of solution 2.2 mg = 0.5 kg = 4.4 ppm Practice 1. Helium gas, 3.0x10-4 g, is dissolved in 200.0 g of water. Express this concentration in ppm. 2. A sample of 300.0 g of drinking water is found to contain 38 mg Pb. What is this concentration in ppm? 3. A solution of lead sulfate contains 0.425 g of lead sulfate in 100.0 g of water. What is this concentration in ppm? 4. A 900.0 g sample of sea water is found to contain 6.7x10-3 g Zn. Express this concentration in ppm. 5. A 365.0 g sample of water contains 23 mg Au. How much gold is present in the sample in ppm? 6. A 650.0 g hard-water sample contains 101 mg Ca. What is this concentration in ppm? 7. An 870.0 g river water sample contains 2 mg of cadmium. Express the concentration of cadmium in ppm. Answers Pg 461 1) 2) 3) 4) 5) 6) 7) 1.5 ppm 130 ppm 4250 ppm 7.4 ppm 63 ppm 155 ppm 2.3 ppm Molarity (M) mol Solute molarity (M) L Solution Example What is the molarity of a potassium chloride, KCl, solution that has a volume of 400.0 mL and contains 85.0 g KCl? Solution msolute = 85.0 g KCl Vsolution = 400.0 mL MKCl = ? Convert 85.0 g KCl to moles: 1 mol KCl 85.0 g KCl 1.14 mol KCl 74.55 g KCl Convert 400.0 mL to L: 1L 400.0 mL 0.400 L 1000 mL Calculate the molarity: mol KCl 1.14 mol KCl M= 2.85 mol/L L Solution 0.4000 L Example How many grams of solute are needed to make 2.50 L of a 1.75 M solution of Ba(NO3)2? Solution to Example 1.75 mol n 1.75 M= 1L 2.5 L 1 L n (1.75 mol) (2.5 L) n=4.38 mol 261.3 g Ba(NO3 ) 2 4.38 mol Ba(NO3 ) 2 1 mol Ba(NO3 ) 2 1144 g Ba(NO3 ) 2 Practice 1. Vinegar contains 5.0 g of acetic acid, CH3COOH in 100.0 mL of solution. Calculate the molarity of acetic acid in vinegar 2. If 18.25 g HCl is dissolved in enough water to make 500.0 mL of solution, what is the molarity of the HCl solution? 3. If 20.0 g H2SO4 is dissolved in enough water to make 250.0 mL of solution, what is the molarity of the sulfuric acid solution? 4. A solution of AgNO3 contains 29.66 g of solute in 100.0 mL of solution. What is the molarity of the solution? 5. A solution of barium hydroxide, Ba(OH)2 contains 4.285 g of barium hydroxide in 100.0 mL of solution. What is the molarity of the solution? 6. What mass of KBr is present in 25 mL of 0.85 M solution of potassium bromide? 7. If all the water in 430.0 mL of a 0.45 M NaCl solution evaporates, what mass of NaCl will remain? Answers 1) 2) 3) 4) 5) 6) 7) 0.83 M 1.001 M 0.816 M 1.75 M 0.2501 M 2.5 g KBr 11 g NaCl Molarity (M) moles of solute molarity (M) = liters of solution % Concentration (By Mass) mass solute % by mass= 100 mass solution Masssolution = dsolution x vsolution Example What mass of lithium nitrate, LiNO3, would have to be dissolved in 30.0 g of water in order to make an 18.0% solution? Solution to Example msolute 100 msolution msolute 100=18 msolute 30.0 msolute 0.18 msolute 30.0 msolute 0.18( msolute 30.0) msolute 0.18msolute 5.4 0.18msolute 0.18msolute 0.82msolute 5.4 msolute 5.4 6.59( g ) 0.82 PRACTICE 1. A student wants to make a 5.00% of rubidium chloride, RbCl using 0.377 g of the substance. What mass of water will be needed to make the solution? 2. A chemist dissolves 3.50 g of potassium iodate and 0.23 g of potassium hydroxide in 805.05 g of water. What is the percentage concentration of each solution? 3. What is the percentage concentration of 75.0 g of ethanol dissolved in 500.0 g of water? Answers 1. 7.16 g H2O 2. 0.430% potassium iodate 0.0284% potassium hydroxide 3. 13.0% ethanol Molality (m) mol Solute molality = kg Solvent Do NOT capitalize “m”. Example A NaCl (molar mass = 58 g/mol) solution is made by dissolving 29 g of NaCl in 200 g of water. What is the molality of the solution? Mole Fraction mol of solute mole fraction, solute totoal mol of solution mol of solvent solvent totoal mol of solution solute solvent 1 Example What is the mole fraction of sulfur dioxide in an industrial exhaust gas containing 128.0 g of sulfur dioxide dissolved in every 1500. g of carbon dioxide? Solution Stoichiometry Mass, Volume, Concentration of Substance A Moles of Substance A Moles of Substance B Mass, Volume, Concentration of Substance B Example What volume, in mL, of a 0.500 M solution of copper (II) sulfate, CuSO4, is needed to react with an excess of aluminum to provide 11.0 g of copper? 3CuSO4(aq) + 2Al(s) → 3Cu(s) + Al2(SO4)3(aq) Solution to Example CuSO4 solution = 0.500 M mCu = 11.0 g Vsolution = ? 1 mol Cu 3 mol CuSO4 11.0 g Cu 0.173 mol CuSO4 63.55 g Cu 3 mol Cu 0.500 mol CuSO 4 0.173 mol CuSO 4 0.500 M = 1 L Solution V (0.500) V=(0.173) 1 L 0.173 L 1L V= 0.346 L 346 mL 0.500 1000 mL Practice Pg 467 1) Commercial hydrochloric acid, HCl, is 12.0 molar. Calculate the mass of HCl in 250.0 mL of the solution. 2) An excess of zinc is added to 125 mL of 0.100 M HCl solution. What mass of zinc chloride is formed? Zn + 2HCl → ZnCl2 + H2 3) Yellow CdS pigment is prepared by reacting ammonium sulfide with cadmium nitrate. What mass of CdS can be prepared by mixing 2.50 L of a 1.25 M Cd(NO3)2 solution with an excess of (NH4)2S? Cd(NO3)2(aq) + (NH4)2S(aq) → CdS(aq) + 2NH4NO3(aq) Answers 1) 109 g HCl 2) 0.852 g ZnCl2 3) 451 g CdS
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