Aqueous Solutions and Solubility Equilibria

CHEMISTRY 12

Chapter 9 Aqueous Solutions and Solubility Equilibria

Solutions for Practice Problems Student Textbook page 424

1. Problem Predict whether an aqueous solution of each salt is neutral, acidic, or basic. (a) NaCN (b) LiF (c) Mg(NO3)2 (d) NH4I What Is Required? Predict whether each aqueous solution is acidic, basic, or neutral. What Is Given? The formula of each salt is given. Plan Your Strategy Determine whether the cation is from a strong or weak base, and whether the anion is from a strong or weak acid. Ions derived from weak bases or weak acids react with water and affect the pH of the solution. Act on Your Strategy (a) Sodium cyanide, NaCN, is the salt of a strong base (NaOH) and a weak acid (HCN). Only the cyanide ions react with water. The solution is basic. (b) Lithium ﬂuoride, LiF, is the salt of a strong base (LiOH) and a weak acid (HF). Only the ﬂuoride ions react with water. The solution is basic. (c) Magnesium nitrate, Mg(NO3)2, is the salt of a strong base (Mg(OH)2) and a strong acid (HNO3(aq)). Neither ion reacts with water, so the solution is neutral. (d) Ammonium iodide, NH4I, is the salt of a weak base (NH3(aq)) and a strong acid (HI(aq)). Only the ammonium ions react with water. The solution is acidic. Check Your Solution Check that you have correctly identified whether the cation is from a strong or weak base, and whether the anion is from a strong or weak acid.

2. Problem Is the solution of each salt acidic, basic, or neutral? For solutions that are not neutral, write equations that support your predictions. (a) NH4BrO4 (b) NaBrO4 (c) NaOBr (d) NH4Br What Is Required? Predict whether each aqueous solution is acidic, basic, or neutral. Write equations that represent the solutions that are acidic or basic.

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 158 CHEMISTRY 12

What Is Given? The formula of each salt is given. Plan Your Strategy Determine whether the cation is from a strong or weak base, and whether the anion is from a strong or weak acid. Ions derived from weak bases or weak acids react with water and affect the pH of the solution. Act on Your Strategy (a) Ammonium perbromate, NH4BrO4 , is the salt of a weak base (NH3(aq)) and a strong acid (HBrO4). Only the ammonium ions react with water. + + NH4 (aq) + H2O() NH3(aq) + H3O (aq) The solution is acidic. (b) Sodium perbromate, NaBrO4, is the salt of a strong base (NaOH) and a strong acid (HBrO4). Neither ion reacts with water, so the solution is neutral. (c) Sodium hypobromite, NaOBr, is the salt of a strong base (NaOH) and a weak acid (HOBr(aq)). Only the hypobromite ions react with water. − − OBr (aq) + H2O() HOBr(aq) + OH (aq) The solution is basic. (d) Ammonium bromide, NH4Br, is the salt of a weak base (NH3(aq)) and a strong acid (HBr(aq)). Only the ammonium ions react with water. + + NH4 (aq) + H2O() NH3(aq) + H3O (aq) The solution is acidic. Check Your Solution The equations that represent the reactions with water support the prediction that NH4BrO4 and NH4Br dissolve to form an acidic solution and NaOBr dissolves to form a basic solution. Sodium hypobromite is the salt of a strong base-strong acid, so neither ion reacts with water and the solution is neutral.

3. Problem −5 Ka for benzoic acid, C6H5COOH, is 6.3 × 10 . Ka for phenol, C6H5OH, is −10 − − 1.3 × 10 . Which is the stronger base, C6H5COO (aq) or C6H5O (aq)? Explain your answer. What Is Required? − − You must determine which ion, C6H5COO (aq) or C6H5O (aq), is the stronger base. What Is Given? The Ka of each conjugate acid is given. Plan Your Strategy −14 For a conjugate acid-base pair, KaKb = 1.0 × 10 . Therefore, a small value of Ka for an acid results in a large value of Kb for the conjugate base. Act on Your Strategy Because Ka for phenol is smaller than Ka for benzoic acid, Kb for the phenolate − ion must be larger than Kb for the benzoate ion. Consequently, C6H5O (aq), is the stronger base. Check Your Solution −14 Using the equation KaKb = 1.0 × 10 , you can calculate the Kb for each conjugate − −10 − −5 base. Kb for C6H5COO (aq) is 1.6 × 10 . Kb for C6H5O (aq) is 7.7 × 10 . − This supports the reasoning that C6H5O (aq) is the stronger base.

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 159 CHEMISTRY 12

4. Problem Sodium hydrogen sulfite, NaHSO3, is a preservative that is used to prevent the discolouration of dried fruit. In aqueous solution, the hydrogen sulfite ion can act as either an acid or a base. Predict whether NaHSO3 dissolves to form an acidic solution or a basic solution. (Refer to Appendix E for ionization data.) What Is Required? You must decide whether NaHSO3 dissolves to form an acidic solution or a basic solution. What Is Given? −2 Ka for H2SO3(aq) = 1.4 × 10 − −8 Ka for HSO3 (aq) = 6.3 × 10 Plan Your Strategy The hydrogen sulfite ion is amphoteric. Compare the equilibrium constants Ka and Kb − for HSO3 (aq) acting as an acid and a base, to determine which reaction goes farthest to completion. Act on Your Strategy The ion reactions with water are − 2− + −8 HSO3 (aq) + H2O() SO3 (aq) + H3O (aq) Ka = 6.3 × 10 − − HSO3 (aq) + H2O() H2SO3(aq) + OH (aq) Kb = ? Kb can be calculated using the value for Ka of H2SO3(aq). Kw Kb = Ka −14 = 1.0 × 10 1.4 × 10−2 − = 7.1 × 10 13 The equilibrium constant (Ka) for the hydrogen sulfite ion acting as an acid is greater than the equilibrium constant (Kb) for the ion acting as a base. Therefore, the solution is acidic. Check Your Solution It is a common difficulty in this type of problem to identify correctly the base reaction and the corresponding value for Kb. Kb must be calculated using the Ka value for − H2SO3(aq), because HSO3 (aq) is the conjugate base of H2SO3(aq).

Solutions for Practice Problems Student Textbook page 428

5. Problem After titrating sodium hydroxide with hydroﬂuoric acid, a chemist determined that the reaction had formed an aqueous solution of 0.020 mol/L sodium ﬂuoride. Determine the pH of the solution. What Is Required? You need to calculate the pH of the solution. What Is Given? [NaF] = 0.020 mol/L −4 From Appendix E, Ka for HF(aq) = 6.3 × 10 Plan Your Strategy Step 1 Decide which ion reacts with water. Write the equation that represents the hydrolysis reaction.

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 160 CHEMISTRY 12

Step 2 Determine the equilibrium constant for the ion involved in the hydrolysis reaction. Step 3 Divide the ion concentration by the appropriate equilibrium constant to determine whether the change in concentration of the ion can be ignored. Step 4 Set up an ICE table for the ion that is involved in the reaction with water. Let x represent the change in the concentration of the ion that reacts. Step 5 Write the equilibrium expression. Substitute the equilibrium concentrations into the expression, and solve for x. + Step 6 Use the value of x to determine [H3O ]. Then calculate the pH of the solution. Act on Your Strategy Step 1 Sodium fluoride is the salt of a strong base (NaOH) and a weak acid (HF). Thus, only the fluoride ion reacts with water. − − F (aq) + H2O() HF(aq) + OH (aq) Step 2 Kb for the fluoride ion can be calculated using Ka for HF(aq) . Kw Kb = Ka −14 = 1.0 × 10 6.3 × 10−4 − = 1.6 × 10 11 + − Step 3 NaF(aq) Na (aq) + F (aq) − ∴[F ] = 0.020 mol/L − F = 0.020 −11 Kb 1.6 × 10 = 1.2 × 109 Because this value is much greater than 500, the change in concentration of − F (aq) can be ignored compared with its initial concentration. − − Step 4 Concentration (mol/L) F (aq) ++H2O() 2HF(aq) OH (aq) Initial 0.020 0 ~0 Change −x +x +x Equilibrium 0.020 − x ≈ 0.020 x x

[HF][OH−] Step 5 K = − b [F ] − 1.6 × 10 11 = (x)(x) 0.020 x = 3.2 × 10−13 − = 5.6 × 10 7 mol/L − − Step 6 x = [OH ] = 5.6 × 10 7 mol/L − pOH =−log[OH] − =−log(5.6 × 10 7) = 6.25 pH = 14.00 − pOH = 14.00 − 6.25 = 7.75 Check Your Solution The pH of the solution is weakly basic. This is consistent with a dilute aqueous solution of the salt of a strong base and a weak acid.

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 161 CHEMISTRY 12

6. Problem Part way through a titration, 2.0 × 101 mL of 0.10 mol/L sodium hydroxide has been added to 3.0 × 101 mL of 0.10 mol/L hydrochloric acid. What is the pH of the solution? What Is Required? You need to calculate the pH of the solution. What Is Given? 20 mL of 0.10 mol/L NaOH(aq) has been added to 30 mL of 0.10 mol/L HCl(aq). Plan Your Strategy Step 1 Calculate the amount of each reagent using the following equation: Amount (mol) = concentration (mol/L) × volume (L) Step 2 Write the chemical equation for the reaction and determine the excess reagent. Step 3 Calculate the concentration of the excess reagent using the following equation: Concentration (mol/L) = amount in excess (mol) total volume (L) Step 4 Calculate the pH of the solution. Act on Your Strategy −2 Step 1 Amount NaOH(aq) = (0.10 mol/L) × (2.0 × 10 L) − = 2.0 × 10 3 mol −2 Amount HCl(aq) = (0.10 mol/L) × (3.0 × 10 L) − = 3.0 × 10 3 mol Step 2 NaOH(aq) + HCl(aq) → NaCl(aq) + H2O() Because the acid and base react in a 1:1 ratio, the HCl(aq) is in excess. −3 −3 Step 3 Amount of excess HCl(aq) = (3.0 × 10 mol) − (2.0 × 10 mol) − = 1.0 × 10 3 mol Total volume = (2.0 × 10−2 L) + (3.0 × 10−2 L) = 5.0 × 10−2 L × −3 [HCl] = 10 10 mol 5.0 × 10−2 L − = 2.0 × 10 2 mol/L + −2 Step 4 HCl(aq) is a strong acid. Therefore, [H3O ] = 2.0 × 10 mol/L. + pH =−log[H3O ] − =−log(2.0 × 10 2) = 1.70 Check Your Solution The solutions have the same molar concentration. Because the volume of hydrochloric acid is greater, the solution should be acidic and the pH less than 7.

7. Problem 0.025 mol/L benzoic acid, C6H5COOH, is titrated with 0.025 mol/L sodium hydroxide solution. Calculate the pH at equivalence. What Is Required? You need to calculate the pH at equivalence. What Is Given? You know that 0.025 mol/L C6H5COOH(aq) reacts with 0.025 mol/L NaOH(aq). Tables of Ka and Kb values are in Appendix E. Plan Your Strategy Step 1 Write the balanced chemical equation for the reaction. Step 2 Calculate the concentration of the salt formed, based on the amount (in mol) and the total volume of the solution.

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 162 CHEMISTRY 12

Step 3 Decide which ion reacts with water. Write the equation that represents the reaction. Step 4 Determine the equilibrium constant for the ion that is involved in the hydrolysis reaction. Step 5 Divide the concentration of the ion identiﬁed in Step 3 by the appropriate ionization constant to determine whether the change in concentration of the ion can be ignored. Step 6 Set up an ICE table for the ion that is involved in the reaction with water. Let x represent the change in the concentration of the ion that reacts. Step 7 Write the equilibrium expression. Substitute the equilibrium concentrations into the expression, and solve for x. + Step 8 Use the value of x to determine [H3O ]. Then calculate the pH of the solution. Act on Your Strategy Step 1 The following chemical equation represents the reaction. C6H5COOH(aq) + NaOH(aq) → NaC6H5COO(aq) + H2O() Step 2 The acid and base react in a 1:1 ratio, and the concentrations are the same. Therefore, the volume of each reagent must be the same and the total volume will be double the initial volume. Therefore, [NaC H COO] = 0.025 mol/L 6 5 2 = 0.0125 mol/L + − + Step 3 The salt forms Na (aq) and C6H5COO (aq) in solution. Na (aq) is the cation − of a strong base, so it does not react with water. C6H5COO (aq) is the conjugate base of a weak acid, so it does react with water. The pH of the solution is therefore determined by the extent of the following reaction. − − C6H5COO (aq) + H2O() → C6H5COOH(aq) + OH (aq) Step 4 Therefore, Kw Kb for C6H5COO(aq) = Ka −14 = 1.0 × 10 6.3 × 10−5 − = 1.6 × 10 10 − [C6H5COO ] = 0.0125 Step 5 −10 Kb 1.6 × 10 = 7.8 × 107 − This is well above 500, so the change in [C6H5COO ] can be ignored. Step 6 − − Concentration (mol/L) C6H5COOH (aq) ++H2O() C6H5COOH(aq) OH (aq) Initial 0.0125 0 ~0 Change −x +x +x Equilibrium 0.0125 − x ≈ 0.0125 x x

− [C6H5COOH][OH ] Step 7 Kb = − [C6H5COO ] − 1.6 × 10 10 = (x)(x) √0.0125 x = 2.0 × 10−12 − = 1.4 × 10 6 mol/L

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 163 CHEMISTRY 12

− − Step 8 x = [OH ] = 1.4 × 10 6 mol/L − pOH =−log[OH ] − =−log(1.4 × 10 6) = 5.85 pH = 14.00 − 5.85 = 8.15 Check Your Solution The titration forms an aqueous solution of a salt derived from a strong base and a weak acid. The solution should be basic, which is supported by the calculation of the pH.

8. Problem 50.0 mL of 0.10 mol/L hydrobromic acid is titrated with 0.10 mol/L aqueous ammonia. Determine the pH at equivalence. What Is Required? You need to calculate the pH at equivalence. What Is Given? You know that 50 mL of 0.10 mol/L HBr(aq) reacts with 0.10 mol/L NH3(aq). Tables of Ka and Kb values are in Appendix E. Plan Your Strategy Step 1 Write the balanced chemical equation for the reaction. Step 2 Find the volume of NH3(aq) added to neutralize the hydrobromic acid based on the amount and concentration of HBr(aq). Step 3 Calculate the concentration of the salt formed, based on the amount (in mol) and the total volume of the solution. Step 4 Decide which ion reacts with water. Write the equation that represents the reaction. Step 5 Determine the equilibrium constant for the ion that is involved in the hydrolysis reaction. Step 6 Divide the concentration of the ion identified in Step 4 by the appropriate ionization constant to determine whether the change in concentration of the ion can be ignored. Step 7 Set up an ICE table for the ion that is involved in the reaction with water. Let x represent the change in the concentration of the ion that reacts. Step 8 Write the equilibrium expression. Substitute the equilibrium concentrations into the expression, and solve for x. + Step 9 Use the value of x to determine [H3O ]. Then calculate the pH of the solution. Act on Your Strategy Step 1 The following chemical equation represents the reaction. NH3(aq) + HBr(aq) → NH4Br(aq) Step 2 Amount (in mol) of NH3(aq) = 0.20 mol/L × 0.020 L − = 4.0 × 10 3 mol Step 3 NH3(aq) and HBr(aq) react in a 1:1 ratio. Because [NH3] = [HBr], the volume of aqueous ammonia added must be the same as the volume of hydrobromic acid. Therefore, the volume of NH3(aq) is 50 mL, and the total volume of the solution is twice the volume of HBr(aq) used. Thus, the [NH4Br] must be half the initial [HBr]. Therefore, [NH4Br] = 0.050 mol/L

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 164 CHEMISTRY 12

Step 4 The titration results in an aqueous solution of ammonium bromide, NH4Br(aq). + − NH4 (aq) is the conjugate acid of a weak base, so it reacts with water. Br (aq) is the conjugate base of a strong acid, so it does not react with water. The solution will be acidic, and the pH is determined by the extent of the following reaction. + + NH4 (aq) + H2O() → NH3(aq) + H3O (aq) −5 Step 5 From Appendix E, Kb for NH3(aq) is 1.8 × 10 . Ka for the conjugate base, NH4(aq), can be calculated using the relationship KaKb = Kw. Kw Ka = Kb −14 = 1.0 × 10 1.8 × 10−5 − = 5.6 × 10 10 + [NH4 ] = 0.050 Step 6 −10 Ka 5.6 × 10 = 8.9 × 107 + This is well above 500, so the change in [NH4] can be ignored. + + Step 7 Concentration (mol/L) NH4 (aq) ++H2O() NH3(aq) H3O (aq) Initial 0.050 0 ~0 Change −x +x +x Equilibrium 0.050 − x ≈ 0.050 x x

+ [NH3][H3O ] Step 8 Ka = + [NH4 ] − 5.6 × 10 10 = (x)(x) √0.050 x = 2.8 × 10−11 − = 5.3 × 10 6 mol/L + −6 Step 9 x = [H3O ] = 5.3 × 10 mol/L + pH =−log[H3O ] − =−log(5.3 × 10 6) = 5.28 Check Your Solution The titration forms an aqueous solution of a salt derived from a weak base and a strong acid. The solution should be acidic, which is supported by the calculation of the pH.

Solutions for Practice Problems Student Textbook page 432

9. Problem Write the balanced chemical equation that represents the dissociation of each compound in water. Then write the corresponding solubility product expression. (a) copper(I) chloride (b) barium ﬂuoride (c) silver sulfate (d) calcium phosphate

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 165 CHEMISTRY 12

Solution First, write a balanced equation for the equilibrium between excess solid and dissolved ions in a saturated aqueous solution. Then use the balanced equation to write the expression for Ksp. + − (a) CuCl(s) Cu (aq) + Cl (aq) + − Ksp = [Cu ][Cl ] 2+ − (b) BaF2(s) Ba (aq) + 2F (aq) 2+ − 2 Ksp = [Ba ][F ] + 2− (c) Ag2SO4(s) 2Ag (aq) + SO4 (aq) + 2 2− Ksp = [Ag ] [SO4 ] 2+ 3− (d) Ca3(PO4)2(s) 3Ca (aq) + 2PO4 (aq) 2+ 3 3− 2 Ksp = [Ca ] [PO4 ] Check Your Solution The Ksp expressions are based on balanced equations for saturated solutions of slightly soluble ionic compounds. The exponents in the Ksp expressions match the corresponding coefficients in the chemical equations. The coefficient 1 is not written, following chemistry conventions.

10. Problem Write a balanced dissolution equation and solubility product expression for silver carbonate, Ag2CO3 . Solution First, write a balanced equation for the equilibrium between excess solid and dissolved ions in a saturated aqueous solution. Then use the balanced equation to write the expression for Ksp. + − Ag2CO3(s) 2Ag (aq) + CO3 (aq) + 2 − Ksp = [Ag ] [CO3 ] Check Your Solution The Ksp expression is based on the balanced equation between excess solid and dissolved ions. The exponents in the Ksp expression match the corresponding coefﬁcients in the chemical equation.

11. Problem Write a balanced dissolution equation and solubility product expression for magnesium ammonium phosphate, MgNH4PO4. Solution First, write a balanced equation for the equilibrium between excess solid and dissolved ions in a saturated aqueous solution. Then use the balanced equation to write the expression for Ksp. 2+ + 3− MgNH4PO4(s) Mg (aq) + NH4 (aq) + PO4 (aq) 2+ + 3− Ksp = [Mg ][NH4 ][PO4 ] Check Your Solution The Ksp expression is based on the balanced equation between excess solid and dissolved ions. The exponents in the Ksp expression match the corresponding coefﬁcients in the chemical equation.

12. Problem Iron (III) nitrate has very low solubility. (a) Write the solubility product expression for iron(III) nitrate. (b) Do you expect the value of Ksp of iron(III) nitrate to be larger or smaller than the Ksp for aluminum hydroxide, which has a slightly higher solubility?

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 166 CHEMISTRY 12

Solution (a) Write a balanced equation for the equilibrium between excess solid iron(III) nitrate and dissolved ions in a saturated aqueous solution. Then use the balanced equation to write the expression for Ksp. 3+ − Fe(NO3)3(s) Fe (aq) + 3NO3 (aq) 3+ − 3 Ksp = [Fe ][NO3 ] (b) Write a balanced equation for the equilibrium between excess solid aluminum hydroxide and dissolved ions in a saturated aqueous solution. Then use the balanced equation to write the expression for Ksp. Finally, compare the two equilibrium expressions and decide which expression is larger. 3+ − Al(OH)3(s) Al (aq) + 3OH (aq) 3+ − 3 Ksp = [Al ][OH ] The equilibrium expressions have the same form. In both cases, if x mol/L of salt 4 dissolves, Ksp is given by x . Therefore, the most soluble salt will have the larger value of Ksp. The value of Ksp for iron(III) nitrate is smaller than Ksp for aluminum hydroxide. Check Your Solution Each Ksp expression is based on the balanced equation between excess solid and dissolved ions. The exponents in the Ksp expressions match the corresponding coefﬁcients in the chemical equations. The equilibrium expressions are of the same form, so the least soluble salt has the lower value for Ksp.

Solutions for Practice Problems Student Textbook page 433

13. Problem The maximum solubility of silver cyanide, AgCN, is 1.5 × 10−8 mol/L at 25˚C. Calculate Ksp for silver cyanide. What Is Required? You need to ﬁnd the value of Ksp for AgCN at 25˚C. What Is Given? You know the solubility of AgCN at 25˚C. Plan Your Strategy Step 1 Write an equation for the dissolution of AgCN. Step 2 Use the equation to write the solubility product expression. Step 3 Find the concentration (in mol/L) of each ion. Step 4 Substitute the concentrations into the solubility product expression, and calculate Ksp. Act on Your Strategy + − Step 1 AgCN(s) Ag (aq) + CN (aq) + − Step 2 Ksp = [Ag ][CN ] + − − Step 3 [Ag ] = [CN ] = [AgCN] = 1.5 × 10 8 mol/L + − Step 4 Ksp = [Ag ][CN ] − − = (1.5 × 10 8)(1.5 × 10 8) − = 2.2 × 10 16 −16 Ksp for silver cyanide is 2.2 × 10 at 25˚C. Check Your Solution The value of Ksp is less than the concentration of the salt, as expected. It has the correct number of signiﬁcant digits.

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 167 CHEMISTRY 12

14. Problem A saturated solution of copper(II) phosphate, Cu3(PO4)2, has a concentration of −7 2 6.1 × 10 gCu3(PO4)2 per 1.00 × 10 mL of solution at 25˚C. What is Ksp for Cu3(PO4)2 at 25˚C ? What Is Required? You need to ﬁnd the value of Ksp for Cu3(PO4)2 at 25˚C. What Is Given? −7 2 You know 6.1 × 10 g of Cu3(PO4)2 is dissolved in 1.00 × 10 mL of solution at 25˚C. Plan Your Strategy Step 1 Write an equation for the dissolution of Cu3(PO4)2. Step 2 Use the equation to write the solubility product expression. Step 3 Find the concentration (in mol/L) of copper(II) phosphate. Next, ﬁnd the concentration (in mol/L) of each ion. Step 4 Substitute the concentrations into the solubility product expression, and calculate Ksp. Act on Your Strategy 2+ 3− Step 1 Cu3(PO4)2(s) 3Cu (aq) + 2PO4 (aq) 2+ 3 3− 2 Step 2 Ksp = [Cu ] [PO4 ] Step 3 The molar mass of Cu3(PO4)2 is 380.6 g/mol. 6.1 × 10−7g Amount of Cu (PO ) = 3 4 2 380.6 g/mol − = 1.6 × 10 9 mol × −9 [Cu (PO ) ] = 1.6 10 mol 3 4 2 0.100 L − = 1.6 × 10 8 mol/L 2+ [Cu ] = 3 × [Cu3(PO4)2] − = 3 × (1.6 × 10 8 mol/L) − = 4.8 × 10 8 mol/L 3− −8 [PO4 ] = 2 × [Cu3(PO4)2] = 3.2 × 10 mol/L 2+ 3 3− 2 Step 4 Ksp = [Cu ] [PO4 ] − − = (4.8 × 10 8)3(3.2 × 10 8)2 − = 1.1 × 10 37 −37 Ksp for copper(II) phosphate is 1.1 × 10 at 25˚C. Check Your Solution Check that you wrote the balanced chemical equation and the corresponding Ksp equation correctly. Pay attention to molar relationships and to the exponent of each term. Check the concentration calculations carefully.

15. Problem 20 A saturated solution of CaF2 contains 1.2 × 10 formula units of calcium ﬂuoride per litre of solution. Calculate Ksp for CaF2. What Is Required? You need to ﬁnd the value of Ksp for CaF2. What Is Given? 20 You know one litre of solution contains 1.2 × 10 formula units of CaF2.

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 168 CHEMISTRY 12

Plan Your Strategy Step 1 Write an equation for the dissolution of CaF2. Step 2 Use the equation to write the solubility product expression. Step 3 Find the concentration (in mol/L) of calcium fluoride. Next, find the concentration (in mol/L) of each ion. Step 4 Substitute the concentrations into the solubility product expression, and calculate Ksp. Act on Your Strategy 2+ − Step 1 CaF2(s) Ca (aq) + 2F (aq) 2+ − 2 Step 2 Ksp = [Ca ][F ] Step 3 The amount of CaF2 is given by the following formula: Amount = number of formula units Avogadro’s constant 20 = 1.2 × 10 formula units 6.0 × 1023 formula units/mol − = 2.0 × 10 4 mol −4 [CaF2] = 2.0 × 10 mol/L 2+ −4 [Ca ] = [CaF2] = 2.0 × 10 mol/L − [F ] = 2 × [CaF2] − = 2 × (2.0 × 10 4 mol/L) − = 4.0 × 10 4 mol/L 2+ − 2 Step 4 Ksp = [Ca ][F ] − − = (2.0 × 10 4)(4.0 × 10 4)2 − = 3.2 × 10 11 −11 Ksp for calcium fluoride is 3.2 × 10 . Check Your Solution Check that you wrote the balanced chemical equation and the corresponding Ksp equation correctly. Pay attention to molar relationships and to the exponent of each term. Check the concentration calculations carefully.

16. Problem The concentration of mercury(I) iodide, Hg2I2, in a saturated solution at 25˚C is 1.5 × 10−4 ppm. (a) Calculate Ksp for Hg2I2. The solubility equilibrium is written as follows: 2+ − Hg2I2 Hg2 + 2I (b) State any assumptions that you made when you converted ppm to mol/L. What Is Required? You need to ﬁnd the value of Ksp for Hg2I2 at 25˚C. What Is Given? You know the concentration of Hg2I2 in a saturated solution at 25˚C is 1.5 × 10−4 ppm. You are given the dissolution equation. Plan Your Strategy Step 1 Use the dissolution equation to write the solubility product expression. Step 2 Find the concentration (in mol/L) of mercury(I) iodide. Then, ﬁnd the concentration (in mol/L) of each ion. Step 3 Substitute the concentrations into the solubility product expression, and calculate Ksp.

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 169 CHEMISTRY 12

Act on Your Strategy 2+ − 2 (a) Step 1 Ksp = [Hg2 ][I ] Step 2 1 ppm is equivalent to 1 mg of solute dissolved in 1 kg of solution. 1 kg −4 of water at 4˚C has a volume of 1 L. Therefore, 1.5 × 10 mg of Hg2I2 is dissolved in 1.0 L of solution. 1.5 × 10−4 mg = 1.5 × 10−7g The molar mass of Hg2I2 is 655 g/mol. 1.5 × 10−7 g Amount of Hg I = 2 2 655 g/mol − = 2.3 × 10 10 mol −10 [Hg2I2] = 2.3 × 10 mol/L 2+ −10 [Hg2 ] = [Hg2I2] = 2.3 × 10 mol/L − [I ] = 2 × [Hg2I2] − = 2 × (2.3 × 10 10 mol/L) − = 4.6 × 10 10 mol/L 2+ − 2 Step 3 Ksp = [Hg2 ][I ] − − = (2.3 × 10 10)(4.6 × 10 10)2 − = 4.9 × 10 29 −29 Ksp for mercury(I) iodide is 4.9 × 10 at 25˚C. (b) 1 ppm is equivalent to 1 mg of solute dissolved in 1 kg of solution. This leads to two assumptions in this problem. Firstly, that the solution is so dilute it can be treated as 1 kg of water. Secondly, you assume that 1 kg of water has a volume of 1 L. This is only strictly true at 4˚C. Check Your Solution Check that you wrote the Ksp equation correctly. Pay attention to molar relationships and to the exponent of each term. Check the concentration calculations carefully.

Solutions for Practice Problems Student Textbook page 436

17. Problem −10 Ksp for silver chloride, AgCl, is 1.8 × 10 at 25˚C. (a) Calculate the molar solubility of AgCl in a saturated solution at 25˚C. (b) How many formula units of AgCl are dissolved in 1.0 L of saturated silver chloride solution? (c) What is the percent (m/v) of AgCl in a saturated solution at 25˚C? What Is Required? You need to determine the solubility (in mol/L, in formula units, and expressed as a mass/volume percent) of AgCl at 25˚C. What Is Given? −10 At 25˚C, Ksp for AgCl is 1.8 × 10 . Plan Your Strategy (a) Step 1 Write the dissolution equilibrium equation. Step 2 Use the equilibrium equation to write an expression for Ksp. Step 3 Set up an ICE table. Let x represent molar solubility. Use the stoichiometry of the equilibrium equation to write expressions for the equilibrium concentrations of the ions. Step 4 Substitute your expressions into the expression for Ksp, and solve for x. (b) Convert the molar solubility to formula units per litre. (c) Convert the molar solubility to a percent (m/v).

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 170 CHEMISTRY 12

Act on Your Strategy 2+ − (a) Step 1 AgCl(s) Ag (aq) + Cl (aq) + − Step 2 Ksp = [Ag ][Cl ] + − Step 3 Concentration (mol/L) AgCl(s) Ag (aq) + Cl (aq) Initial 0 0 Change +x +x Equilibrium x x + − Step 4 Ksp = [Ag ][Cl ] = (x)(x) × −10 = 2 Therefore, 1.8 10 √x x = 1.8 × 10−10 − = 1.3 × 10 5 mol/L The molar solubility of AgCl in water is 1.3 × 10−5 mol/L. − − (b) 1.3 × 10 5 mol = (1.3 × 10 5) × (6.0 × 1023 formula units/mol) = 7.8 × 1018 formula units The solubility of AgCl in water is 7.8 × 1018 formula units/L mass of solute (g) (c) Mass/volume percent = × 100% volume of solution (mL) Molar mass of AgCl = 143.3 g/mol − Mass in 1 L of solution = (1.3 × 10 5 mol/L) × (143.3 g/mol) − = 1.9 × 10 3 g/L 1.9 × 10−3 g Mass/volume percent = × 100% 1000 mL The solubility of AgCl at 25˚C is 1.9 × 10−4% (m/v). Check Your Solution + − Substitute the values of [Ag ] and [Cl ] into the Ksp equation. You should get the given Ksp. Check the number of signiﬁcant digits given in the question, and the number of digits in the answers.

18. Problem Iron(III) hydroxide, Fe(OH)3, is an extremely insoluble compound. Ksp for Fe(OH)3 −39 is 2.8 × 10 at 25˚C. Calculate the molar solubility of Fe(OH)3 at 25˚C. What Is Required? You need to determine the solubility (in mol/L) of Fe(OH)3 at 25˚C. What Is Given? −39 At 25˚C, Ksp for Fe(OH)3 is 2.8 × 10 . Plan Your Strategy Step 1 Write the dissolution equilibrium equation. Step 2 Use the equilibrium equation to write an expression for Ksp. Step 3 Set up an ICE table. Let x represent molar solubility. Use the stoichiometry of the equilibrium equation to write expressions for the equilibrium concentrations of the ions. Step 4 Substitute your expressions into the expression for Ksp, and solve for x.

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 171 CHEMISTRY 12

Act on Your Strategy 3+ − Step 1 Fe(OH)3(s) Fe (aq) + 3OH (aq) 3+ − 3 Step 2 Ksp = [Fe ][OH ] 3+ − Step 3 Concentration (mol/L) Fe(OH)3(s) Fe (aq) + 3OH (aq) Initial 0 0 Change +x +3x Equilibrium x 3x

3+ − 3 Step 4 Ksp = [Fe ][OH ] = (x) × (3x)3 = 27x4 − 2.8 × 10 39 = 27x4 × −39 x = 4 2.8 10 27 − = 1.0 × 10 10 mol/L −10 The molar solubility of Fe(OH)3 in water is 1.0 × 10 mol/L. Check Your Solution 3+ − Recall that x = [Fe ]eq and 3x = [OH ]eq. Substitute these values into the Ksp equation. You should get the given Ksp.

19. Problem −6 Ksp for zinc iodate, Zn(IO3)2 , is 3.9 × 10 at 25˚C. Calculate the solubility (in mol and in g/L) of Zn(IO3)2 in a saturated solution. What Is Required? You need to determine the solubility (in mol/L and in g/L) of Zn(IO3)2 at 25˚C. What Is Given? −6 At 25˚C, Ksp for Zn(IO3)2 is 3.9 × 10 . Plan Your Strategy Step 1 Write the dissolution equilibrium equation. Step 2 Use the equilibrium equation to write an expression for Ksp. Step 3 Set up an ICE table. Let x represent molar solubility. Use the stoichiometry of the equilibrium equation to write expressions for the equilibrium concentrations of the ions. Step 4 Substitute your expressions into the expression for Ksp, and solve for x. Step 5 Determine the solubility in g/L. Act on Your Strategy 2+ − Step 1 Zn(IO3)2(s) Zn (aq) + 2IO3 (aq) 2+ − 2 Step 2 Ksp = [Zn ][IO3 ] 2+ − Step 3 Concentration (mol/L) Zn(IO3)2(s) Zn (aq) + 2IO3 (aq) Initial 0 0 Change +x +2x Equilibrium x 2x

2+ − 2 Step 4 Ksp = [Zn ][IO3 ] = (x) × (2x)2 = 4x3 3.9 × 10−6 = 4x3

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 172 CHEMISTRY 12

× −6 x = 3 3.9 10 4 − = 9.9 × 10 3 mol/L −3 The molar solubility of Zn(IO3)2 in water is 9.9 × 10 mol/L. Step 5 Molar mass of Zn(IO3)2 = 415.2 g/mol − − 9.9 × 10 3 mol/L = (9.9 × 10 3mol/L) × (415.2 g/mol) = 4.1 g/L The solubility of Zn(IO3)2 in water is 4.1 g/L. Check Your Solution 2+ − Recall that x = [Zn ]eq and 2x = [IO3 ]eq. Substitute these values into the Ksp equation. You should get the given Ksp. The answers have the correct number of signiﬁcant digits.

20. Problem What is the maximum number of formula units of zinc sulﬁde, ZnS, that can dissolve −22 in 1.0 L of solution at 25˚C? Ksp for ZnS is 2.0 × 10 . What Is Required? You need to determine the solubility (in formula units) of ZnS at 25˚C. What Is Given? −22 At 25˚C, Ksp for ZnS is 2.0 × 10 . Plan Your Strategy Step 1 Write the dissolution equilibrium equation. Step 2 Use the equilibrium equation to write an expression for Ksp. Step 3 Set up an ICE table. Let x represent molar solubility. Use the stoichiometry of the equilibrium equation to write expressions for the equilibrium concentrations of the ions. Step 4 Substitute your expressions into the expression for Ksp, and solve for x. Step 5 Determine the number of formula units dissolved in 1.0 L. Act on Your Strategy 2+ 2− Step 1 ZnS(s) Zn (aq) + S (aq) 2+ 2− Step 2 Ksp = [Zn ][S ] 2+ 2− Step 3 Concentration (mol/L) ZnS(s) Zn (aq) + S (aq) Initial 0 0

Change +x x Equilibrium x x

2+ 2− 2 Step 4 Ksp = [Zn ][S ] = x × −22 = 2 Therefore,√ 2.0 10 x x = 2.0 × 10−22 − = 1.4 × 10 11 mol/L The molar solubility of ZnS in water is 1.4 × 10−11 mol/L. − − Step 5 1.4 × 10 11 mol = (1.4 × 10 11 mol) × (6.0 × 1023 formula units/mol) = 8.4 × 1012 formula units The solubility of ZnS in water is 8.4 × 1012 formula units/L. Check Your Solution 2+ 2− Recall that x = [Zn ]eq = [S ]eq . Substitute these values into the Ksp equation. You should get the given Ksp.

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 173 CHEMISTRY 12

Solutions for Practice Problems Student Textbook page 439

21. Problem Determine the molar solubility of AgCl (a) in pure water (b) in 0.15 mol/L NaCl What Is Required? You need determine the solubility of AgCl (in mol/L) in water, and then in a solution of NaCl. What Is Given? −10 From Appendix E, Ksp for AgCl = 1.77 × 10 . You know the concentration of the solution of NaCl. Plan Your Strategy (a) Step 1 Write the dissolution equilibrium equation. Step 2 Use the equilibrium equation to write an expression for Ksp. Step 3 Set up an ICE table. Let x represent molar solubility. Use the stoichiometry of the equilibrium equation to write expressions for the equilibrium concentrations of the ions. Step 4 Substitute your expressions into the expression for Ksp, and solve for x. (b) Step 1 Set up an ICE table. The initial concentrations are based on the solution of NaCl. Let x represent the concentration of chloride (the common ion) that is contributed by AgCl. Step 2 Solve for x. Act on Your Strategy + − (a) Step 1 AgCl(s) Ag (aq) + Cl (aq) + − −10 Step 2 Ksp = [Ag ][Cl ] = 1.77 × 10 + − Step 3 Concentration (mol/L) AgCl(s) Ag (aq) + Cl (aq) Initial 0 0 Change +x +x Equilibrium x x + − Step 4 Ksp = [Ag ][Cl ] × −10 = 2 1.77 10 x x = 1.77 × 10−10 − x =±1.33 × 10 5 The negative root has no physical meaning. The solubility of AgCl is 1.33 × 10−5 mol/L. + − (b) Step 1 Concentration (mol/L) AgCl(s) Ag (aq) + Cl (aq) Initial 0 0.15 Change +x +x Equilibrium x 0.15 + x

Step 2 Since Ksp is very small, you can assume that x is much smaller than 0.15. To check the validity of this assumption, determine whether or not 0.15 is more than 500 times greater than Ksp, as shown on the next page.

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 174 CHEMISTRY 12

0.15 = 0.15 −10 Ksp 1.77 × 10 = 8.5 × 108 > 500 Therefore, in the (0.15 + x) term, x can be ignored. In other words, (0.15 + x) is approximately equal to 0.15. As a result, you can simplify the equation as follows: + − Ksp = [Ag ][Cl ] − 1.77 × 10 10 = (x)(0.15 + x) ≈ (x)(0.15) Therefore, 0.15x ≈ 1.77 × 10−10 x ≈ 1.18 × 10−9 mol/L The solubility of AgCl in a solution of 0.15 mol/L NaCl is 1.18 × 10−9 mol/L. Check Your Solution Your approximation in Step 6 was reasonable. The solubility x is much smaller than 0.15. Le Châtelier’s principle predicts a smaller solubility in a solution containing a common ion, and this is conﬁrmed by the calculations.

22. Problem Determine the molar solubility of lead(II) iodide, PbI2, in 0.050 mol/L NaI. What Is Required? You need to determine the solubility of PbI2 (in mol/L) in a solution of NaI. What Is Given? −9 From Appendix E, Ksp for PbI2 = 9.8 × 10 . You know the concentration of the solution of NaI. Plan Your Strategy Step 1 Write the dissolution equilibrium equation. Step 2 Use the equilibrium equation to write an expression for Ksp. Step 3 Set up an ICE table. The initial concentrations are based on the solution of NaI. Let x represent the concentration of iodide (the common ion) that is contributed by NaI. Step 4 Substitute your expressions into the expression for Ksp, and solve for x. Act on Your Strategy 2+ − Step 1 PbI2(s) Pb (aq) + 2I (aq) 2+ − 2 Step 2 Ksp = [Pb ][I ] 2+ − Step 3 Concentration (mol/L) PbI2(s) Pb (aq) + 2I (aq) Initial 0 0.050 Change +x +2x Equilibrium x 0.050 + 2x

Step 4 Since Ksp is very small, you can assume that 2x is much smaller than 0.050. To check the validity of this assumption, determine whether or not 0.050 is more than 500 times greater than Ksp, as shown. 0.050 = 0.050 −9 Ksp 9.8 × 10 = 5.1 × 106 > 500

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 175 CHEMISTRY 12

Therefore, in the (0.050 + 2x) term, 2x can be ignored. In other words, (0.050 + 2x) is approximately equal to 0.050. As a result, you can simplify the equation as follows: 2+ − 2 Ksp = [Pb ][I ] − 9.8 × 10 9 = (x)(0.050 + 2x)2 ≈ (x)(0.050)2 − − Therefore, (2.5 × 10 3)(x) ≈ 9.8 × 10 9 − x ≈ 3.9 × 10 6 mol/L −6 The solubility of PbI2 in a solution of 0.050 mol/L NaI is 3.9 × 10 mol/L. Check Your Solution Your approximation in Step 4 was reasonable. The solubility 2x is much smaller than 0.050. You can substitute your calculated values of equilibrium concentrations into the expression you wrote for Ksp. The calculated value should equal the given value for Ksp, within the errors introduced by mathematical rounding.

23. Problem Calculate the solubility of calcium sulfate, CaSO4, (a) in pure water (b) in 0.25 mol/L Na2SO4 What Is Required? You need determine the solubility of CaSO4 (in mol/L) in water, and then in a solution of Na2SO4. What Is Given? −5 From Appendix E, Ksp for CaSO4 = 4.93 × 10 . You know the concentration of the solution of Na2SO4. Plan Your Strategy (a) Step 1 Write the dissolution equilibrium equation. Step 2 Use the equilibrium equation to write an expression for Ksp. Step 3 Set up an ICE table. Let x represent molar solubility. Use the stoichiometry of the equilibrium equation to write expressions for the equilibrium concentrations of the ions. Step 4 Substitute your expressions into the expression for Ksp, and solve for x. (b) Step 1 Set up an ICE table. The initial concentrations are based on the solution of Na2SO4. Let x represent the concentration of sulfate (the common ion) that is contributed by Na2SO4. Step 2 Solve for x. Act on Your Strategy 2+ 2− (a) Step 1 CaSO4(s) Ca (aq) + SO4 (aq) 2+ 2− Step 2 Ksp = [Ca ][SO4 ] − = 4.93 × 10 5 2+ 2− Step 3 Concentration (mol/L) CaSO4(s) Ca (aq) + SO4 (aq) Initial 0 0 Change +x +x Equilibrium x x

+ 2− 2 Step 4 Ksp = [Ca ][SO4 ] = x × −5 = 2 4.93 10 x x = 4.93 × 10−5 − x =±7.02 × 10 3

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 176 CHEMISTRY 12

The negative root has no physical meaning. The solubility of CaSO4 is 7.02 × 10−3 mol/L.

2+ 2− (b) Step 1 Concentration (mol/L) CaSO4(s) Ca (aq) + SO4 (aq) Initial 0 0.25 Change +x +x Equilibrium x 0.25 + x

Step 2 Since Ksp is very small, you can assume that x is much smaller than 0.25. To check the validity of this assumption, determine whether or not 0.25 is more than 500 times greater than Ksp, as shown below. 0.25 = 0.25 −5 Ksp 4.93 × 10 = 5.1 × 103 > 500 Therefore, in the (0.25 + x) term, x can be ignored. In other words, (0.25 + x) is approximately equal to 0.25. As a result, you can simplify the equation as follows: 2+ 2− Ksp = [Ca ][SO4 ] − 4.93 × 10 5 = (x)(0.25 + x) ≈ (x)(0.25) Therefore, 0.25x ≈ 4.93 × 10−5 x ≈ 1.97 × 10−4 mol/L The solubility of AgCl in a solution of 0.15 mol/L NaCl is 1.97 × 10−4 mol/L. Check Your Solution Your approximation in Step 6 was reasonable. The solubility x is a much smaller than 0.25. Le Châtelier’s principle predicts a smaller solubility in a solution containing a common ion, and this is conﬁrmed by the calculations.

24. Problem −5 Ksp for lead(II) chloride, PbCl2, is 1.6 × 10 . Calculate the molar solubility of PbCl2. (a) in pure water (b) in 0.10 mol/L CaCl2 What Is Required? You need to determine the solubility of PbCl2 (in mol/L) in water, and then in a solution of CaCl2. What Is Given? −5 You know that Ksp for PbCl2 is 1.6 × 10 , and you know the concentration of the solution of CaCl2. Plan Your Strategy (a) Step 1 Write the dissolution equilibrium equation. Step 2 Use the equilibrium equation to write an expression for Ksp. Step 3 Set up an ICE table. Let x represent molar solubility. Use the stoichiometry of the equilibrium equation to write expressions for the equilibrium concentrations of the ions. Step 4 Substitute your expressions into the expression for Ksp, and solve for x. (b) Step 1 Set up an ICE table. The initial concentrations are based on the solution of CaCl2. Let x represent the concentration of chloride (the common ion) that is contributed by CaCl2. Step 2 Solve for x.

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 177 CHEMISTRY 12

Act on Your Strategy 2+ − (a) Step 1 PbCl2(s) Pb (aq) + 2Cl (aq) 2+ − 2 Step 2 Ksp = [Pb ][Cl ] − = 1.6 × 10 5 2+ − Step 3 Concentration (mol/L) PbCl2(s) Pb (aq) + 2Cl (aq) Initial 0 0 Change +x +2x Equilibrium x 2x

2+ − 2 Step 4 Ksp = [Pb ][Cl ] − 1.6 × 10 5 = x(2x)2 × −5 = 3 1.6 10 √4x 3 x = 4.0 × 10−6 − x = 1.6 × 10 2 −2 The solubility of PbCl2 is 1.6 × 10 mol/L. (b) Step 1 Each formula unit of calcium chloride dissociates to form two chloride ions. − Therefore, 0.10 mol/L CaCl2 dissociates to give [Cl ] = 0.20 mol/L.

2+ − Concentration (mol/L) PbCl2(s) Pb (aq) + 2Cl (aq) Initial 0 0.20 Change +x +2x Equilibrium x 0.20 + 2x

Step 2 Since Ksp is very small, you can assume that x is much smaller than 0.20. To check the validity of this assumption, determine whether or not 0.20 is more than 500 times greater than Ksp, as shown below. 0.20 = 0.20 −5 Ksp 1.6 × 10 = 1.2 × 104 > 500 Therefore, in the (0.20 + 2x) term, 2x can be ignored. In other words, (0.20 + 2x) is approximately equal to 0.20. As a result, you can simplify the equation as follows: 2+ − 2 Ksp = [Pb ][Cl ] − 1.6 × 10 5 = (x)(0.20 + 2x)2 ≈ (x)(0.20)2 Therefore, (4.0 × 10−2)(x) ≈ 1.6 × 10−5 x ≈ 4.0 × 10−4 mol/L The solubility of PbCl2 in a solution of 0.10 mol/L CaCl2 is 4.0 × 10−4 mol/L. Check Your Solution Your approximation in Step 6 was reasonable. The solubility 2x is a much smaller than 0.20. Le Châtelier’s principle predicts a smaller solubility in a solution containing a common ion, and this is confirmed by the calculations.

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 178 CHEMISTRY 12

Solutions for Practice Problems Student Textbook Page 441

25. Problem A buffer solution is made by mixing 250 mL of 0.200 mol/L aqueous ammonia and 400 mL of 0.150 mol/L ammonium chloride. Calculate the pH of the buffer solution. What Is Required? You need to determine the pH of the buffer solution. What Is Given? 250 mL of 0.200 mol/L aqueous ammonia and 400 mL of 0.150 mol/L ammonium −5 chloride are mixed. From Appendix E, Kb for NH3(aq) is 1.8 × 10 . Plan Your Strategy + Step 1 Determine the initial concentrations of NH3(aq) and NH4 (aq) in the buffer solution. Step 2 Write the reaction for the dissociation of NH3(aq). Set up an ICE table, + including the initial concentration of NH4 (aq) in the buffer. Step 3 Write the equation for Kb, and substitute equilibrium terms into the equation. Step 4 Solve the equation for x. Assume that x is small compared with the initial concentrations. Check the validity of this assumption when you ﬁnd the value of x. Step 5 Use the following equations: pOH =−log[OH−] pH = 14.00 = pOH Act on Your Strategy Step 1 When the solutions are mixed, the total volume is: (250 mL + 400 mL) = 650 mL × [NH ] = 0.250 L 0.200 mol/L 3 0.650 L − = 7.69 × 10 2 mol/L + × [NH ] = 0.400 L 0.150 mol/L 4 0.650 L − = 9.23 × 10 2 mol/L + − Step 2 Concentration (mol/L) NH3(aq) + H2O() NH4 (aq) + OH (aq) − − Initial 7.69 × 10 2 9.23 × 10 2 ~0

Change −x +x +x − − Equilibrium 7.69 × 10 2 − x 9.23 × 10 2 + x x

+ − [NH4 ][OH ] Step 3 Kb = [NH3] −2 = (9.23 × 10 + x)(x) (7.69 × 10−2 − x) − − Step 4 Assume that (7.69 × 10 2 − x) ≈ 7.69 × 10 2 and (9.23 × 10−2 + x) ≈ 9.23 × 10−2 . × −2 K = (9.23 10 )(x) = 1.8 × 10−5 b (7.69 × 10−2) Therefore, x = 1.5 × 10−5 mol/L = [OH−]

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 179 CHEMISTRY 12

− Step 5 pOH =−log(1.5 × 10 5) = 4.82 pH = 14.00 − 4.82 = 9.18 The pH of the buffer solution is 9.18. Check Your Solution The value of x (1.5 × 10−5) is negligible compared with the initial concentration of each component. A buffer that is made using a weak base and its conjugate acid should have a pH that is greater than 7.

26. Problem A buffer solution is made by mixing 200 mL of 0.200 mol/L aqueous ammonia and 450 mL of 0.150 mol/L ammonium chloride. Calculate the pH of the buffer solution. What Is Required? You need to determine the pH of the buffer solution. What Is Given? 200 mL of 0.200 mol/L aqueous ammonia and 450 mL of 0.150 mol/L ammonium −5 chloride are mixed. From Appendix E, Kb for NH3(aq) is 1.8 × 10 . Plan Your Strategy + Step 1 Determine the initial concentrations of NH3(aq) and NH4 (aq) in the buffer solution. Step 2 Write the reaction for the dissociation of NH3(aq). Set up an ICE table, + including the initial concentration of NH4 (aq) in the buffer. Step 3 Write the equation for Kb, and substitute equilibrium terms into the equation. Step 4 Solve the equation for x. Assume that x is small compared with the initial concentrations. Check the validity of this assumption when you ﬁnd the value of x. Step 5 Use the following equations: pOH =−log[OH−] pH = 14.00 − pOH Act on Your Strategy Step 1 When the solutions are mixed, the total volume is (250 + 400) = 650 mL. × [NH ] = 0.200 L 0.0200 mol/L 3 0.650 L − = 6.15 × 10 2 mol/L + × [NH ] = 0.450 L 0.150 mol/L 4 0.650 L − = 1.04 × 10 1 mol/L + − Step 2 Concentration (mol/L) NH3(aq) + H2O() NH4 (aq) + OH (aq) − − Initial 6.15 × 10 2 1.04 × 10 1 ~0 Change −x +x +x − − Equilibrium 6.15 × 10 2 − x 1.04 × 10 1 + x x

+ − [NH4 ][OH ] Step 3 Kb = [NH3] −1 = (1.04 × 10 + x)(x) (6.15 × 10−2 − x)

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 180 CHEMISTRY 12

− − Step 4 Assume that (6.15 × 10 2 − x) ≈ 6.15 × 10 2 and (1.04 × 10−1 + x) ≈ 1.04 × 10−1 . × −1 + K = (1.04 10 x)(x) = 1.8 × 10−5 b (6.15 × 10−2) Therefore, x = 1.1 × 10−5 mol/L = [OH−] − Step 5 pOH =−log(1.1 × 10 5) = 4.96 pH = 14.00 − 4.96 = 9.04 The pH of the buffer solution is 9.04. Check Your Solution The value of x (1.1 × 10−5) is negligible compared with the initial concentration of each component. A buffer that is made using a weak base and its conjugate acid should have a pH that is greater than 7.

27. Problem A buffer solution contains 0.200 mol/L nitrous acid, HNO2(aq), and 0.140 mol/L potassium nitrite, KNO2(aq). What is the pH of the buffer solution? What Is Required? You need to determine the pH of the buffer solution. What Is Given? 0.200 mol/L HNO2(aq) is mixed with 0.140 mol/L potassium nitrite, KNO2(aq). −4 From Appendix E, Ka for HNO2 is 5.6 × 10 . Plan Your Strategy Step 1 Write the reaction for the dissociation of HNO2. Set up an ICE table, − including the initial concentration of NO2 in the buffer. Step 2 Write the equation for Ka, and substitute equilibrium terms into the equation. Step 3 Solve the equation for x. Assume that x is small compared with the initial concentrations. Check the validity of this assumption when you ﬁnd the value of x. + Step 4 Use the equation pH =−log[H3O ]. Act on Your Strategy + − Step 1 HNO2 + H2O() H3O (aq) + NO2 (aq)

− + Concentration (mol/L) HNO2(aq) + H2O() NO2 (aq) + H3O (aq) Initial 0.200 0.140 ~0 Change −x +x +x Equilibrium 0.200 − x 0.140 + x x

− + [NO2][H3O ] Step 2 Ka = [HNO2] = (0.140 + x)(x) (0.200 − x) Step 3 Assume that (0.140 + x) ≈ 0.140 and (0.200 − x) ≈ 0.200. K = (1.40)(x) = 5.6 × 10−4 a (0.200) − ∴x = 8.0 × 10 4 mol/L − Step 4 pH =−log(8.00 × 10 4) = 3.10 The pH of the buffer solution is 3.10.

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 181 CHEMISTRY 12

Check Your Solution The value of x (8.0 × 10−4) is negligible compared with the initial concentration of each component. A buffer that is made using a weak acid and its conjugate base should have a pH that is less than 7.

28. Problem A buffer solution is prepared by dissolving 1.80 g of benzoic acid, C6H5COOH, and 1.95 g of sodium benzoate, NaC6H5COO, in 800 mL of water. Calculate the pH of the buffer solution. What Is Required? You need to determine the pH of the buffer solution. What Is Given? 1.8 g of C6H5COOH and 1.95 g of NaC6H5COO are dissolved in 800 mL of water. −5 From Appendix E, Ka for C6H5COOH is 6.3 × 10 . Plan Your Strategy − Step 1 Determine the initial concentrations of C6H5COOH and C6H5COO in the buffer solution. Step 2 Write the reaction for the dissociation of C6H5COOH. Set up an ICE table, − including the initial concentration of C6H5COO in the buffer. Step 3 Write the equation for Ka, and substitute equilibrium terms into the equation. Step 4 Solve the equation for x. Assume that x is small compared with the initial concentrations. Check the validity of this assumption when you ﬁnd the value of x. Step 5 Use the following equation: + pH =−log[H3O ] Act on Your Strategy Step 1 The molar mass of C6H5COOH is 122.1 g/mol. 1.80 g [C H COOH] = 6 5 122.1 g/mol − = 1.47 × 10 2 mol The molar mass of NaC6H5COO is 144.1 g/mol. 1.95 g [NaC H COO] = 6 5 144.1 g/mol − = 1.35 × 10 2 mol Step 2 − + Concentration (mol/L) C6H5COOH(aq) + H2O() C6H5COOH (aq) + H3O (aq) − − Initial 1.47 × 10 2 1.35 × 10 2 ~0 Change −x +x +x − − Equilibrium 1.47 × 10 2 − x 1.35 × 10 2 + x x

− 3 [C6H5COO ][H3O ] Step 3 Ka = [C6H5COOH] −2 = (1.35 × 10 + x)(x) (1.47 × 10−2 − x) − − Step 4 Assume that (1.35 × 10 2 + x) ≈ 1.35 × 10 2 and (1.47 × 10−2 − x) ≈ 1.47 × 10−2 . × −2 + K = (1.35 10 x)(x) = 6.3 × 10−5 a (1.47 × 10−2) − ∴x = 6.9 × 10 5mol/L

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 182 CHEMISTRY 12

− Step 5 pH =−log(6.9 × 10 5) = 4.16 The pH of the buffer solution is 4.16. Check Your Solution The value of x (6.9 × 10−5) is negligible compared with the initial concentration of each component. A buffer that is made using a weak acid and its conjugate base should have a pH that is less than 7.

Solutions for Practice Problems Student Textbook page 446

29. Problem A solution contains 0.15 mol/L of NaCl and 0.0034 mol/L Pb(NO3)2. Does a precipitate form? Include a balanced chemical equation for the formation of the −5 possible precipitate. Ksp for PbCl2 is 1.7 × 10 . Solution Write the ion product expression for lead(II) chloride. Then substitute the ion concentrations into the expression to determine Qsp. Compare Qsp with Ksp, and predict whether or not a precipitate will form. 2+ − PbCl2(s) Pb (aq) + 2Cl (aq) Ion product expression = [Pb2+][Cl−]2 2 Qsp = (0.0034) × (0.15) − = 7.6 × 10 5 Because Qsp is greater than Ksp, a precipitate will form. Check Your Solution It seems reasonable that a precipitate formed, since Ksp for lead(II) chloride is small compared with the concentration of chloride ions and lead ions.

30. Problem One drop (0.050 mL) of 1.5 mol/L potassium chromate, K2CrO4, is added to 250 mL of 0.10 mol/L AgNO3. Does a precipitate form? Include a balanced chemical equation for the formation of the possible precipitate. Ksp for Ag2CrO4 is 2.6 × 10−12. What Is Required? Will silver chromate precipitate under the given conditions? What Is Given? You know the concentration and volume of the potassium chromate and silver nitrate solutions. For K2CrO4, c = 1.5 mol/L and V = 0.050 mL. For AgNO3, c = 0.10 mol/L and V = 250 mL. You also know Ksp for silver chromate. Plan Your Strategy Step 1 Determine the concentrations of silver ions and chromate ions in the reaction mixture. Step 2 Substitute the ion concentrations into the ion product expression for silver chromate to determine Qsp. Step 3 Compare Qsp with Ksp, and predict whether or not a precipitate will form.

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 183 CHEMISTRY 12

Act on Your Strategy Step 1 Because the volume of the potassium chromate solution is much smaller than the volume of the silver nitrate solution, you can ignore it. In other words, assume that the ﬁnal volume of the reaction mixture is 0.250 L. + − AgNO3(aq) → Ag (aq) + NO3 (aq) + ∴[Ag ] = [AgNO3] = 0.10 mol/L + 2− K2CrO4(aq) → 2K (aq) + CrO4 (aq) 2− c × Vinitial [CrO4 ] = [K2CrO4] = Vﬁnal −3 = (1.5 mol/L)(0.050 mL × 1.0 × 10 L/mL) 0.250 L − = 3.0 × 10 4 mol/L Step 2 The possible precipitate is silver chromate. + 2− 2Ag (aq) + CrO4 → Ag2CrO4(s) + 2 2− Qsp = [Ag ] [CrO4 ] − = (0.10)2 × (3.0 × 10 4) − = 3.0 × 10 6 −12 Step 3 Since Qsp > Ksp , Ag2CrO4 will precipitate until Qsp = 2.6 × 10 . Check Your Solution The units are correct in the calculation of the concentration of silver ions. It seems reasonable that a precipitate formed, since Ksp for silver chromate is very small compared with the concentration of chromate ions and silver ions.

31. Problem 2 A chemist adds 0.010 g of CaCl2 to 5.0 × 10 mL of 0.0015 mol/L sodium carbonate, Na2CO3. Does a precipitate of calcium carbonate form? Include a balanced chemical equation for the formation of the possible precipitate. What Is Required? Will calcium carbonate precipitate under the given conditions? What Is Given? You know the concentration and volume of the sodium carbonate solution. 2 For Na2CO3, c = 0.0015 mol/L and V = 5.0 × 10 mL. For CaCl2, 0.010 g is added to the solution. From Appendix E, you also know that Ksp for calcium carbonate is 3.36 × 10−9. Plan Your Strategy Step 1 Determine the concentrations of calcium ions and carbonate ions in the reaction mixture. Step 2 Substitute the ion concentrations into the ion product expression for calcium carbonate to determine Qsp. Step 3 Compare Qsp with Ksp, and predict whether or not a precipitate will form.

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 184 CHEMISTRY 12

Act on Your Strategy + 2− Step 1 Na2CO3(aq) → 2Na (aq) + CO3 (aq) 2− ∴[CO3 ] = [Na2CO3] = 0.0015 mol/L The molar mass of CaCl2 is 111.0 g/mol. Therefore, 0.010 g Amount CaCl = 2 111.0 g/mol − = 9.0 × 10 5 mol × −5 [CaCl ] = 9.0 10 mol 2 0.50 L − = 1.8 × 10 4 mol/L 2+ − CaCl2(aq) → Ca (aq) + 2Cl (aq) 2+ −4 Therefore, [Ca ] = [CaCl2] = 1.8 × 10 mol/L Step 2 The possible precipitate is calcium carbonate. 2+ 2− Ca (aq) + CO3 → CaCO3(s) 2+ 2− Qsp = [Ca ][CO3 ] − = (1.8 × 10 4) × (0.0015) − = 2.7 × 10 7 −9 Step 3 Since Qsp > Ksp , CaCO3 will precipitate until Qsp = 3.36 × 10 . Check Your Solution The units are correct in the calculation of the concentration of silver ions. It seems reasonable that a precipitate formed, since Ksp for calcium carbonate is small compared with the concentration of calcium ions and carbonate ions.

32. Problem 2 0.10 mg of magnesium chloride, MgCl2, is added to 2.5 × 10 mL of 0.0010 mol/L NaOH. Does a precipitate of magnesium hydroxide form? Include a balanced chemical equation for the formation of the possible precipitate. What Is Required? Will magnesium hydroxide precipitate under the given conditions? What Is Given? You know the concentration and volume of the sodium hydroxide solution. 2 For NaOH, c = 0.0010 mol/L and V = 2.5 × 10 mL. For MgCl2, 0.10 mg is added to the solution. From Appendix E, you also know that Ksp for magnesium hydroxide is 5.61 × 10−12. Plan Your Strategy Step 1 Determine the concentrations of magnesium ions and hydroxide ions in the reaction mixture. Step 2 Substitute the ion concentrations into the ion product expression for magnesium hydroxide to determine Qsp. Step 3 Compare Qsp with Ksp, and predict whether or not a precipitate will form.

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 185 CHEMISTRY 12

Act on Your Strategy + − Step 1 NaOH(aq) → Na (aq) + OH (aq) Therefore, [OH−] = [NaOH] = 0.0010 mol/L

The molar mass of MgCl2 is 95.2 g/mol. Therefore, 0.10 mg × 10−3 g/mg Amount MgCl = 2 95.2 g/mol − = 1.1 × 10 6 mol × −6 [MgCl ] = 1.1 10 mol 2 0.25 L − = 4.2 × 10 6 mol/L 2+ − MgCl2(aq) → Mg (aq) + 2Cl (aq) 2+ −6 Therefore, [Mg ] = [MgCl2] = 4.2 × 10 mol/L. Step 2 The possible precipitate is magnesium hydroxide. 2+ − Mg (aq) + 2OH (aq) → Mg(OH)2(s) 2+ − 2 Qsp = [Mg ][OH ] − = (4.2 × 10 6) × (0.0010)2 − = 4.2 × 10 12 Step 3 Since Qsp < Ksp , no precipitate forms. Check Your Solution The units are correct in the calculation of the concentration of silver ions. It seems reasonable that no precipitate formed, since the concentration of magnesium ion is small.

Solutions for Practice Problems Student Textbook page 447

33. Problem 2 −3 1.0 × 10 mL of 1.0 × 10 mol/L Pb(NO3)2 is added to 40 mL of 0.040 mol/L NaCl. Does a precipitate form? Include a balanced chemical equation for the formation of the possible precipitate. What Is Required? Decide whether a precipitate will form under the given circumstances. Write a balanced chemical equation for the formation of the possible precipitate. What Is Given? You know the initial concentration and volume of the lead(II) nitrate and sodium −3 2 chloride solutions. For Pb(NO3)2, c = 1.0 × 10 mol/L and V = 1.0 × 10 mL. For NaCl, c = 0.040 mol/L and V = 40 mL. As well, you have the solubility guidelines in Table 9.3. Plan Your Strategy Step 1 Decide whether a compound with low solubility forms when lead(II) nitrate and sodium chloride are mixed, using the solubility guidelines. Step 2 If an insoluble compound forms, write an equation that represents the reaction. Look up Ksp for the compound in Appendix E. Step 3 Determine the concentrations of the ions that make up the compound. Step 4 Substitute the concentrations of the ions into the ion product expression for the compound to determine Qsp. Step 5 Compare Qsp with Ksp, and predict whether or not a precipitate forms.

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 186 CHEMISTRY 12

Act on Your Strategy Step 1 When you mix the solutions, lead(II) chloride, PbCl2, and sodium nitrate, NaNO3, can also form. Sodium nitrate is soluble, but lead(II) chloride is not, according to guideline 2. Step 2 Pb(NO3)2(aq) + NaCl(aq) PbCl2(s) + NaNO3(aq) −5 From Table 9.2, Ksp for PbCl2 is 1.7 × 10 . Step 3 When calculating the volume of the reaction mixture, assume that it is equal to the sum of the volumes of the solutions. This is a reasonable assumption, because both solutions are dilute. 2+ c × Vinitial [Pb ] = [Pb(NO3)2] = Vﬁnal −3 = (1.0 × 10 mol/L)(100 mL) (100 mL + 40 mL) − = 7.1 × 10 4 mol/L − × [Cl ] = [NaCl] = c Vinitial Vﬁnal = (0.040 mol/L)(40 mL) (100 mL + 40 mL) − = 1.1 × 10 2 mol 2+ − 2 Step 4 Qsp = [Pb ][Cl ] − − = (7.1 × 10 4)(1.1 × 10 2)2 − = 8.6 × 10 8 Step 5 Since Qsp < Ksp , no precipitate forms. Check Your Solution The units in the calculation of the concentrations of the ions are correct. It seems reasonable that no precipitate forms, since the concentrations of ions are relatively small.

34. Problem 2 2 2.3 × 10 mL of 0.0015 mol/L AgNO3 is added to 1.3 × 10 mL of 0.010 mol/L calcium acetate, Ca(CH3COO)2. Does a precipitate form? Include a balanced chemical equation for the formation of the possible precipitate. Ksp for AgCH3COO is 2.0 × 10−3 . What Is Required? Decide whether a precipitate will form under the given circumstances. Write a balanced chemical equation for the formation of the possible precipitate. What Is Given? You know the initial concentration and volume of the silver nitrate and calcium acetate solutions. For AgNO3, c = 0.0015 mol/L and V = 230 mL. For Ca(CH3COO)2, c = 0.010 mol/L and V = 130 mL. As well, you have the solubility guidelines in Table 9.3. Plan Your Strategy Step 1 The Ksp value for silver acetate indicates that it is not very soluble in water. Write an equation that represents the reaction to form silver acetate. Step 2 Determine the concentrations of the ions that make up the compound. Step 3 Substitute the concentrations of the ions into the ion product expression for the compound to determine Qsp. Step 4 Compare Ksp with Qsp, and predict whether or not a precipitate forms.

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 187 CHEMISTRY 12

Act on Your Strategy Step 1 2AgNO3(aq) + Ca(CH3COO)2(aq) 2AgCH3COO(s) + Ca(NO3)2(aq) Step 2 When calculating the volume of the reaction mixture, assume that it is equal to the sum of the volumes of the solutions. This is a reasonable assumption, because both solutions are dilute. + c × Vinitial [Ag ] = [AgNO3] = Vﬁnal = (0.0015 mol/L)(230 mL) (230 mL + 130 mL) − = 9.6 × 10 4 mol/L

− c × Vinitial [CH3COO ] = 2 × [Ca(CH3COO)2] = 2 × Vﬁnal = 2 × (0.010 mol/L)(130 mL) (130 mL + 230 mL) − = 7.2 × 10 3 mol/L + − Step 3 Qsp = [Ag ][CH3COO ] − − = (9.6 × 10 4)(7.2 × 10 3) − = 6.9 × 10 6 Step 4 Since Qsp < Ksp , no precipitate is formed. Check Your Solution The units in the calculation of the concentrations of the ions are correct. It seems reasonable that no precipitate formed, since Ksp for silver acetate is a relatively large value compared with the concentrations of the silver ions and acetate ions.

35. Problem 25 mL of 0.10 mol/L NaOH is added to 5.0 × 102 mL of 0.00010 mol/L cobalt(II) chloride, CoCl2. Does a precipitate form? Include a balanced chemical equation for the formation of the possible precipitate. What Is Given? You know the initial concentration and volume of the sodium hydroxide and cobalt(II) chloride solutions. For NaOH, c = 0.10 mol/L and V = 25 mL. For CoCl2, c = 0.00010 mol/L and V = 500 mL. As well, you have the solubility guidelines in Table 9.3. Plan Your Strategy Step 1 Decide whether a compound with low solubility forms when sodium hydroxide and cobalt(II) chloride are mixed, using the solubility guidelines. Step 2 If an insoluble compound forms, write an equation that represents the reaction. Look up Ksp for the compound in Appendix E. Step 3 Determine the concentrations of the ions that make up the compound. Step 4 Substitute the concentrations of the ions into the ion product expression for the compound to determine Qsp. Step 5 Compare Qsp with Ksp, and predict whether or not a precipitate forms. Act on Your Strategy Step 1 When you mix the sodium hydroxide and cobalt(II) chloride solutions, sodium chloride, NaCl, and cobalt(II) hydroxide, Co(OH)2, can also form. Sodium chloride is soluble, but cobalt(II) hydroxide is probably not, according to guideline 2. Step 2 NaOH(aq) + CoCl2(aq) → NaCl(aq) + Co(OH)2(s) −15 From Appendix E, Ksp for Co(OH)2 is 5.92 × 10 .

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 188 CHEMISTRY 12

Step 3 When calculating the volume of the reaction mixture, assume that it is equal to the sum of the volumes of the solutions. This is a reasonable assumption, because both solutions are dilute. 2+ c × Vinitial [Co ] = [CoCl2] = Vﬁnal = (0.00010 mol/L)(500 mL) (500 mL + 25 mL) − = 9.5 × 10 5 mol/L

− × [OH ] = [NaOH] = c Vinitial Vﬁnal = (0.10 mol/L)(25 mL) (25 mL + 500 mL) − = 4.8 × 10 3 mol/L 2+ − 2 Step 4 Qsp = [Co ][Cl ] − − = (9.5 × 10 5)(4.8 × 10 3)2 − = 2.2 × 10 9 −15 Step 5 Since Qsp > Ksp , Co(OH)2 precipitates until Qsp = 5.92 × 10 . Check Your Solution The units in the calculation of the concentrations of the ions are correct. It seems reasonable that a precipitate formed, since Ksp for cobalt hydroxide is very small compared with the concentrations of the cobalt ions and hydroxide ions.

36. Problem 250 mL of 0.0011 mol/L Al2(SO4)3 is added to 50 mL of 0.022 mol/L BaCl2. Does a precipitate form? Include a balanced chemical equation for the formation of the possible precipitate. What Is Given? You know the initial concentration and volume of the aluminum sulfate and barium chloride solutions. For Al2(SO4)3, c = 0.0011 mol/L and V = 250 mL. For BaCl2, c = 0.022 mol/L and V = 50 mL. As well, you have the solubility guidelines in Table 9.3. Plan Your Strategy Step 1 Decide whether a compound with low solubility forms when calcium nitrate and sodium ﬂuoride are mixed, using the solubility guidelines. Step 2 If an insoluble compound forms, write an equation that represents the reaction. Look up Ksp for the compound in Appendix E. Step 3 Determine the concentrations of the ions that make up the compound. Step 4 Substitute the concentrations of the ions into the ion product expression for the compound to determine Qsp. Step 5 Compare Qsp with Ksp, and predict whether or not a precipitate forms. Act on Your Strategy Step 1 When you mix the aluminum sulfate and barium chloride solutions, aluminum chloride, AlCl3, and barium sulfate, BaSO4, can also form. Aluminum chloride is soluble, but barium sulfate is not, according to guideline 4. Step 2 Al2(SO4)3(aq) + 3BaCl2(aq) → 2AlCl3(aq) + 3BaSO4(s) −10 From Appendix E, Ksp for BaSO4 is 1.08 × 10 .

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 189 CHEMISTRY 12

Step 3 When calculating the volume of the reaction mixture, assume that it is equal to the sum of the volumes of the solutions. This is a reasonable assumption, because both solutions are dilute. 2+ c × Vinitial [Ba ] = [BaCl2] = Vﬁnal = (0.022 mol/L)(50 mL) (50 mL + 250 mL) − = 3.7 × 10 3 mol/L

2− c × Vinitial [SO4 ] = 3 × [Al2(SO4)3] = 3 × Vﬁnal = 3 × (0.0011 mol/L)(250 mL) (250 mL + 50 mL) − = 2.8 × 10 3 mol/L 2+ 2− Step 4 Qsp = [Ba ][SO4 ] − − = (3.7 × 10 3)(2.8 × 10 3) − = 1.0 × 10 5 −10 Step 5 Since Qsp > Ksp , BaSO4 precipitates until Qsp = 1.08 × 10 . Check Your Solution The units in the calculation of the concentrations of the ions are correct. It seems reasonable that a precipitate formed, since Ksp for barium sulfate is very small compared with the concentrations of the barium ions and sulfate ions.

Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR 190