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.