AcidBase Calculations Answers 1. In each of the following solutions, determine: i) hydronium ion concentration iii) ph ii) hydroxide ion concentration iv) poh a) 0.030 mol/dm 3 HCl HCl is a strong acid HCl + H 2 O H 3 O + + Cl i) HCl completely breaks down, therefore [H 3 O + ] = 0.030 mol/l ii) [OH ] = 10 poh = 3.3 x 10 13 mol/l iii) ph = log [H 3 O + ] = 1.52 iv) poh = 14 ph = 12.48 b) 0.125 mol/dm 3 HBr HBr is a strong acid HCl + H 2 O H 3 O + + Br i) HBr completely breaks down, therefore [H 3 O + ] = 0.125 mol/l ii) [OH ] = 10 poh = 8.00 x 10 14 mol/l iii) ph = log [H 3 O + ] = 0.903 iv) poh = 14 ph = 13.097 c) 0.025 mol/dm 3 NaOH NaOH is a strong base NaOH Na + + OH i) [H 3 O + ] = 10 ph = 4.00 x 10 13 mol/l ii) [OH ] = 0.025 mol/l iii) ph = 14 poh = 12.40 iv) poh = log [OH ] = 1.60 d) 0.050 mol/dm 3 Ca(OH) 2 Ca(OH) 2 is a strong base Ca(OH) 2 Ca 2+ + 2OH i) [H 3 O + ] = 10 ph = 1.00 x 10 13 mol/l ii) [OH ] = 2(0.050 mol/l) = 0.10 mol/l iii) ph = 14 poh = 13.00 iv) poh = log [OH ] = 1.00 L. Farrell Chemistry 12 AcidBase Calculations Answers Page 1 of 13
e) 0.100 mol/dm 3 H 2 CO 3 H 2 CO 3 is a weak acid: H 2 CO 3 + H 2 O H 3 O + + HCO 3 H 2 CO 3 + H 2 O H 3 O + + HCO 3 [Initial] 0.100 0 0 [Equilibrium] 0.100 x x x K a = H 3O + HCO 3 H 2 CO 3 4.4 x 10 7 = x x 0.100x Shortcut check, [HA] K a > 500 4.4 x 10 7 = x x 0.100 (4.4 x 10 7 )(0.100) = x 2 4.4 x 10 8 = x 2 x = 2.1 x 10 4 i) [H 3 O + ] = 2.1 x 10 4 mol/l ii) [OH ] = 10 poh = 4.8 x 10 11 mol/l iii) ph = log [H 3 O + ] = 3.68 iv) poh = 14 ph = 10.32 L. Farrell Chemistry 12 AcidBase Calculations Answers Page 2 of 13
f) 0.250 mol/dm 3 HCN HCN is a weak acid: HCN + H 2 O H 3 O + + CN HCN + H 2 O H 3 O + + CN [Initial] 0.250 0 0 [Equilibrium] 0.250 x x x K a = H 3O + CN HCN 6.2 x 10 10 = x x 0.250x Shortcut check, [HA] K a > 500 6.2 x 10 10 = x x 0.250 (6.2 x 10 10 )(0.250) = x 2 1.55 x 10 10 = x 2 x = 1.2 x 10 5 i) [H 3 O + ] = 1.2 x 10 5 mol/l ii) [OH ] = 10 poh = 8.0 x 10 10 mol/l iii) ph = log [H 3 O + ] = 4.90 iv) poh = 14 ph = 9.10 2. Calculate [H 3 O + ], [ClO 4 ] and [OH ] in an aqueous solution that is 0.150 M in HClO 4 (aq). Is the solution acidic or basic? HClO 4 is a strong acid: HClO 4 + H 2 O H 3 O + + ClO 4 [H 3 O + ] = 0.150 mol/l [ClO 4 ] = 0.150 mol/l K w = [H 3 O + ][OH ] [OH ] = K w [H 3 O + ] [OH ] = 1.00 X 10 14 (0.150 mol/l) = 6.67 x 10 14 mol/l The solution is acidic, [H 3 O + ] > [OH ] L. Farrell Chemistry 12 AcidBase Calculations Answers Page 3 of 13
3. Calculate, [OH ], [K + ] and [H 3 O + ] in an aqueous solution that is 0.250 M in KOH(aq). Is the solution acidic or basic? KOH is a strong base: KOH K + + OH [OH ] = 0.250 mol/l [K + ] = 0.250 mol/l K w = [H 3 O + ][OH ] [H 3 O + ] = K w [OH ] [H 3 O + ] = 1.00 X 10 14 (0.250 mol/l) = 4.00 x 10 14 mol/l The solution is basic, [H 3 O + ] < [OH ] 4. Compute [Ca 2+ ], [OH ] and [H 3 O + ] for a solution that is prepared by dissolving 0.600 g of Ca(OH) 2 in enough water to make 0.500 dm 3 of solution. Ca(OH) 2 is a strong base: Ca(OH) 2 Ca 2+ + 2OH [Ca(OH) 2 ] = n V n = mass molar mass n = 0.600 g 74.09268 g/mol = 0.00810 mol [Ca(OH) 2 ] = n V [Ca(OH) 2 ] = 0.00810 mol 0.500 L = 0.0162 mol/l [OH ] = 2(0.0162 mol/l) = 0.0324 mol/l [Ca 2+ ] = 0.0162 mol/l K w = [H 3 O + ][OH ] [H 3 O + ] = K w [OH ] [H 3 O + ] = 1.00 X 10 14 (0.0324 mol/l) = 3.09 x 10 13 mol/l The solution is basic, [H 3 O + ] < [OH ] L. Farrell Chemistry 12 AcidBase Calculations Answers Page 4 of 13
5. The value of K a in water at 25 C for benzoic acid, HC 7 H 5 O 2 is 6.46 x 10 5. Calculate the ph of an aqueous solution of 0.0200 M HC 7 H 5 O 2. HC 7 H 5 O 2 is a weak acid: HC 7 H 5 O 2 + H 2 O H 3 O + + C 7 H 5 O 2 HC 7 H 5 O 2 + H 2 O H 3 O + + C 7 H 5 O 2 [Initial] 0.0200 0 0 [Equilibrium] 0.0200 x x x K a = H 3O + C 7 H 5 O 2 HC 7 H 5 O 2 6.46 x 10 5 = x x 0.0200x Shortcut check, [HA] K a < 500, therefore must keep x (6.46 x 10 5 )(0.0200 x) = x 2 1.292 x 10 6 6.46 x 10 5 x = x 2 0 = x 2 + 6.46 x 10 5 x 1.292 x 10 6 6.46 x 10 6.46 x 10 4 1 1.292 x 10 2 1 6.46 x 10 5.172 x 10 2 6.46 x 10 0.00227 2 x = 0.001169 or +0.001105 x = [H 3 O + ] = 0.00110 mol/l ph = log(0.00110) = 2.957 L. Farrell Chemistry 12 AcidBase Calculations Answers Page 5 of 13
6. Sulfamic acid, HO 3 SNH 2, is used as a stabilizer for chlorine in swimming pools. Calculate the ph of a 0.0400 M sulfamic acid solution given that K a = 0.100. HO 3 SNH 2 is a weak acid: HA + H 2 O H 3 O + + A HA + H 2 O H 3 O + + A [Initial] 0.0400 0 0 [Equilibrium] 0.0400 x x x K a = H 3O + A HA 0.100 = x x 0.0400x Shortcut check, [HA] K a < 500, therefore must keep x (0.100)(0.0400 x) = x 2 0.00400 0.100x = x 2 0 = x 2 + 0.100x 0.00400 0.100 0.100 4 1 0.00400 2 1 0.100 0.0260 2 x = 0.131 or +0.0306 x = [H 3 O + ] = 0.00306 mol/l ph = log(0.0306) = 1.514 L. Farrell Chemistry 12 AcidBase Calculations Answers Page 6 of 13
7. Calculate the ph of an aqueous solution of 0.150 M HClO. HClO is a weak acid: HClO + H 2 O H 3 O + + ClO HClO + H 2 O H 3 O + + ClO [Initial] 0.150 0 0 [Equilibrium] 0.150 x x x K a = H 3O + ClO HClO 2.9 x 10 8 = x x 0.150x Shortcut check, [HA] K a > 500 2.9 x 10 8 = x x 0.150 (2.9 x 10 8 )(0.150) = x 2 4.35 x 10 9 = x 2 x = 6.6 x 10 5 [H 3 O + ] = 6.6 x 10 5 mol/l ph = log [H 3 O + ] = 4.18 L. Farrell Chemistry 12 AcidBase Calculations Answers Page 7 of 13
8. Suppose two aspirin tablets (1 tablet = 324 mg) are dissolved in enough water to make 500.0 cm 3 of solution. Compute the ph of the resulting solution if K a = 2.75 x 10 5 and the molar mass for aspirin (acetylsalicylic acid) is 180.15 g/mol. 2 tablets = (2)(324 mg) = 648 mg = 0.648 g mol = mass molar mass mol = (0.648 g) (180.15 g/mol) = 0.00360 mol aspirin C = n V C = (0.00360 mol) (0.5000 L) = 0.00719 mol/l = concentration of aspirin solution aspirin is a weak acid: HA + H 2 O H 3 O + + A HA + H 2 O H 3 O + + A [Initial] 0.00719 0 0 [Equilibrium] 0.00719 x x x K a = H 3O + A HA 2.75 x 10 = x x 0.00719x Shortcut check, [HA] K a < 500, therefore must keep x (2.75 x 10 5 )(0.00719 x) = x 2 1.98 x 10 7 2.75 x 10 5 x = x 2 0 = x 2 + 2.75 x 10 5 x 1.98 x 10 7 2.75 x 10 2.75 x 10 4 1 1.98 x 10 2 1 2.75 x 10 7.92 x 10 2 2.75 x 10 8.90 x 10 2 x = 4.59 x 10 4 or +4.31 x 10 4 x = [H 3 O + ] = 4.31 x 10 4 mol/l ph = log(4.31 x 10 4 ) = 3.365 L. Farrell Chemistry 12 AcidBase Calculations Answers Page 8 of 13
9. The ph of a 0.72 mol/l solution of benzoic acid, HC 7 H 5 O 2, is 2.68. Calculate the numerical value of the K a for this acid. HC 7 H 5 O 2 is a weak acid: HC 7 H 5 O 2 + H 2 O H 3 O + + C 7 H 5 O 2 HC 7 H 5 O 2 + H 2 O H 3 O + + C 7 H 5 O 2 [Initial] 0.72 0 0 [Equilibrium] 0.72 x x x K a = H 3O C 7 H 5 O 2 HC 7 H 5 O 2 K = x x 0.0200x ph = 2.68 [H 3 O + ] = 10 ph = 2.1 x 10 3 mol/l = x K a = 0.0021 0.0021 0.720.0021 = 6.1 x 10 6 10. The value of K a for acetic acid is 1.8 x 10 5 at 25 C. Calculate the [H 3 O + ] in a 0.10 mol/l solution of acetic acid. What is the percent ionization of the acetic acid? HC 2 H 3 O 2 is a weak acid: HC 2 H 3 O 2 + H 2 O H 3 O + + C 2 H 3 O 2 HC 2 H 3 O 2 + H 2 O H 3 O + + C 2 H 3 O 2 [Initial] 0.10 0 0 [Equilibrium] 0.10 x x x K a = H 3O + C 2 H 3 O 2 HC 2 H 3 O 2 1.8 x 10 5 = x x 0.10 x Shortcut check, [HA] K a > 500 1.8 x 10 5 = x x 0.10 (1.8 x 10 5 )(0.10) = x 2 1.8 x 10 6 = x 2 x = 1.3 x 10 3 [H 3 O + ] = 1.3 x 10 3 mol/l percent ionization= amount ionized original amount x 100 percent ionized= H 3O + HA x 100 percent ionized = 0.0013 0.10 x 100 = 1.3% ionized L. Farrell Chemistry 12 AcidBase Calculations Answers Page 9 of 13
11. The ionization constant, K b, for the weak base ammonia, NH 3, is 1.7 x 10 5. Calculate the [OH ], ph, and percent ionization of ammonia in a 0.082 mol/l solution of ammonia. NH 3 is a weak base: NH 3 + H 2 O NH 4 + + OH NH 3 + H 2 O + NH 4 + OH [Initial] 0.082 0 0 [Equilibrium] 0.082 x x x K b = NH 4 + OH NH 3 1.7 x 10 5 = x x 0.082 x Shortcut check, [B] K b > 500 1.7 x 10 5 = x x 0.082 (1.7 x 10 5 )(0.082) = x 2 1.394 x 10 6 = x 2 x = 1.18 x 10 3 [OH ] = 1.2 x 10 3 mol/l percent ionization= amount ionized original amount x 100 percent ionized= OH x 100 percent ionized = 0.00118 0.082 x 100 = 14% ionized L. Farrell Chemistry 12 AcidBase Calculations Answers Page 10 of 13
12. The hydroxide ion concentration in a 0.157 mol/l solution of sodium propanoate, NaC 2 H 5 COO(aq), is found to be 1.1 x 10 5 mol/l. Calculate the base ionization constant for the propanoate ion. The propanoate ion is a weak base: B + H 2 O BH + + OH B + H 2 O BH + + OH [Initial] 0.157 0 0 [Equilibrium] 0.157 x x x K b = BH+ OH K = x x 0.157x [OH ] = 1.1 x 10 5 mol/l = x K b = 1.1 x 1.1 x 0.157 1.1 x = 7.7 x 10 10 13. Aniline, C 6 H 5 NH 2, is closely related to ammonia and is also a weak base. If the ph of a 0.10 mol/l aniline solution was found to be 8.81, what is its K b? Alinine is a weak base: C 6 H 5 NH 2 + H 2 O C 6 H 5 NH 3 + + OH B + H 2 O BH + + OH [Initial] 0.10 0 0 [Equilibrium] 0.10 x x x K b = BH+ OH K = x x 0.10 x ph = 8.81 poh = 14 8.81 = 5.19 [OH ] = 10 poh = 10 5.19 = 6.5 x 10 6 mol/l = x K b = 6.5 x 6.5 x 0.10 6.5 x = 4.2 x 10 10 L. Farrell Chemistry 12 AcidBase Calculations Answers Page 11 of 13
14. Codeine has a K b of 1.73 x 10 6. Calculate the ph of a 0.020 mol/l codeine solution. Codeine is a weak base: B + H 2 O BH + + OH B + H 2 O BH + + OH [Initial] 0.020 0 0 [Equilibrium] 0.020 x x x K b = BH+ OH 1.73 x 10 6 = x x 0.020 x Shortcut check, [B] K b > 500 1.73 x 10 6 = x x 0.020 (1.73 x 10 6 )(0.020) = x 2 3.46 x 10 8 = x 2 x = 1.86 x 10 4 [OH ] = 1.9 x 10 4 mol/l poh = log(1.9 x 10 4 ) = 3.73 ph = 14 poh = 10.27 15. Codeine (C 18 H 21 NO 3 ) is a weak organic base. A 5.0 x 10ˉ3 M solution of codeine has a ph of 9.95. Calculate the value of K b for this substance under the conditions in this question. Codeine is a weak base: C 18 H 21 NO 3 + H 2 O C 18 H 21 NO 3 H + + OH B + H 2 O BH + + OH [Initial] 0.0050 0 0 [Equilibrium] 0.0050 x x x K = K b = BH+ OH x x 0.0050 x ph = 9.95 poh = 14 9.95 = 4.05 [OH ] = 10 poh = 10 4.05 = 8.9 x 10 5 mol/l = x K b = 8.9 x 105 8.9 x 10 5 0.0050 8.9 x 10 5 = 1.5 x 10 6 L. Farrell Chemistry 12 AcidBase Calculations Answers Page 12 of 13
16. A 0.0135 M solution of a weak base (generic formula = B) has a ph of 8.39. Calculate the K b for this weak base. Weak Base: B + H 2 O BH + + OH B + H 2 O BH + + OH [Initial] 0.0135 0 0 [Equilibrium] 0.0135 x x x K = K b = BH+ OH x x 0.0135 x ph = 8.39 poh = 14 8.39 = 5.61 [OH ] = 10 poh = 10 5.61 = 2.5 x 10 6 mol/l = x K b = 2.5 x 106 2.5 x 10 6 0.0135 2.5 x 10 6 = 4.5 x 10 10 17. Calculate the ph in a 3.0 M Pyridine solution with K b = 1.5 x 10 9. Pyridine is a weak base: B + H 2 O BH + + OH B + H 2 O BH + + OH [Initial] 3.0 0 0 [Equilibrium] 3.0 x x x K b = BH+ OH 1.5 x 10 9 = x x 3.0 x Shortcut check, [B] K b > 500 1.5 x 10 9 = x x 3.0 (1.5 x 10 9 )(3.0) = x 2 4.5 x 10 9 = x 2 x = 6.7 x 10 5 [OH ] = 6.7 x 10 5 mol/l poh = log(6.7 x 10 5 ) = 4.17 ph = 14 poh = 9.83 L. Farrell Chemistry 12 AcidBase Calculations Answers Page 13 of 13