Chemistry 19 Problem Set 3 Spring, 018 Solutions 1. Problem 3, page 78, textbook Answer (a) (b) (c) (d) HOBr (acid 1) + HSO 4 (acid 1) + HS (base 1) + C 6 H 5 NH + 3 (acid 1) + H O (base ) H 3O + (acid ) + H O (base ) H 3O + (acid ) + H O (acid ) H S (acid 1) + OH (base ) C 6H 5 NH (base 1) + OBr (base 1) SO 4 (base 1) OH (base ) H O (acid ). Calculate the ph and poh of the following solutions of strong acids or bases: (a) 0.015 M HCl (aq) [H 3 O + ] = 1.5 10 ph = log 10 (1.5 10 ) = 1.903 poh = 14.00 ph = 1.10 1
(b) 0.0460 M NaOH (aq) [OH ] = 4.60 10 poh = log 10 (4.60 10 ) = 1.337 ph = 14.00 poh = 1.66 (c) 1.75 10 8 M HCl (aq) There are two equilibrium processes to consider and HCl (aq) + H O (l) H 3 O + (aq) + Cl (aq) H O (l) + H O (l) H 3 O + (aq) + OH (aq) All the OH (aq) results from the self dissociation of water. We let x = [OH ]. The H 3 O + concentration has contributions from both equilibria. Then [H 3 O + ] = x + 1.75 10 8. Using the ion product of water or Using the quadratic formula x = 1.75 10 8 ± [H 3 O + ][OH ] = x(x + 1.75 10 8 ) = 1.00 10 14 + 1.75 10 8 x 1.00 10 14 = 0 (1.75 10 8 ) + 4 10 14 Only the positive solution for x is physical and = 9.0 10 8, 1.10 10 7. poh = log 10 (9.0 10 8 ) = 7.04 ph = 14.00 poh = 6.96 (d) 1.75 10 8 M NaOH (aq) The method of solution is identical to the solution to problem c, except that we set x = [H 3 O + ]. Then using similar algebra ph = log 10 (9.0 10 8 ) = 7.04 poh = 14.00 ph = 6.96 3. Problem 14, page 78, textbook n OH = (0.15 L)(0.606 mol L 1 ) = 0.0758 mol [OH 0.0758 mol ] = = 00.00505 M 15.0 L poh = log 10 (0.00505) =.97 ph = 14.00 poh = 11.70
4. Problem 19, page 78, textbook Before mixing n H + = (0.05000 L)(0.0155 mol L 1 ) = 0.000775 mol After mixing n OH = (0.07500 L)(0.0106 mol L 1 ) = 0.000795 mol n OH = 0.000795 mol 0.000775 mol = 0.000000 mol [OH ] = 0.000000mol = 0.000160 M 0.15 L poh = log 10 (0.000160) = 3.796 ph = 14.00 poh = 10.0 5. Problem 0, Page 78, textbook [H 3 O + ] = 10.1 = 7.58 10 3 M After mixing n H3 O + = (7.58 10 3 mol L 1 )(0.0500 L) = 1.94 10 4 mol [OH ] = 10 (14 1.65) = 4.47 10 M n OH = (0.0500 L)(4.47 10 mol L 1 ) = 1.17 10 3 mol n OH = 1.17 10 3 mol 1.94 10 4 mol = 9.76 10 4 mol [OH ] = 9.76 10 4 mol = 1.95 10 M 0.0500 L poh = log 10 (1.95 10 ) = 1.71 ph = 14.00 poh = 1.9 6. Calculate the volume of a 0.100 M aqueous KOH solution that must be added to 100. ml of a 0.100 M aqueous HCl solution to obtain a solution having a ph=.000. Before mixing n H3 O + = (0.100 mol L 1 )(0.100 L) = 0.0100 mol Let V = volume of the KOH solution to be added. Then the final H 3 O + concentration is n H3 O + = (0.0100 mol (0.100 mol L 1 )V = 10 mol L 1 V final V + 0.100 L (10 mol L 1 )V + 1.00 10 3 mol = 0.0100 mol (0.100 mol L 1 )V (0.110 mol L 1 )V = 0.00900 mol V = 0.0818 L = 81.8 ml 3
7. Given that sodium hydroxide is a strong base, calculate the ph of an aqueous 7.5 10 8 M sodium hydroxide solution. Approximations do not work for this problem. [H 3 O + ][OH ] = 1.0 10 14 Let y = [H 3 O + ] [OH ] = y + 7.5 10 8 y(y + 7.5 10 8 ) = 1.0 10 14 y + 7.5 10 8 y 1.0 10 14 = 0 y = 7.5 10 8 ± [(7.5 10 8 ) + 4.0 10 14 ] 1/ Keeping the positive solution y = [H 3 O + ] = 6.9 10 8 M ph = log 10 (6.9 10 8 ) = 7.16 8. Calculate the ph and percent ionization of a.00 M aqueous HCN solution given that for the reaction HCN (aq) + H O (l) H 3 O + (aq) + CN (aq), K a = 4.1 10 10. Answer K a = [H 3O + ][CN ] [HCN] [HCN] [H 3 O + ] [CN ] initial.00 M 0 M 0 M change -x M x M x M equilibrium (.00-x) M x M x M % ionized =.00 x = 4.1 10 10 = 8. 10 10 x =.9 10 5.9 10 5.00 100. = 1.4 10 3 % ph = log 10 (.9 10 5 ) = 4.54 4
9. The pk a of phenol (C 6 H 5 OH) when acting as a weak acid in aqueous solution is 9.99. Calculate the ph of a.0 M phenol solution. Approximations work for this system. K a = 10 9.99 = 1.0 10 10 C 6 H 5 OH (aq) + H O (l) C 6 H 5 O (aq) + H 3O + (aq) [C 6 H 5 OH] [C 6 H 5 O ] [H 3 O + ] initial.0 M 0 M 0 M change -y y y equilibrium (.0 -y) M y M y M y.0 y y.0 = 1.0 10 10 y =.0 10 10 = 1.4 10 5 M = [H 3 O + ] ph = log 10 (1.4 10 5 ) = 4.84 10. When a 0.07 M aqueous solution of benzoic acid reacts C 6 H 5 COOH (aq) + H O (l) H 3 O + (aq) + C 6H 5 COO (aq) the ph is found to be.68. Calculate the pk a for the reaction. Answer Let BA C 6 H 5 COOH and A C 6 H 5 COO. Then [H 3 O + ] = 10.68 =.1 10 3 = [A ] [BA] [H 3 O + ] [A ] initial 0.07 M 0 M 0 M change -x M x M x M equilibrium (0.07-x) M x M x M K a = [H 3O + ][A ] [BA] = x =.1 10 3 0.07 x = (.1 10 3 ) = 6.3 10 5 0.07.1 10 3 pk a = log 10 (6.3 10 5 ) = 4.0 5
11. Formic acid (HCOOH) is a monoprotic acid with acid dissociation reaction HCOOH (aq) + H O (l) HCOO (aq) + H 3O + (aq). The ph of a 1.0 10 3 M formic acid solution is found to be 3.41. Calculate K a and pk a for formic acid. [H 3 O + ] = 10 3.41 = 3.9 10 4 [HCOOH] [H 3 O + ] [HCOO ] initial 1.0 10 3 M 0 M 0 M change -3.9 10 4 M 3.9 10 4 M 3.9 10 4 M equilibrium 8.1 10 4 M 3.9 10 4 M 3.9 10 4 M K a = [HCOO ][H 3 O + ] [HCOOH] = (3.9 10 4 ) 8.1 10 4 = 1.9 10 4 pk a = log 10 K a = log 10 (1.9 10 4 ) = 3.73 1. Dimethylamine ionizes according to the reaction (CH 3 ) NH (aq) + H O (l) (CH 3 ) NH + (aq) + OH (aq) A 0.95 M solution of dimethylamine is found to have a ph of 1.3. Calculate the pk b of the reaction. K b = [(CH 3) NH + ][OH ] [(CH 3 ) NH] poh = 14.00 ph = 1.68 [OH ] = 10 1.68 =.1 10 [(CH 3 ) NH] [(CH 3 ) NH + ] [OH ] initial 0.95 M 0 M 0 M change -x M x M x M equilibrium (0.95-x) M x M x M 0.95 x = (.1 10 ) = 4.7 10 4 0.95.1 10 pk b = log 10 (4.7 10 4 ) = 3.3 6
13. Hydrofluoric acid (HF (aq) ) is formed when gas-phase hydrogen fluoride (HF (g) ) is dissolved in water according to the reaction HF (aq) + H O (l ) F (aq) + H 3O + (aq) The pk a of hydrofluoric acid is 3.18. Calculate the ph of a solution that forms when 0.5 grams of gas-phase HF are dissolved in water to make 1. L of solution. Approximations do not work for this problem. K a = 10 3.18 = 6.6 10 4 Initially ( ) mol (0.5 g) (19.0 + 1.0) g [HF] = = 1.0 10 M 1. L [HF] [H 3 O + ] [F ] initial 1.0 10 M 0 M 0 M change -y M y M y M equilibrium (1.0 10 y) M y M y M 6.6 10 4 = [F ][H 3 O + ] [HF] = y 1.0 10 y Ignoring the negative solution y + 6.6 10 4 y 6.6 10 6 = 0 y = 6.6 10 4 + [(6.6 10 4 ) + 4(6.6 10 6 )] 1/ =. 10 3 [H 3 O + ] =. 10 3 M ph = log 10 (. 10 3 ) =.66 14. Ethylamine, C H 5 NH, is a weak base in aqueous solution. The ph of a 1.0 M aqueous ethylamine solution is measured to be 1.3. Calculate the pk b of ethylamine. C H 5 NH (aq) + H O (l) C H 5 NH + 3(aq) + OH (aq) poh = 14.00 ph = 1.68 [OH ] = 10 1.68 =.1 10 [C H 5 NH ] [C H 5 NH + 3 ] [OH ] initial 1.0 M 0 M 0 M change.1 10 M.1 10 M.1 10 M equilibrium 0.98 M.1 10 M.1 10 M 7
K b = [C H 5 NH + 3 ][OH ] [C H 5 NH ] = (.1 10 ) 0.98 = 4.5 10 4 pk b = log 10 (4.5 10 4 ) = 3.35 15. The two values for the pk a of hydrosulfuric acid according to the reactions H S (aq) + H O (l) H 3 O + (aq) + HS (aq) HS (aq) + H O (l) H 3 O + (aq) + S (aq) are respectively 7.00 and 19.0. Calculate the concentrations of all species present in a 0.0100 and 0.00100 M aqueous solution of hydrosulfuric acid. [H 3 O + ][HS ] = 1.00 10 7 [H S] (a) 0.0100 M [H 3 O + ][S ] [HS ] = 1.00 10 19 [H S] [HS ] [H 3 O + ] initial 0.0100 M 0 M 0 M equilibrium (0.0100 - x) M x M x M 0.0100 x = 1.00 10 7 = 0.0100 = 1.00 10 9 x = [H 3 O + ] = [HS ] = 1.00 10 9 = 3.16 10 5 M [H S] = 0.0100 M [HS ] [S ] [H 3 O + ] initial 3.16 10 5 M 0 M 3.16 10 5 M equilibrium (3.16 10 5 - x) M x M (3.16 10 5 + x) M x(3.16 10 5 + x) 3.16 10 5 x x = [S ] = 1.00 10 19 (b) 0.00100 M. In this case the approximations used in solving the previous case are a bit large, and full solution to the quadratic equations are needed. 8
[H S] [HS ] [H 3 O + ] initial 0.00100 M 0 M 0 M equilibrium (0.00100 - x) M x M x M 0.00100 x = 1.00 10 7 + 1.00 10 7 x 1.00 10 10 = 0 x = 1.00 10 7 ± (1.00 10 7 ) + 4 1.00 10 10 Keeping only the positive solution x = [H 3 O + ] = [HS ] = 9.95 10 6 [H S] = 0.00100 M [HS ] [S ] [H 3 O + ] initial 9.95 10 6 M 0 M 9.95 10 6 M equilibrium (9.95 10 6 - x) M x M (9.95 10 6 + x) M x(9.95 10 6 + x) 9.95 10 6 x x = [S ] = 1.00 10 19 16. When phthalic acid combines with water, two equilibrium reactions occur H C 8 H 4 O 4(aq) + H O (l) H 3 O + (aq) + HC 8H 4 O 4(aq) K a1 = 1.1 10 3 and HC 8 H 4 O 4(aq) + H O (l) H 3 O + (aq) + C 8H 4 O 4(aq) K a = 3.9 10 6 Calculate the ph and concentrations of all species present in a 0.010 M phthalic acid solution. K a1 = [H 3O + ][HC 8 H 4 O 4 ] = 1.1 10 3 [H C 8 H 4 O 4 ] [H C 8 H 4 O 4 ] [HC 8 H 4 O 4 ] [H 3 O + ] initial 0.010 M 0 M 0 M equilibrium (0.010 - x) M x M x M 9
0.010 x = 1.1 10 3 + 1.1 10 3 x 1.1 10 5 = 0 x = 1.1 10 3 ± (1.1 10 3 ) + 4 1.1 10 5 x = [H 3 O + ] = [HC 8 H 4 O 4 ] =.8 10 3 M K a = [H 3O + ][C 8 H 4 O 4 ] [HC 8 H 4 O 4 ] = 3.9 10 6 [HC 8 H 4 O 4 ] [C 8 H 4 O 4 ] [H 3 O + ] initial.8 10 3 M 0 M.8 10 3 M equilibrium (.8 10 3 x) M x M (.8 10 3 + x) M x(.8 10 3 + x).8 10 3 x x = 3.9 10 6 = [C 8 H 4 O 4 ] ph = log 10.8 10 3 =.55 17. Germanic acid (H GeO 3 ) is a diprotic acid with pk a s for the two protons in aqueous solution given by pk a,1 = 9.01 and pk a, = 1.30. Calculate the equilibrium concentrations of HGeO 3(aq), GeO 3(aq) and H 3O + (aq) and the ph for a 1.50 M aqueous solution of germanic acid. Approximations work for this problem. K a1 = 10 9.01 = 9.8 10 10 K a = 10 1.30 = 5.0 10 13 H GeO 3(aq) + H O (l) HGeO 3(aq) + H 3O + (aq) [H GeO 3 ] [HGeO 3 ] [H 3 O + ] initial 1.50 M 0 M 0 M change -y y y equilibrium (1.50 -y) M y M y M 9.8 10 10 = [HGeO 3 ][H 3 O + ] [H GeO 3 ] = y 1.50 y y 1.50 y = [H 3 O + ] = [HGeO 3 ] = 3.8 10 5 M ph = log 10 (3.8 10 5 ) = 4.4 HGeO 3(aq) + H O (l) GeO 3(aq) + H 3O + (aq) 10
[HGeO 3 ] [GeO 3 ] [H 3 O + ] initial 3.8 10 5 M 0 M 3.8 10 5 M change -y y y equilibrium (3.8 10 5 -y) M y M (3.8 10 5 + y) M 5.0 10 13 = [GeO 3 ][H 3 O + ] [HGeO 3 ] = y(3.8 10 5 + y) 3.8 10 5 y y [GeO 3 ] = 5.0 10 13 M 18. Selenic acid (H SeO 4 ) is a diprotic acid. Like sulfuric acid, the first proton is essentially 100% ionized in aqueous solution. The second proton dissociates according to the reaction HSeO 4(aq) + H O (l) SeO 4(aq) + H 3O + (aq) with a pk a = 1.9. Consider a 0.10 M solution of selenic acid. Calculate the equilibrium concentrations of HSeO 4 and SeO 4, and the ph of the solution at equilibrium. Approximations work for the dissociation equilibrium. [HSeO 4 ] [SeO 4 ] [H 3 O + ] initial 0.10 M 0 M 0.10 M change -y y y equilibrium (0.10 -y) M y M (0.10+y) M K a = 10 1.9 = 1. 10 = [SeO 4 ][H 3 O + ] [HSeO 4 ] = y(0.10 + y) 0.10 y y [SeO 4 ] = 1. 10 [HSeO 4 ] = 0.10 1. 10 = 0.10 (to two significant figures) [H 3 O + ] = 0.10 + 1. 10 = 0.11 M ph = log 10 (0.11) = 0.96 11