Chemistry 192 Problem Set 3 Spring, 2018 Solutions

Transcription

1 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) M HCl (aq) [H 3 O + ] = ph = log 10 ( ) = poh = ph =

2 (b) M NaOH (aq) [OH ] = poh = log 10 ( ) = ph = poh = 1.66 (c) 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 Using the ion product of water or Using the quadratic formula x = ± [H 3 O + ][OH ] = x(x ) = x = 0 ( ) Only the positive solution for x is physical and = , poh = log 10 ( ) = 7.04 ph = poh = 6.96 (d) 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 ( ) = 7.04 poh = ph = Problem 14, page 78, textbook n OH = (0.15 L)(0.606 mol L 1 ) = mol [OH mol ] = = M 15.0 L poh = log 10 ( ) =.97 ph = poh = 11.70

3 4. Problem 19, page 78, textbook Before mixing n H + = ( L)( mol L 1 ) = mol After mixing n OH = ( L)( mol L 1 ) = mol n OH = mol mol = mol [OH ] = mol = M 0.15 L poh = log 10 ( ) = ph = poh = Problem 0, Page 78, textbook [H 3 O + ] = 10.1 = M After mixing n H3 O + = ( mol L 1 )( L) = mol [OH ] = 10 ( ) = M n OH = ( L)( mol L 1 ) = mol n OH = mol mol = mol [OH ] = mol = M L poh = log 10 ( ) = 1.71 ph = poh = Calculate the volume of a M aqueous KOH solution that must be added to 100. ml of a 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) = mol Let V = volume of the KOH solution to be added. Then the final H 3 O + concentration is n H3 O + = ( mol (0.100 mol L 1 )V = 10 mol L 1 V final V L (10 mol L 1 )V mol = mol (0.100 mol L 1 )V (0.110 mol L 1 )V = mol V = L = 81.8 ml 3

4 7. Given that sodium hydroxide is a strong base, calculate the ph of an aqueous M sodium hydroxide solution. Approximations do not work for this problem. [H 3 O + ][OH ] = Let y = [H 3 O + ] [OH ] = y y(y ) = y y = 0 y = ± [( ) ] 1/ Keeping the positive solution y = [H 3 O + ] = M ph = log 10 ( ) = 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 = 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 = = x = = % ph = log 10 ( ) =

5 9. The pk a of phenol (C 6 H 5 OH) when acting as a weak acid in aqueous solution is Calculate the ph of a.0 M phenol solution. Approximations work for this system. K a = = 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 = y = = M = [H 3 O + ] ph = log 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 + ] = = = [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 = x = ( ) = pk a = log 10 ( ) = 4.0 5

6 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 M formic acid solution is found to be Calculate K a and pk a for formic acid. [H 3 O + ] = = [HCOOH] [H 3 O + ] [HCOO ] initial M 0 M 0 M change M M M equilibrium M M M K a = [HCOO ][H 3 O + ] [HCOOH] = ( ) = pk a = log 10 K a = log 10 ( ) = 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 = ph = 1.68 [OH ] = =.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 ) = pk b = log 10 ( ) = 3.3 6

7 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 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 = = Initially ( ) mol (0.5 g) ( ) g [HF] = = M 1. L [HF] [H 3 O + ] [F ] initial M 0 M 0 M change -y M y M y M equilibrium ( y) M y M y M = [F ][H 3 O + ] [HF] = y y Ignoring the negative solution y y = 0 y = [( ) + 4( )] 1/ = [H 3 O + ] = M ph = log 10 ( ) = 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 = ph = 1.68 [OH ] = =.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

8 K b = [C H 5 NH + 3 ][OH ] [C H 5 NH ] = (.1 10 ) 0.98 = pk b = log 10 ( ) = 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 Calculate the concentrations of all species present in a and M aqueous solution of hydrosulfuric acid. [H 3 O + ][HS ] = [H S] (a) M [H 3 O + ][S ] [HS ] = [H S] [HS ] [H 3 O + ] initial M 0 M 0 M equilibrium ( x) M x M x M x = = = x = [H 3 O + ] = [HS ] = = M [H S] = M [HS ] [S ] [H 3 O + ] initial M 0 M M equilibrium ( x) M x M ( x) M x( x) x x = [S ] = (b) 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

9 [H S] [HS ] [H 3 O + ] initial M 0 M 0 M equilibrium ( x) M x M x M x = x = 0 x = ± ( ) Keeping only the positive solution x = [H 3 O + ] = [HS ] = [H S] = M [HS ] [S ] [H 3 O + ] initial M 0 M M equilibrium ( x) M x M ( x) M x( x) x x = [S ] = 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 = and HC 8 H 4 O 4(aq) + H O (l) H 3 O + (aq) + C 8H 4 O 4(aq) K a = Calculate the ph and concentrations of all species present in a M phthalic acid solution. K a1 = [H 3O + ][HC 8 H 4 O 4 ] = [H C 8 H 4 O 4 ] [H C 8 H 4 O 4 ] [HC 8 H 4 O 4 ] [H 3 O + ] initial M 0 M 0 M equilibrium ( x) M x M x M 9

10 0.010 x = x = 0 x = ± ( ) x = [H 3 O + ] = [HC 8 H 4 O 4 ] = M K a = [H 3O + ][C 8 H 4 O 4 ] [HC 8 H 4 O 4 ] = [HC 8 H 4 O 4 ] [C 8 H 4 O 4 ] [H 3 O + ] initial M 0 M M equilibrium ( x) M x M ( x) M x( x) x x = = [C 8 H 4 O 4 ] ph = log = 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, = 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 = = K a = = 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 = [HGeO 3 ][H 3 O + ] [H GeO 3 ] = y 1.50 y y 1.50 y = [H 3 O + ] = [HGeO 3 ] = M ph = log 10 ( ) = 4.4 HGeO 3(aq) + H O (l) GeO 3(aq) + H 3O + (aq) 10

11 [HGeO 3 ] [GeO 3 ] [H 3 O + ] initial M 0 M M change -y y y equilibrium ( y) M y M ( y) M = [GeO 3 ][H 3 O + ] [HGeO 3 ] = y( y) y y [GeO 3 ] = 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 = = = [SeO 4 ][H 3 O + ] [HSeO 4 ] = y( y) 0.10 y y [SeO 4 ] = [HSeO 4 ] = = 0.10 (to two significant figures) [H 3 O + ] = = 0.11 M ph = log 10 (0.11) =