Question

(a) Calculate the percent ionization of 0.00710 M butanoic acid (Ka = 1.5e-05). % ionization = % (b) Calculate the percent ionization of 0.00710 M butanoic acid in a solution containing 0.0210 M sodium butanoate. % ionization = %

Answer 1

a)

butanoic acid ---------------> butanoate + H+

0.00710                                  0             0

0.00710 - x                             x              x

Ka = x^2 / 0.00710 - x

1.5 x 10^-5 = x^2 / 0.00710 - x

x = 3.19 x 10^-4

[H+] = 3.19 x 10^-4 M

% ionization = (3.19 x 10^-4 / 0.00710 ) x 100

% ionization = 4.49 %

b)

HA -------------> A- + H+

Ka = [A-][H+] / [HA]

1.5 x 10^-5 = [0.0210][H+] / 0.00710

[H+] = 5.07 x 10^-6 M

% ionization = (5.07 x 10^-6 / 0.0071 ) x 100

% ionization = 0.0714 %