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Question

1(a) Fluoroacetic acid, HC2H2FO2, occurs in Gifblaar, one of the most poisonous of all plants: HC2H2FO2(aq) + H2O (l) ------> H3O+(aq) + C2H2FO2–(aq). A 0.50 M solution of fluoroacetic acid has a pH of 1.46. Determine Ka. 

(b)i Determine Kb for the fluoroacetate ion. 

(ii) Determine the [OH–]eq in a 0.50 M solution of the fluoroacetate ion. 

(iii) Determine the pH of the 0.50 M solution of fluoroacetate ion.

Answer 1

(a) Ka = [H3O+][C2H2FO2–]/[HC2H2FO2]

Find [H3O+] from pH: pH = -log [H3O+] therefore [H3O+] = 1x10^-1.46 = 3.47x10^-2. This is also the concentration of C2H2FO2-. Thus, Ka = (1.46x10^-2)^2/0.50 = 4.26x10^-4

(b) Kb x Ka = Kw = 1x10^-14

Kb = 1x10^-14/4.26x10^-4 = 2.35x10^-11

(ii) C2H2FO2^- + H2O ==> HC2H2FO2 + OH^-

Kb = [ HC2H2FO2][OH^-]/[ C2H2FO2^- ] = 2.35x10^-11 = (x)(x)/0.50 - x

Ignoring x in denominator to avoid quadratic, we get: x^2 = 1.175x10^-11

x = 3.43x10^-6 = [OH^-]

(iii) pOH = -log [OH-] = 5.46

pH = 14 - pOH = 8.54