Question

Part A Find the pH of a 0.290M HF solution.

Part B Find the percent dissociation of a 0.290M HF solution.

Part C Find the pH of each of the following solutions of mixtures of acids.

0.100M in HBr and 0.150M in HCHO₂

0.150M in HNO₂ and 8.5x10^-2M in HNO₃

Answer 1

Part A - [HF] = 0.290M

HF   $\rightleftharpoons$   H⁺   + F^-

Ka of HF = 3.5 x 10^-4

Ka = [H+][F-] / [HF]

3.5 x 10^-4 = [H+]² / 0.290

[H+]² = 1.015x10^-4

[H+] = 0.010 M

[H⁺] = 0.010 M

pH = - log[H⁺]

= - log 0.010

= 2

Part B -

[H+] = 0.010M

[HF] = 0.290M

% dissociation = 0.010 / 0.290 * 100

= 3.44 %