Question

An aqueous solution contains a mixture of .175 M HF (K_a = 7.2 x 10^-4) and .250 M HNO₂ (K_a = 4.0 x 10^-4). Calculate the pH of this solution (express your pH to the hundredth position).

Answer 1

-log K_a = pK_a

For HF, pK_a = - log (7.2 x 10^-4) = 3.142

For HNO₂, pK_a = - log (4 x 10^-4) = 3.397

HF is the deciding reagent for pH.

             HF       ---->      H⁺     +    F^-

Initial         0.175 M                 0                0

Change      -x                      +x               +x

Equilibrium        0.175 - x                 x                  x

K_a = [H⁺][F^-] / [HF]

7.2 x 10^-4 = x² / (0.175 - x)

0.00072 (0.175 - x) = x²

x = 0.0112 M

[H⁺] = 0.0112 M

pH = -log [H⁺] = 1.95