What is the pH of 0.0050 HF (Ka = 6.8 x 10-4) ?

2.73

11.70

11.27

2.30

The correct answer is 2.73.

HF is a weak acid which partially dissociates to release H + and F-

HF → H⁺ + F⁻

Initial 0.0050 0 0

Change - x + x + x

Equilibrium 0.0050-x + x + x

Solve by using the equilibrium expression: = [H⁺] [F⁻] / [HF]

6.8 x 10⁻⁴ = x. x / 0.0050 - x

6.8 x 10⁻⁴ = x² / 0.0050

x² = 6.8 x 10⁻⁴ x 0.0050

x² = 3.4 x 10⁻⁶

x = 3.4 x 10⁻⁶

[H⁺] = 1.84 x 10⁻³

pH = - log [H⁺] = - log (1.84 x 10⁻³)

pH = 2.73