The Ka of hydrofluoric acid (HF) at 25.0 °C is 6.8 × 10-4. What is the pH of a 0.25 M aqueous solution of HF?

Question

A. 0.60

B. 1.88

C. 1.28

D. 3.64

E. 4.34

Answer 1

HF <==> H+ + F-

Ka = 6.8x10^-4 = [H+][F-] / [HF]

6.8x10^-4 = (x)(x) / 0.25 - x (assume x is small relative to 0.25 and ignore it to avoid using quadratic)

6.8x10^-4 = x² / 0.25

x² = 1.7x10^-4

x = 1.3x10^-2 M (this is ~ 5.2% of 0.25 so our assumption above was not valid. We must use quadratic)

6.8x10^-4 = x2 /0.25 - x

1.7x10^-4 - 6.8x10^-4 x = x²

Use quadratic to find x = 0.0127 M = [H+]

pH = -log H+ = -log 0.0127

pH = 1.90 so answer is probably 1.88(B)