What is the pH of a 0.0740 M solution of hydrocyanic acid, HCN (Ka = 4.9 × 10⁻¹⁰)?

Answer 1

pH = -log[H⁺] so we need to find the [H⁺]

HCN ==> H⁺ + CN^-

Ka = 4.9x10^-10 = [H⁺][CN^-]/[HCN]

4.9x10^-10 = (x)(x)/(0.0740 - x) and if we assume x is small relative to 0.074 we can ignore it (see below)

4.9x10-10 = x2/0.0740

x² = 3.6x10^-11

x = 6x10^-6 and this is very small relative to 0.074 so our assumption was correct.

[H+] = 6x10^-6

pH = -log 6x10^-6

pH = 5.2