What is the pH of a 0.810 M solution of NaCN (Ka of HCN is 4.9 × 10⁻¹⁰)?

Question

What is the pH of a 0.810 M solution of NaCN (Ka of HCN is 4.9 × 10⁻¹⁰)?

Answer 1

When given a salt, and asked to find the pH, you must look at the hydrolysis of the salt. Since NaCN is the salt formed from a weak acid (HCN) and a strong base (NaOH), the pH of the salt solution will be basic (>7).

Hydrolysis:

CN^- + H2O ==> HCN + OH^-

Here, CN^- is acting as a base, so we use the Kb for CN^-

Kb = 1x10^-14/Ka = 1x10^-14 / 4.9x10^-9 = 2.04x10^-6

Kb = 2.04x10^-6 = [HCN][OH^-] / [CN^-]

2.04x10^-6 = (x)(x) / 0.810

1.65x10^-6 = x²

x = 1.29x10^-3 M = [OH^-]

pOH = -log [OH^-] = 2.89

pH = 14 - pOH

pH = 14 - 2.89

pH = 11.1