Question

Cyanic acid (HCN) is a weak acid with a K_a value of 6.2 ✕ 10^-10. Calculate the pH of a 1.26 M solution of sodium cyanide - NaCN(aq).

Answer 1

Since this is a salt solution (aq. NaCN) which could be considered to have formed from the neutralization of a strong base (NaOH) and a weak acid (HCN), the Na⁺ will have no effect on the pH of the solution while the CN^- ion will undergo hydrolysis:

CN^- + H₂O --> HCN + OH^- (eq.1)

Kb = [HA} [OH^-] / [A^-] (eq.2)

where [A^-] represents the concentration of CN^- (1.26 M)

Kb = Kw / Ka = 1 x 10^-14 / 6.2 x 10^-10  = 1.6 x 10^-5

Based on equation (eq.2), the quantities of HCN and OH^- produced must be the same and therefore [HCN] = [OH^-]. We will set this equal to x.

Plugging into the original equation 2 yields:

1.6 x 10^-5 = x² / 1.26 M

Solving for x yields 4.5 x 10^-3 which is equal to the [OH^-]

The pOH = -log[OH^-]

The pOH then is equal to -log (4.5 x 10^-3) = 2.3

The pH of the solution would be 14 - 2.3 = 11.7

pH = 11.7.