If Ka HCN=2.9 X 10^-10 and Kb NH3 = 1.8 x 10^-5, calculate the pH of a .600M NH4CN salt solution.

Answer 1

Since NH4CN is clearly the salt of a weak acid (HCN) and a weak base (NH3), we can use the following:

pH = 1/2 (pKa + pKaB) where pKaB is the acid dissociation constant of the conjugate acid of the base.

So, let us first find pKa and pKaB:

Ka = 2.9x10^-10; pKa = - log 2.9x10^-10 = 9.54

Kb - 1.8x10^-5; KaB = 1x10^-14/1.8x10^-5 = 5.56x10^-10; pKaB = 10.3

pH = 1/2 (9.54 + 10.3) = 1/2 (19.84)

pH = 9.9

(NOTE: This is an approximation and is calculated from an equation using assumptions and approximations so as to avoid using a 4th degree polynomial).