Question

HCN is a weak acid (Ka = 6.20 × 10–10) and so the salt, KCN, acts as a weak base. What is the pH of a solution that is 0.0720 M in KCN at 25 °C?

Answer 1

First we write the hydrolysis equation of KCN as follows:

KCN + H2O → HCN + KOH

CN- + H2O → HCN + OH- [[net ionic]]
so
Kb = [HCN] [OH-] / [CN-]

We know that Ka*Kb = Kw

Then Kb = 1.0*10^-14 / 6.20 × 10^–10

= 1.61*10 ^-5

from the first equation above
if we let [OH-] = X then [HCN] = X also
and {KCN] = 0.0720 - X
so
1.613*10^-5 = X * X / (0.0720 - X)

1.16 *10^-6 - 1.61*10^-5 X = X^2

X^2 + 1.61 *10^-5 X - 1.16*10^ -6 = 0

X = 0.001069 (the positive root)
so
pOH = -log(0.001069) = 2.97
then
pH = 14.00 - 2.97 = 11.03