Calculate the pOH of a 0.0987 M aqueous sodium cyanide solution at 25.0 °C. Kb for CN- is 4.9 x 10-10.

Question

A. 9.33

B. 5.16

C. 8.84

D. 1.01

E. 10.00

Answer 1

Look at the hydrolysis of CN^-

CN^- + H₂O ==> HCN + OH^- ... note how CN^- is acting as a base so we need to use Kb for CN^-

Kb = [HCN][OH-] / [HCN]

4.9x10^-10 = (x)(x) / 0.0987 - x (assume x is small and ignore so don't use quadratic)

x² = 4.84x10^-11

x = 6.96x10^-6 = [OH-]

pOH = - log [OH-]

pOH = 5.16 (B)