Question

What is [H₃O+] and pH in a 0.10 M solution of HCN at 25

Answer 1

We have , 0.10M HCN solution at 25^oc & Ka for HCN = 4.0 * 10^-10

                               HCN + H2O -----------> H3O⁺ + CN^-

Intial concentration     0.1 M                         0 M       0 M

Change in concentration   -x                           +X         +X

Equilibrium concentration    (0.1-x)                   X             X

Ka = [H3O⁺][CN^-] / [HCN]

4.0 * 10^-10 = (X)(X) / (0.1-X)

4.0 * 10^-10 = (X)(X) / (0.1)

        X²     = (0.1) * (4.0 * 10^-10)

[H3O⁺] = X = 6.3 * 10^-6 M

[H3O⁺] = 6.3 * 10^-6 M

pH = -log[H3O⁺]

      = -log (6.3 * 10-6)

pH = 5.2