Question

What is the ph of a 0.023 m solution of hcn (ka = 4.9 x 10-10)?

Answer 1

The ionic equation is simply:

HCN --> H+ + CN-

Therefore the formula for Ka is:

Ka = [H+] [CN-] / [HCN]
[H+] = [CN-] = x
4.9 x 10^-10 = x^2 / 0.023
x^2 = 1.127 x 10^-11
x = 3.357 x 10^-6 = [H+] = [CN-]

pH = -log [H+]
pH = -log 3.357 x 10^-6
pH = 5.47