Please ANswer?

Question

Hydrocyanic acid is a very weak acid.

a) Write an equilibrium expression for the ionization of 0.10 mol/L HCN(aq). Include the percent ionization at SATP.

b) Calculate [H+(aq)] and the pH of a 0.10 mol/L solution of HCN (aq) at SATP.

Answer 1

Ka = 4 x 10^-10
HCN ===> H+ + CN-
Let [H+] = x

Ka = [H+][CN-]/[HCN] = (x)(x)/(0.10) = 4 x 10^-10

x^2/(0.10) = 4 x 10^-10

x^2 = 4 x 10^-8

x = 2.41 x 10^-4 = [H+]