A 0.250 M solution of HCN has a pH = 4.87.

a) Determine the [H3O+] in the solution.

b) Calculate the percent ionization of HCN in this solution.

a. [H+] = 1.35x10^-5 M

b. 5.4x10^-3%

Explanation:

a) pH = - log [H+]

[H+] = 10^-pH [H+] = 10^-4.87 [H+] = 1.35x10^-5 M

b) HCN - > H + CN at equilibrium [H+]=[CN-]

0.250 M - > 100%

1.35x10^-5 M - >x

x = (1.35x10^-5 M * 100) / 0.250M

x=5.4x10^-3%