Question

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.

Answer 1

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

b.5.4x10^-3%

Step-by-step 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%