Question

H2SO4 is a strong acid because the first proton ionizes 100%. The Ka of the second proton is 1.1x10-2. What would be the pH of a solution that is 0.100 M H2SO4? Account for the ionization of both protons.

A. 0.955

B. 2.05

C. 0.963

D. 1.00

Answer 1

The pH will be due to both the dissociation of strong acid

1) Due to first dissociation (almost 100%) concentration of hydrogen ion = [H+ ]= [H2SO4 = 0.100 M

H2SO4 ----> HSO₄^- + H⁺

2) second dissociation

                                HSO₄^- ----->SO₄^-2 + H⁺

Initial                        0.1                0           0.1

Change                   -x                  +x          +x

Equilibrium           0.1-x                   x            0.1+ x

Ka = [H+][SO4^-2] / [HSO4^-]

0.011 = x (0.1+x) / 0.1-x

0.0011 - 0.011x = 0.1x + x²

x² + 0.111x - 0.0011 = 0

On solving for x

x = 0.00916

Total [H+] = 0.1 + 0.00916 = 0.10916

pH = -log[H+] = 0.962