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
                              Â
HSO4- ----->SO4-2 +
H+
Initial                      Â
0.1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â
0.1
Change                 Â
-x                Â
+x         +x
Equilibrium         Â
0.1-x                 Â
x          Â
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
