Equilibrium

Ans4

If we use the relation, pH = – log [H3O+], we get pH equal to 8. But this is not correct because an acidic solution cannot have pH greater than 7. It may be noted that in very dilute acidic solution, when H+ concentrations from acid and water are comparable, the concentration of H+ from water cannot be neglected.
Therefore,
[H+] total = [H+] acid + [H+] water
Since HCl is a strong acid and is completely ionized
[H+] HCl = 1.0 x 108
The concentration of H+ from ionization is equal to the [OH–] from water,
[H+] H2O = [OH–] H2O
= x (say)
[H+] total = 1.0 x 108 + x
But
[H+] [OH–] = 1.0 x 1014
(1.0 x 108 + x) (x) = 1.0 x 1014
X2 + 108 x – 1014 = 0
Solving for x, we get x = 9.5 x 108
Therefore,
[H+] = 1.0 x 108 + 9.5 x 108
= 10.5 x 108
= 1.05 x 107
pH = – log [H+] = – log (1.05 x 107) = 6.98
