since it is very dilute acidic solution, so H+ concentrations from acid and water are comparable, nd 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 10-8
The concentration of H+ from ionization is equal to the [OH-] from water,
[H+] H2O = [OH-] H2O
= x (assume)
[H+] total = 1.0 x 10-8 + x
But
[H+] [OH-] = 1.0 x 10-14
(1.0 x 10-8 + x) (x) = 1.0 x 10-14
X2 + 10-8 x – 10-14 = 0
Solving for x, we get x = 9.5 x 10-8
Therefore,
[H+] = 1.0 x 10-8 + 9.5 x 10-8
= 10.5 x 10-8
= 1.05 x 10-7
pH = – log [H+]
= – log (1.05 x 10-7)
= 6.98