**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}

X^{2} + 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**