Explanation

(c)
If the concentration of acid is less than$10^{- 7}$, then the concentration of $H^{+}$ present in water is not neglected.
Total $\left\lbrack H^{+} \right\rbrack = \left\lbrack H^{+} \right\rbrack from\ HCl + \left\lbrack H^{+} \right\rbrack from\ H_{2}O$
$= 10^{- 8} + 10^{- 7}$
$= 10^{- 7}(1.1)$
$= 1.1 \times 10^{- 7}mol/L$
$pH = \ - \log{\lbrack H^{+}\rbrack}$
$= - \log\left( 1.1 \times 10^{- 7} \right)$
$= 6.96$