Explanation

(c)
From $H_{2}O,\ \left\lbrack H^{+} \right\rbrack = 1 \times 10^{- 7}M$
From $HCl\left\lbrack H^{+} \right\rbrack = 1 \times 10^{- 8}M$
Total $\left\lbrack H^{+} \right\rbrack = \left( 1 \times 10^{- 7} + 1 \times 10^{- 8} \right)M$
$\ \ = \left( 1 \times 10^{- 7} + 0.1 \times 10^{- 7} \right)M$
$= 1.1 \times 10^{- 7}M$
$pH = \ - \log{(1.1 \times 10^{- 7})} = 6.9586$