Explanation

(c)
$pH = - \log{\lbrack H^{+}\rbrack}$
pH=14$-$pOH
Calculate pH of all the solution to find which will have maximum pH.
(A)$\frac{M}{10}\ HCl$
$\therefore$ $\left\lbrack H^{+} \right\rbrack = 10^{- 1}$
$pH = - log\ \lbrack H^{+}\rbrack$
$= - \log\left\lbrack 10^{- 1} \right\rbrack$
$= 1$
(B)$\frac{M}{100}\ HCl$
$\therefore$ $\left\lbrack H^{+} \right\rbrack = 10^{- 2}$
$pH = - \log{\lbrack H^{+}\rbrack}$
$= - \log{\lbrack 10^{- 2}\rbrack}$
$= 2$
(C )$\frac{M}{10}\ NaOH$
$\therefore$ $\left\lbrack OH^{-} \right\rbrack = 10^{- 1}$
$pH = 14 - pOH$
$= 14\lbrack - \log{10^{- 1}\rbrack}$
$= 14 - 1$
$= 13$
(D )$\frac{M}{100}\ NaOH$
$\therefore$ $\left\lbrack OH^{-} \right\rbrack = 10^{- 2}$
$pH = 14 - pOH$
$= 14 - \lbrack\log{10^{- 2}\rbrack}$
$= 14 - 2 = 12$
$\therefore$ $\frac{M}{100}\ NaOH$ solution has highest pH.