Explanation

(d)
Given, concentration of $NaOH = 10^{- 10}M$
$NaOH\ \ \ \ \ \rightarrow Na^{+}\ \ \ \ \ \ + \ \ \ OH^{-}$
$10^{- 10}$ $10^{- 10}$ $10^{- 10}$
$\therefore\left\lbrack OH^{-} \right\rbrack from\ NaOH = 10^{- 10}$
$\therefore\ \ \ \ \ \ \ \ \ \ Total\ \left\lbrack OH^{-} \right\rbrack = 10^{- 7} + 10^{- 10}$
$= 10^{- 7}(1 + 0.001)$
$\ = 10^{- 7}\left( \frac{1001}{1000} \right)$
$= 10^{- 10} \times 1001$
$pOH = - \log{\lbrack OH^{-}\rbrack}$
$\ = - \log{(1001 \times 10^{- 10})}$
$= - 3.004 + 10$
$= 6.9996$
$\because\ \ \ \ \ \ \ \ \ \ \ \ \ pH + pOH = 14$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ pH = 14 - 6.9996$
=7.0004=7