(d)
For basic buffer,
$pOH = pK_{b} + \log\frac{\lbrack salt\rbrack}{\lbrack base\rbrack}$
Given, [salt]=[base]=1
$K_{b} = 2 \times 10^{- 5}$
$\therefore$ $pOH = pK_{b} = - \log{(2 \times 10^{- 5})}$
$= 5 - \log 2$
$\ = 5 - 0.30 = 4.70$
$pH = 14 - pOH$
$= 14 - 4.70 = 9.30$