(d)
From Henderson equation
$pOH = pK_{b} + \log\frac{\lbrack salt\rbrack}{\lbrack base\rbrack}$
$pH + pOH = 14$
$pOH = 5.0 + \log\frac{\lbrack 1.0\rbrack}{\lbrack 0.1\rbrack} = 5 + \log{10 = 5 + 1}$
$pOH = 6$
$pH + pOH = 14$
$pH + 6 = 14$
$pH = 14 - 6 = 8$