Question

$BOH + H^{+} \rightleftharpoons B^{+} + H_{2}O$

$a\text{ mole}b\text{ mole}0$

$\frac{1}{5}\text{th run}a - \frac{a}{5}0\frac{a}{5}$

Now, for 1/5th reaction, $P^{OH} = P^{K_{b}} + \log\frac{\left\lbrack B^{-} \right\rbrack}{\lbrack BOH\rbrack}$

Or, $(14 - 9) = P^{K_{b}} + \log\frac{a/5}{4a/5}$

$\therefore P^{K_{b}} = 5.6 \Rightarrow K_{b} = 2.5 \times 10^{- 6}$

At equivalent point: $P^{H} \Rightarrow - \frac{1}{2}\left( P^{K_{b}} + \log C \right)$

Or, $4.5 \Rightarrow - \frac{1}{2}\left( 5.6 + \log C \right) \Rightarrow C = 0.25\text{ M}$

Now, $n_{HCl\text{ used}} = n_{B^{+}\text{ formed}}$

Or, $\frac{V \times 0.5}{1000} = \frac{(100 + V) \times 0.25}{1000} \Rightarrow V_{HCl} = 100\text{ ml}$

Finally, $n_{BOH\text{ takes}} = n_{HCl\text{ used for equivalent point}}$

Or, $\frac{w}{45} = \frac{100 \times 0.5}{1000} \Rightarrow w = 2.25\text{ gm}$

∴ Percentage purity of base = $\frac{2.25}{2.5} \times 100 = 90\%$