Question

A $40 \,ml$. solution of a weak base, $BOH$ is titrated with $0. 1\,N HCl$ solution. The $pH$ of the solution is found to be $10.04$ and $9.14$ after the addition of $5.0 \,mL $ and $20.0 \,mL$ of the acid respectively. Find out the dissociation constant of the base.

Answer 1

Let $40 \,mL$ of base contain a $m\,mol$ of $BOH$

When $pH$ is $10.04, pOH =3.96$ and
when $pH$ is $9.14, pOH$ is $4.86$.
Therefore, $3.96=p K_{b}+\log \frac{0.50}{x-0.5} \ldots$ (i)
$3.96=p K_{b}+\log \frac{2.0}{x-2}\,\,\,\,....$(ii)
Subtracting Eq. (i) from Eq. (ii) gives
$0.92=\log \left(\frac{x-0.5}{0.5} \times \frac{2.0}{x-2}\right) $
$\Rightarrow 28=\frac{4(x-0.5)}{x-2} $
$\Rightarrow a =3.5,$
substituting in equation (i) gives
$3.96=p K_{b}+\log \frac{0.5}{3} $
$K_{b}=1.8 \times 10^{-5}$