Question

A $40ml$. solution of a weak base, $BOH$ is titrated with $0.1NHCl$ solution. The $pH$ of the solution is found to be $10.04$ and $9.14$ after the addition of $5.0mL$ and $20.0mL$ of the acid respectively. Find out the dissociation constant of the base.

Answer 1

Let $40mL$ of base contain a $mmol$ of $BOH$

When $pH$ is $10.04,pOH=3.96$ and
when $pH$ is $9.14,pOH$ is $4.86$.
Therefore, $3.96=p{K}_b+\log \frac{0.50}{x-0.5}\dots$ (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
$3.96=p{K}_b+\log \frac{0.5}{3}$
$K_b=1.8\times 10^{-5}$