Chemical and Ionic Equilibrium 2 Question 67
67. A $40 \mathrm{~mL}$ solution of a weak base, $B \mathrm{OH}$ is titrated with $0.1 \mathrm{~N}$ $\mathrm{HCl}$ solution. The $\mathrm{pH}$ of the solution is found to be 10.04 and 9.14 after the addition of $5.0 \mathrm{~mL}$ and $20.0 \mathrm{~mL}$ of the acid respectively. Find out the dissociation constant of the base.
$(1991,6 M)$
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Solution:
- Let $40 \mathrm{~mL}$ of base contain $x \mathrm{mmol}$ of $\mathrm{BOH}$.
$$ \begin{array}{ccc} \mathrm{BOH}+\mathrm{HCl} & \longrightarrow \mathrm{BCl}+\mathrm{H}_{2} \mathrm{O} \ x-0.5 & 0.5 & \text { When } 5 \mathrm{~mL} \text { acid is added } \ x-2 & 2.0 & \text { When } 20 \mathrm{~mL} \text { of acid is added } \end{array} $$
When $\mathrm{pH}$ is $10.04, \mathrm{pOH}=3.96$ and when $\mathrm{pH}$ is $9.14, \mathrm{pOH}$ is 4.86. Therefore,
$$ \begin{aligned} & 3.96=\mathrm{p} K_{b}+\log \frac{0.50}{x-0.5} \ & 3.96=\mathrm{p} K_{b}+\log \frac{2.0}{x-2} \end{aligned} $$
Subtracting Eq. (i) from Eq. (ii) gives
$$ \begin{aligned}
