Chemical and Ionic Equilibrium - Result Question 123
####67. A $40 mL$ solution of a weak base, $B OH$ 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.
$(1991,6 M)$
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Solution:
- Let $40 mL$ of base contain $x mmol$ of $BOH$.
$$ \begin{array}{ccc} BOH+HCl & \longrightarrow BCl+H _2 O \ x-0.5 & 0.5 & \text { When } 5 mL \text { acid is added } \ x-2 & 2.0 & \text { When } 20 mL \text { of acid is added } \end{array} $$
When $pH$ is $10.04, pOH=3.96$ and when $pH$ is $9.14, pOH$ is 4.86. Therefore,
$$ \begin{aligned} & 3.96=p K _b+\log \frac{0.50}{x-0.5} \ & 3.96=p K _b+\log \frac{2.0}{x-2} \end{aligned} $$
Subtracting Eq. (i) from Eq. (ii) gives
$$ \begin{aligned}
