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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Answer:
Correct Answer: 67. $\left(1.8 \times 10^{-5}\right)$
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,
$\begin{aligned} & 3.96=p K _b+\log \frac{0.50}{x-0.5} \hspace{10mm}…(i)\\ & 3.96=p K _b+\log \frac{2.0}{x-2}\hspace{10mm}…(ii) \end{aligned}$
Subtracting Eq. (i) from Eq. (ii) gives
$\begin{aligned} 0.90 & =\log \left(\frac{2}{x-2} \times \frac{x-0.5}{0.5}\right) \\ \Rightarrow \quad 28 & =\frac{4(x-0.5)}{x-2} \end{aligned}$
$\Rightarrow \quad x=3.5$, substituting in equation (i) gives
$\begin{aligned} 3.96 & =\mathrm{p} K_b+\log \frac{0.5}{3} \\ K_b & =1.8 \times 10^{-5} \end{aligned}$
BETA
