Chemical and Ionic Equilibrium - Result Question 123

Answer:

Correct Answer: 67. $\left(1.8 \times 10^{-5}\right)$

Solution:

67. 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

$\begin{aligned} 3.96 & =\mathrm{p} K_b+\log \frac{0.5}{3} \\ K_b & =1.8 \times 10^{-5} \end{aligned}$