Chemical and Ionic Equilibrium - Result Question 128

Answer:

Correct Answer: 71. $(0.177) $

Solution:

71. $HCN$ for buffer will be formed by the reaction

$ NaCN+HCl \longrightarrow NaCl+HCN $

$mmol$ of $NaCN$ present initially $=\frac{0.01}{49} \times 1000=0.2$

Let $x mmol$ of $HCl$ is added so that $x mmol$ of $NaCN$ will be neutralised forming $x$ mmol of $HCN$.

$pH =p K _a+\log \frac{[NaCN]}{[HCN]} $

$\begin{aligned} 8.5 & =-\log \left(4.1 \times 10^{-13}\right)+\log \frac{0.2-x}{x} \\ x & =0.177 \mathrm{mmol}\end{aligned}$