Chemical and Ionic Equilibrium - Result Question 128
71. How many gram-mole of $HCl$ will be required to prepare one litre of buffer solution (containing $NaCN$ and $HCl$ ) of $pH $ $8 .5$ using $0.01 g$ formula weight of $NaCN$ ?
$K _{HCN}=4.1 \times 10^{-10}$
(1988,4 M)
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Answer:
Correct Answer: 71. $(0.177) $
Solution:
- $HCN$ 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]} $
$\begin{aligned} 8.5 & =-\log \left(4.1 \times 10^{-13}\right)+\log \frac{0.2-x}{x} \\ x & =0.177 \mathrm{mmol}\end{aligned}$
BETA
