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