Chemical and Ionic Equilibrium 2 Question 72
71. How many gram-mole of $\mathrm{HCl}$ will be required to prepare one litre of buffer solution (containing $\mathrm{NaCN}$ and $\mathrm{HCl}$ ) of $\mathrm{pH} 8.5$ using $0.01 \mathrm{~g}$ formula weight of $\mathrm{NaCN}$ ?
$$ K_{\mathrm{HCN}}=4.1 \times 10^{-10} $$
$(1988,4 M)$
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Solution:
- $\mathrm{HCN}$ for buffer will be formed by the reaction
$$ \mathrm{NaCN}+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{HCN} $$
$\mathrm{mmol}$ of $\mathrm{NaCN}$ present initially $=\frac{0.01}{49} \times 1000=0.2$
Let $x \mathrm{mmol}$ of $\mathrm{HCl}$ is added so that $x \mathrm{mmol}$ of $\mathrm{NaCN}$ will be neutralised forming $x$ mmol of $\mathrm{HCN}$.
$$ \begin{aligned} \mathrm{pH} & =\mathrm{p} K_{a}+\log \frac{[\mathrm{NaCN}]}{[\mathrm{HCN}]} \
