Question

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} $

Answer 1

$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 = pK_a + log \frac{ [ NaCN ]}{ [ HCN]} $
$8.5 = - \log ( 4.1 \times 10^{ - 10 }) + log \frac{ 0.2 - x}{ x} $
$x = 0.177\, mmol$