What amount of HCl will be required to prepare one litre of a buffer solution of pH 10.4 using 0.01 mole of NaCN ? Given ${\mathrm{K}}_{\mathrm{ion}} \left(\mathrm{HCN}\right) = 4.1 \times {10}^{-10}.$

The addition of HCl converts NaCN into HCN. Let x be the amount of HCl added. We will have

[NaCN] = (0.01- x)

[HCN] = x

Substitution these values along with pH and ${\mathrm{K}}_{\mathrm{a}}$ in the expression.

$\mathrm{pH} = - \log {\mathrm{K}}_{\mathrm{a}} + \log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]}$

We get 10.4 $=-\log \left[4 \times {10}^{-10}\right] + \log \frac{0.01 -\mathrm{x}}{\mathrm{x}}$

$\mathrm{or} 10.4 = 9.4 + \log \frac{0.01 -\mathrm{x}}{\mathrm{x}}$

or  $\log \frac{0.01 -\mathrm{x}}{\mathrm{x}}=1$

or  $\frac{0.01 -\mathrm{x}}{\mathrm{x}}=10 \Rightarrow 11\mathrm{x} = {10}^{-2}$

or   $\mathrm{x} = 9.9 \times {10}^{-4} \mathrm{M}$

So, number of moles  $= 9.9 \times {10}^{-4}$