Question

A buffer of pH $8.50$ is prepared from $0.02$ mole of $KCN$. The desired volume of buffer solution is to be $1$ litre. How will you make this buffer using $HCl$? What is the change after addition of $0.5\times {10}^{-4}$ mole $HCl$ to $100{cm}^3$ and same amount of $NaOH$ in $100{cm}^3$ of buffer?
$(K_{a}HCN=6.2\times {10}^{-10})$

Answer 1

$KCN+HCl⟶KCl+HCN$
$(0.02-x)$ $x$ $x$ $x$
$pH=p{K}_a+\log \frac{\left[Salt\right]}{\left[Acid\right]}$
$8.50=-\log (6.2\times {10}^{-10})+{\log }_{10}\left[\frac{0.02-x}{x}\right]$
$x=0.01672$
Buffer is prepared by adding $0.01672$ mole of $HCl$ in salt $KCN$
Calculation of pH change when $HCl$ is added:
Moles of $HCl$ added in $1$ litre buffer $0.5\times {10}^{-3}$
$HCl$ will convert more salt into acid
$pH=9.2076+{\log }_{10}\frac{0.00382-0.5\times {10}^{-3}}{0.01672+0.5\times {10}^{-3}}=9.4156$
pH change $=9.4156-8.50$
$=0.9156$

Calculation of pH change when $NaOH$ is added:
Moles of $NaOH$ added in $1$ litre buffer $0.5\times {10}^{-3}$
$NaOH$ will convert $HCl$ into salt
$pH=9.2076+{\log }_{10}\frac{0.00382+0.5\times {10}^{-3}}{0.01672-0.5\times {10}^{-3}}=8.515$
pH change $=8.515-8.5$
$=0.015$