Question

$50mL$ of $0.1M$ ${NH}_{4}OH$ is added to $75mL$ of $0.1M$ ${NH}_{4}Cl$ to make a basic buffer. If $p{K}_a$ of ${NH}_4^{+}$ is $9.26$, calculate $pH$.

• A. $4.44$
• B. $7.89$
• C. $9.56$
• D. $10.76$

Answer 1

Answer: (C)

$9.56$

Consider the following reaction:
$K_a$=${10}^{-p{K}_a}$=${10}^{-9.25}$=$5.5\times {10}^{-10}$
In the given buffer solution, following equilibrium exist:
$N{H}_4^{+}\rightleftharpoons N{H}_3+H^{+}$, where $N{H}_3$ comes from $N{H}_{4}OH$ and $N{H}_4^{+}$ comes from $N{H}_{4}Cl$
We need to find new concentrations as the quantities of $N{H}_{4}OH$ and $N{H}_{4}Cl$ mixed are not same.
Hence,
$\left[N{H}_3\right]$=$0.1\times \left(\frac{50}{75+50}\right)$=$0.04mol/d{m}^3$
$\left[N{H}_4^{+}\right]$=$0.1\times \left(\frac{75}{75+50}\right)$=$0.06mol/d{m}^3$
Applying Law of Mass action, we get,
$K_a$=$\frac{\left[N{H}_3\right]\left[H^{+}\right]}{\left[N{H}_4^{+}\right]}$
Therefore,
$\left[H^{+}\right]$=$\frac{\left[N{H}_4^{+}\right]\times K_a}{\left[N{H}_3\right]}$=$\frac{0.06\times 5.5\times {10}^{-10}}{0.04}$=$8.25\times {10}^{-10}$
Hence,
$pH$=$-lo{g}_{10}[H{]}^{+}$=$-lo{g}_{10}(8.25\times {10}^{-10})$=$9.5$