Question

$50mL$ of $0.1MNaOH$ is added to $75mL$ of $0.1MN{H}_{4}Cl$ to make a basic buffer. If $p{K}_a$ of $N{H}_4^{+}$is $9.26$, then the $pH$ of solution is

• A. $4.44$
• B. $9.56$
• C. $4.74$
• D. $5.76$

Answer 1

Answer: (B)

$N{H}_{4}Cl$ reacts with $NaOH$ to form $N{H}_{4}OH$ and
unreacted $N{H}_{4}Cl$ forms a basic buffer.
Millimoles of $N{H}_{4}Cl=75\times 0.1=7.5$
Millimoles of $NaOH=50\times 0.1=5.0$

Given, $p{K}_a\left(N{H}_4^{+}\right)=9.26$
$\therefore p{K}_b\left(N{H}_3\right)=14-9.26=4.74$
$\Rightarrow pOH=p{K}_b+\log \frac{\left[N{H}_4^{+}\right]}{\left[N{H}_{4}OH\right]}$
$=4.74+\log \frac{2.5}{5.0}$
$\Rightarrow pOH=4.44$
Thus, $pH=14-4.44$
$pH=9.56$