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Q.
$50\,mL$ of $0.1 \,M \,NaOH$ is added to $75\, mL$ of $0.1\, M \,NH _{4} Cl$ to make a basic buffer. If $p K_{a}$ of $NH _{4}^{+}$is $9.26$, then the $pH$ of solution is
Equilibrium
Solution:
$NH _{4} Cl$ reacts with $NaOH$ to form $NH _{4} OH$ and
unreacted $NH _{4} Cl$ forms a basic buffer.
Millimoles of $NH _{4} Cl =75 \times 0.1=7.5$
Millimoles of $NaOH =50 \times 0.1=5.0$
Given, $p K_{a}\left( NH _{4}^{+}\right)=9.26$
$\therefore p K_{b}\left( NH _{3}\right)=14-9.26=4.74$
$\Rightarrow pOH = p K_{b}+\log \frac{\left[ NH _{4}^{+}\right]}{\left[ NH _{4} OH \right]}$
$=4.74+\log \frac{2.5}{5.0}$
$ \Rightarrow pOH=4.44 $
Thus, $ pH =14-4.44 $
$ pH =9.56$
