Ionic Equilibrium Question 3
Question 3 - 25 January - Shift 1
A litre of buffer solution contains 0.1 mole of each of $NH_3$ and $NH_4 Cl$. On the addition of 0.02 mole of $HCl$ by dissolving gaseous $HCl$, the $pH$ of the solution is found to be $\times 10^{-3}$ (Nearest integer)
[Given : $pK_b(NH_3)=4.745$
$\log 2=0.301$
$\log 3=0.477$
$T=298 K]$
Show Answer
Answer: (9.079)
Solution:
In resultant solution
$ n _{NH_3}=0.1-0.02=0.08 $
$ n _{NH_4 Cl}=n _{NH_4^{+}}=0.1+0.02=0.12 $
$pOH=pK_b+\log \frac{[NH_4^{+}]}{[NH_3]}$
$ \begin{aligned} & =4.745+\log \frac{0.12}{0.08} \\ & =4.745+\log \frac{3}{2} \\ & =4.745+0.477-0.301 \\ & pOH=4.921 \\ & pH=14-pH \\ & \quad=9.079 \end{aligned} $
