MCQ
$50\, mL$ of $0.2\, M$ ammonia solution is treated with $25\, mL$ of $0.2\, M\, HCl$. If $pK_b$ of ammonia solution is $4.75$, the $pH$ of the mixture will be
- A$3.75$
- B$4.75$
- ✓$8.25$
- D$9. 25$
moles of $HCl = 0.2\,M \times 25 \times {10^{ - 3}}\,L = 0.005$
moles $HCl$ (total consumed)
moles of $N{H_3} = 0.2\,M \times 50 \times {10^{ - 3}}\,L = 0.01$
moles $HCl$
excess $N{H_3} = 0.01 - 0.005 = 0.005$ moles
$1$ mole ammonia $= 1 $ mole $N{H_4}Cl$
$0.005\,N{H_3} = 0.005\,N{H_4}Cl$
Total volume
$ = {V_{HCl}} + {V_{N{H_3}}} = 25 + 50\, = 75\,ml$
$[N{H_3}] = [N{H_4}Cl] = \frac{{0.005\,mole}}{{75 \times {{10}^{ - 3}}\,L}} = 0.066\,M$
$pOH = p{K_b} + \log \frac{{[N{H_4}Cl]}}{{[N{H_3}]}}$
$pOH = 4.75 + \log \frac{{[0.066]}}{{[0.066]}}$
$pOH = 4.75$
$pH = 14 - pOH \Rightarrow pH = 9.25$
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