MCQ
Calculate the $pH$ on mixing $100\, mL$ of $0.2\,M\, HCN$ with $400\, mL$ of $0.05\, M\, KOH$ at $25\,^oC$ ($pK_b$ for $CN^-= 8$)
- A$4.7$
- ✓$9.3$
- C$5.7$
- D$8.3$
✓
Answer
Correct option: B.
$9.3$
b
$HCN + KOH \to KCN + {H_2}O$
$HCN + KOH \to KCN + {H_2}O$
$n_0\,:$ $0.02$ $0.02$ $0$
$n_f\,:$ $0$ $0$ $0.02$
$\therefore[\mathrm{KCN}]=\frac{0.02 \mathrm{M}}{0.5}=\frac{1}{25} \,\mathrm{M}$
$\therefore \mathrm{pH}=7+\frac{1}{2}\left(\mathrm{pK}_{\mathrm{a}}+\log \mathrm{C}\right)$
$=7+\frac{1}{2}\left(6+\log \frac{1}{25}\right)$
$=7+\frac{1}{2}(6-2 \log 5)$
$=7+\frac{1}{2}(6-2 \times 0.7)$
$=7+\frac{1}{2}(4.6)=9.3$
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