$\mathrm{pK}_{\mathrm{a}_{1}}$ and $\mathrm{pK}_{\mathrm{a}_{2}}$ of $\mathrm{H}_{2} \mathrm{SO}_{3}$ are $1…
$\mathrm{pK}_{\mathrm{a}_{1}}$ and $\mathrm{pK}_{\mathrm{a}_{2}}$ of $\mathrm{H}_{2} \mathrm{SO}_{3}$ are $1.82$ and 7.2 respectively. The $\mathrm{pH}$ after $20 \mathrm{~mL}$ of $0.16 \mathrm{M} \mathrm{NaOH}$ is added to $40 \mathrm{~mL}$ of $0.08 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{3}$ is :
$7.0$
$1.82$
$8.8$
$4.51$
Solution
$\underset{40 \times 0.08 \mathrm{meq}}{\mathrm{H}_{2} \mathrm{SO}_{3}}+\underset{20 \times 0.16 \mathrm{meq}}{\mathrm{NaOH}}{\longrightarrow} \mathrm{NaHSO}_{3}+\mathrm{H}_{2} \mathrm{O}$
$\therefore$ at first equivalence $\mathrm{pH}=\frac{\mathrm{PK}_{\mathrm{a}_{1}}+\mathrm{PK}_{\mathrm{a}_{2}}}{2}$
$=\frac{1.82+7.2}{2}=4.51$
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