Q12 / 160 · TS EAMCET 2019 (03 May Shift 2) · Chemistry

A solution of $0.1$ mole of ${CH}_{3} {NH}_{2} (K_{b} = 5 \times 10^{- 4})$ and $0.08 mole$ of $HCl$ is diluted to one litre, then the $pOH$ of the solution is $(\log 1.25 = 0.1)$

Solution

${CH}_{3} {NH}_{2} + HCl \to {CH}_{3} {NH}_{3} Cl 0.1 mol 0.08 mol 0.02 mol 0.08 mol$

It is a case of basic buffer

$pOH = pKb + \log (\frac{mol of salt}{mol of base})$

$pOH = - \log (5 \times 10^{- 4}) + \log (\frac{0.08}{0.02})$

$pOH = 3.9$

Hence, the correct answer is option: 'B'.