Q9 / 160 · AP EAMCET 2018 (22 Apr Shift 2) · Chemistry

The $\mathrm{pH}$ of a buffer solution formed by mixing $30 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}$ and $30 \mathrm{~mL}$ of $1 \mathrm{M}$ $\mathrm{NH}_4 \mathrm{Cl}$ solutions is 8.6. The $\mathrm{p} K_b$ of $\mathrm{NH}_4 \mathrm{OH}$ is

  • A
    5.4
  • Correct answer
    B
    4.4
  • C
    5.6
  • D
    4.2

Solution

Given: $\mathrm{pH}$ of mixture $=8.6$ Also, $\quad \mathrm{pH}=14-\mathrm{pOH}=14-8.6=5.4$ and $\quad \mathrm{pOH}=\mathrm{p} K_b+\log \frac{[\text { Salt }]}{[\text { Base }]}$ Here, $\quad[$ Salt $]=\left[\mathrm{NH}_4 \mathrm{Cl}\right]=1 \mathrm{M}$ [Base $]=\left[\mathrm{NH}_4 \mathrm{OH}\right]=0.1 \mathrm{M}$ $$ \begin{array}{ll} \therefore & \mathrm{pOH}=\mathrm{p} K_b+\log \frac{1}{0.1} \\ & 5.4=\mathrm{p} K_b+\log 10 \Rightarrow 5.4=\mathrm{p} K_b+1 \\ \therefore \quad & \mathrm{p} K_b=5.4-1=4.4 \end{array} $$

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

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