Q15 / 160 · TS EAMCET 2013 · Chemistry

The $\mathrm{pH}$ of a buffer solution made by mixing $25 \mathrm{~mL}$ of $0.02 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}$ and $25 \mathrm{~mL}$ of $0.2 \mathrm{M}$ $\mathrm{NH}_4 \mathrm{Cl}$ at $25^{\circ}$ is $\left(\mathrm{pK}\right.$ of $\mathrm{NH}_4 \mathrm{OH}=4.8$ )

  • A
    $5.8$
  • Correct answer
    B
    $8.2$
  • C
    $4.8$
  • D
    $3.8$

Solution

As a mixture of $\mathrm{NH}_4 \mathrm{OH}$ and $\mathrm{NH}_4 \mathrm{Cl}$ acts as a basic buffer, so its pH must be basic, (i.e., greater than 7), hence the answer must be 2 nd. It can also be find by calculations : $$ \begin{aligned} \mathrm{pOH} & =\mathrm{p} K_b+\log \frac{[\text { salt }]}{[\text { base }]} \\ & =4.8+\log \frac{0.2 \mathrm{M}}{0.02 \mathrm{M}} \\ & =4.8 \times \log 10 \\ \mathrm{pOH} & =4.8+1=5.8 \\ \mathrm{pH} & =14-\mathrm{pOH}=14-5.8=8.2 \end{aligned} $$

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

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