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How much of 0.3 M NH4OH should be mixed with 30 mL of 0.2 M solution of NH4Cl to give buffer solution of pH 8.65?

(Given, pkb of NH4OH = 4.75)

Option: 1

5 L


Option: 2

5 mL


Option: 3

10 mL


Option: 4

6 mL


Answers (1)

best_answer

\begin{array}{l}{\text { It is a basic buffer. }} \\ {\begin{array}{l}{\mathrm{pH}=8.65, \mathrm{pOH}=14-8.65=5.35} \\ {\mathrm{pOH}=\mathrm{pK}_{b}+\log \frac{\left[\mathrm{NH}_{4}\right]}{\left[\mathrm{NH}_{4} \mathrm{OH}\right]}} \\\\ {5.35=4.75+\log \frac{\left[\mathrm{NH}_{4}\right]}{\left[\mathrm{NH}_{4} \mathrm{OH}\right]}} \\\\ {\therefore \frac{\left[\mathrm{NH}_{4}\right]}{\left[\mathrm{NH}_{4} \mathrm{OH}\right]}=\operatorname{antilog}(5.35-4.75)} \\\\ {\therefore \frac{\left[\mathrm{NH}_{4}\right]}{\left[\mathrm{NH}_{4} \mathrm{OH}\right]}=\frac{30 \times 0.2}{0.3 \times V} \approx 4.0}\end{array}} \\ {\therefore V=5 \mathrm{mL}}\end{array}

Therefore,option(2) is correct

Posted by

Irshad Anwar

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