Chemistry · Ionic Equilibrium

JEE Main 2025 — 7 April, Evening Shift — Question 19

One litre buffer solution was prepared by adding 0.10 mol each of $\mathrm{NH}_{3}$ and $\mathrm{NH}_{4} \mathrm{Cl}$ in deionised water. The change in pH on addition of 0.05 mol of HCl to the above solution is $\_\_\_\_\_$ $\times 10^{-2}$ . (Nearest integer) Given: $\mathrm{pK}_{\mathrm{b}}$ of $\mathrm{NH}_{3}=4.745$ and $\log _{10} 3=0.477$

Answer: 48

Numerical answer — enter this value.

Step-by-step solution

Initially $\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{\left[\mathrm{NH}_{4} \mathrm{Cl}\right]}{\left[\mathrm{NH}_{3}\right]}=\mathrm{pK}+\log \frac{0.1}{0.1}$

$\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}$

$\mathrm{pH}=14-\mathrm{pOH} \Rightarrow \mathrm{pH}=9.255$

When 0.05 mol HCl is added

$\mathrm{NH}_{3}+\mathrm{H}^{+} \rightleftharpoons \mathrm{NH}_{4}{ }^{+}$

$0.1 \quad 0.05$ 0.050 0.15

$\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{0.15}{0.05}=5.222$

$\mathrm{pH}=8.778$

Change in $\mathrm{pH}=0.477$ or $47.7 \times 10^{-2}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 7 April, Evening Shift
Subject: Chemistry
Chapter: Ionic Equilibrium
Topic: Buffer solution & pH of Special Buffers - Zwitter ions ++