JEE Main 2023 — Chemistry Physical Chemistry

2023integerhard

A litre of buffer solution contains $0.1$ mole of each of ${\mathrm{NH}}_{3}$ and ${\mathrm{NH}}_{4}\mathrm{Cl}$ . On the addition of $0.02$ mole of $\mathrm{HCl}$ by dissolving gaseous $\mathrm{HCl}$ , the $\mathrm{pH}$ of the solution is found to be $_____\times {10}^{-3}$ (Nearest integer)

Given: ${\mathrm{pK}}_{b}({\mathrm{NH}}_{3})=4.745$

$\mathrm{log}2=0.301$

$\mathrm{log}3=0.477$

$T=298K]$

Official previous-year question

Held on 25 Jan 2023 · Verified 6 Jul 2026.

Did you get this right?

Solution

In resultant solution after addition of $\mathrm{HCl}$ to the basic buffer:

${n}_{{\mathrm{NH}}_{3}}=0.1-0.02=0.08$

${n}_{{\mathrm{NH}}_{4}\mathrm{Cl}}={n}_{{\mathrm{NH}}_{4}^{+}}=0.1+0.02=0.12$

Using Henderson-Hasselbach equation:

$\mathrm{pOH}={\mathrm{pK}}_{b}+\mathrm{log}\frac{[{\mathrm{NH}}_{4}^{+}]}{[{\mathrm{NH}}_{3}]}$

$=4.745+\mathrm{log}\frac{0.12}{0.08}$

$=4.745+\mathrm{log}\frac{3}{2}$

$=4.745+0.477-0.301$

$\mathrm{pOH}=4.921$

We know,

$\mathrm{pH}=14-\mathrm{pH}$

$=9.079$

Your note

JEE Main Physical Chemistry in other years

Physical Chemistry 2026 Physical Chemistry 2025 Physical Chemistry 2024 Physical Chemistry 2022 Physical Chemistry 2021 Physical Chemistry 2020 Physical Chemistry 2019 Physical Chemistry 2018 Physical Chemistry 2017 Physical Chemistry 2016 Physical Chemistry 2015 Physical Chemistry 2014

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2023 — Chemistry Physical Chemistry

2023integerhard

Given: ${\mathrm{pK}}_{b}({\mathrm{NH}}_{3})=4.745$

$\mathrm{log}2=0.301$

$\mathrm{log}3=0.477$

$T=298K]$

Official previous-year question

Held on 25 Jan 2023 · Verified 6 Jul 2026.

Did you get this right?

Solution

In resultant solution after addition of $\mathrm{HCl}$ to the basic buffer:

${n}_{{\mathrm{NH}}_{3}}=0.1-0.02=0.08$

${n}_{{\mathrm{NH}}_{4}\mathrm{Cl}}={n}_{{\mathrm{NH}}_{4}^{+}}=0.1+0.02=0.12$

Using Henderson-Hasselbach equation:

$\mathrm{pOH}={\mathrm{pK}}_{b}+\mathrm{log}\frac{[{\mathrm{NH}}_{4}^{+}]}{[{\mathrm{NH}}_{3}]}$

$=4.745+\mathrm{log}\frac{0.12}{0.08}$

$=4.745+\mathrm{log}\frac{3}{2}$

$=4.745+0.477-0.301$

$\mathrm{pOH}=4.921$

We know,

$\mathrm{pH}=14-\mathrm{pH}$

$=9.079$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2023 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry

JEE Main 2023 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry