Consider the dissociation of the weak acid HX as given below HX ( aq ) rightleftharpoons H + ( aq )+ X - ( aq ), Ka =1.2 × 10 -5 [ K a : dissociation constant] The osmotic pressure of 0.03 M aqueous solution of HX at 300 K is × 10 -2 bar (nearest integer) [Given : R = 0.083 L bar Mol -1 K -1 ]

Chemistry · Ionic Equilibrium

JEE Main 2024 — 6 April, Shift 1 — Question 76

Consider the dissociation of the weak acid HX as given below $\mathrm{HX}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{X}^{-}(\mathrm{aq}), \mathrm{Ka}=1.2 \times 10^{-5}$

[ $\mathrm{K}_{\mathrm{a}}:$ dissociation constant]

The osmotic pressure of 0.03 M aqueous solution of HX at 300 K is $\qquad$ $\times 10^{-2}$ bar (nearest integer)

$\text{[Given : R = 0.083 L bar Mol}^{-1} \text{K}^{-1} \text{]}$

Answer: 76

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{HX} \rightleftharpoons \mathrm{H}^{+}+\mathrm{X}^{-} \quad \mathrm{K}_{\mathrm{a}}=1.2 \times 10^{-5}$ 0.03 M $0.03-\mathrm{x} \quad \mathrm{x} \quad \mathrm{x}$

$\mathrm{K}_{\mathrm{a}}=1.2 \times 10^{-5}=\frac{\mathrm{x}^{2}}{0.03-\mathrm{x}}$

$0.03-\mathrm{x} \approx 0.03\left(\mathrm{~K}_{\mathrm{a}}\right.$ is very small $)$

$\frac{\mathrm{x}^{2}}{0.03}=1.2 \times 10^{-5}$ $\mathrm{x}=6 \times 10^{-4}$

Final solution : $0.03-x+x+x$ $=0.03+\mathrm{x}=0.03+6 \times 10^{-4}$

$\Pi=\left(0.03+\left(6 \times 10^{-4}\right)\right) \times 0.083 \times 300$

$=76.19 \times 10^{-2} \approx 76 \times 10^{-2}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 1
Subject: Chemistry
Chapter: Ionic Equilibrium
Topic: Solutions containing one Acid or Base