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Chemistry · Ionic Equilibrium

JEE Advanced 2023 — Paper 1 — Question 36

On decreasing the pH from 7 to 2, the solubility of a sparingly soluble salt (MX) of a weak acid (HX) increased from 10−4   mol L−110^{-4}\; \mathrm{~mol} \mathrm{~L}^{-1} to 10−3   mol   L−110^{-3} \;\mathrm{~mol} \;\mathrm{~L}^{-1}. The pKa\mathrm{pK}_{\mathrm{a}} of HX is

  1. Option A:

    3

  2. Option B:

    4

    Correct
  3. Option C:

    5

  4. Option D:

    2

Answer: B

Step-by-step solution

Given:

Initial   pH  =7,S1=10−4 mol L−1\text{Initial\; pH\;}=7,\quad S_1=10^{-4}\ \text{mol L}^{-1} Final   pH  =2,S2=10−3 mol L−1\text{Final\; pH\;}=2,\quad S_2=10^{-3}\ \text{mol L}^{-1} Salt: MX (salt   of   weak   acid   HX)\text{Salt: } MX \text{ (salt\; of\; weak\; acid\; } HX)

For a salt of a weak acid:

S=Ksp(1+[H+]Ka)S=\sqrt{K_{sp}\left(1+\frac{[H^+]}{K_a}\right)}

Solubility ratio:

S2S1=10⇒(S2S1)2=100\frac{S_2}{S_1}=10 \Rightarrow \left(\frac{S_2}{S_1}\right)^2=100 1+10−2Ka1+10−7Ka=100\frac{1+\dfrac{10^{-2}}{K_a}}{1+\dfrac{10^{-7}}{K_a}}=100

Since

10−7≪Ka⇒1+10−7Ka≈110^{-7}\ll K_a \Rightarrow 1+\frac{10^{-7}}{K_a}\approx 1 1+10−2Ka=1001+\frac{10^{-2}}{K_a}=100 10−2Ka=99⇒Ka≈10−4\frac{10^{-2}}{K_a}=99 \Rightarrow K_a\approx 10^{-4}
pKa=4\boxed{pK_a=4}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2023
Paper
Paper 1
Subject
Chemistry
Chapter
Ionic Equilibrium
Topic
Sparingly Soluble Salts, Solubility Product & Precipitation Conditions
On decreasing the pH from 7 to 2, the solubility of a sparingly… | JEE Advanced 2023 PYQ with Solution · DhiX AI