JEE Main 2019 — Ionic Equilibrium Question with Solution
From: JEE Main 2019 (Online) 12th April Evening Slot
Question
The molar solubility of Cd(OH)2 is 1.84 × 10–5
M in water. The expected solubility of Cd(OH)2 in a buffer
solution of pH = 12 is :
Choose an option
Show full solutionCorrect option: A
Correct answer
A2.49 × 10–10 M
Step-by-step explanation
| Cd(OH)2(s) | ⇌ | Cd2+(aq) | + | 2OH(aq) |
|---|---|---|---|---|
| S | 2S |
Ksp[Cd(OH)2] = 4(S)3 = 4(1.84 × 10–5)3 [Given S = 1.84 × 10–5]
For buffer solution of pH = 12 :
As pH = 12 so pOH = 2
[OH] = 10-2
Let the solubilty = S1
| Cd(OH)2(s) | ⇌ | Cd2+(aq) | + | 2OH(aq) |
|---|---|---|---|---|
| S1 | 2S1 + 10-2 |
Ksp[Cd(OH)2] = S1(10-2)2 = 4(1.84 × 10–5)3
S1 = 2.49 × 10–10 M
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This is a previous-year question from JEE Main 2019, covering the Ionic Equilibrium chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.