JEE Main 2021ChemistryIonic EquilibriumSolubility Product And Common Ion EffectmediumNumerical
JEE Main 2021 — Ionic Equilibrium Question with Solution
From: JEE Main 2021 (Online) 18th March Evening Shift
Question
The solubility of CdSO4 in water is 8.0 × 10−4 mol L−1. Its solubility in 0.01 M H2SO4 solution is __________ × 10−6 mol L−1. (Round off to the Nearest Integer). (Assume that solubility is much less than 0.01 M)
Enter your answer
▸Show full solutionCorrect answer: 64
Correct answer
64
Step-by-step explanation
In pure water
∴ Ksp = (s)2 = (8 × 10-4)2 = 64 × 10-8
In H2SO4 solution,
As S1 < < 0.01 so, S1 + 0.01 ≃ 0.01
∴ Ksp = [Cd+2] [So4−2]
⇒ 64 × 10-8 = S1× 0.01
⇒ S1 = 64 × 10-6
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Ionic Equilibrium chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
This is a previous-year question from JEE Main 2021, 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.
