JEE Main 2021ChemistryIonic EquilibriumMediumNumerical

JEE Main 2021 — Ionic Equilibrium Question with Solution

JEE Main 2021 (24 Feb Shift 2)

Question

The solubility product of PbI2 is 8.0×10-9. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x×10-6 mol/L. The value of x is _________ (Rounded off to the nearest integer)

[Given 2=1.41]

Enter your answer

Show full solutionCorrect answer: 141
Correct answer
141

Step-by-step explanation

Given: KspPbI2=8×10-9

To calculate : solubility of PbI2 in 0.1M solution of PbNO22

I PbNO32→Pb(aq)+2+2NO3-aq

0.1 M---         0.1 M         0.2 M

II PbI2s⇌Pb+2aq+2I-aq

                         s                2s

[Pb+2]=s+0.1

≃0.1

Now: Ksp=8×10-9=Pb+2I-2

⇒8×10-9=0.1×2 s2

⇒8×10-8=4 s2⇒s=2×10-4

⇒S=141×10-6M

⇒x=141

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About this question

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.