JEE Main 2020ChemistryIonic EquilibriumSolubility Product And Common Ion EffectmediumMCQ

JEE Main 2020Ionic Equilibrium Question with Solution

From: JEE Main 2020 (Online) 9th January Morning Slot

Question

The Ksp for the following dissociation is 1.6 × 10–5



Which of the following choices is correct for a mixture of 300 mL 0.134 M Pb(NO3)2 and 100 mL 0.4 M NaCl ?

Choose an option

Show full solutionCorrect option: A
Correct answer
AQ > Ksp

Step-by-step explanation

[Pb2+] =

= 1.005 × 10–1 M

[Cl-] =

= 10–1 M



Q = [Pb2+] × [Cl]2

= 0.1005 × (0.1)2

= 1.005 × 10–3

Given Ksp = 1.6 × 10–5

Q > Ksp

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.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2020, 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.