Verified Solution • 2025–26 Session moderate

QJEEM20Ch0843: JEE Main 2020 Chemistry - Class 11 JEEM20Ch0843 Solution: The K sp for the following dissociation is 1.6 × 10 –5 P b C l…

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

PbCl2(s)⇋Pb(aq)2++2Cl(aq)−

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 ?

Multiple Choice Options

  • A Q > Ksp
  • B Not enough data provided
  • C Q < Ksp
  • D Q = Ksp

Step-by-Step Verified Solution

[Pb2+] = 300×0.134400

= 1.005 × 10–1 M

[Cl-] = 100×0.4400

= 10–1 M

PbCl2(s)⇋Pb(aq)2++2Cl(aq)−

Q = [Pb2+] × [Cl–]2

= 0.1005 × (0.1)2

= 1.005 × 10–3

Given Ksp = 1.6 × 10–5

∴ Q > Ksp

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