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JEE Main 20209 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
9 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2020 · 9 January · Morning Shift · Chemistry · Ionic Equilibrium

The K _ sp for the following dissociation is 1.6 × 10 ^ - 5 PbCl _...

The Ksp{K}_{sp} for the following dissociation is 1.6×10−51.6 \times {10}^{- 5}
PbCl2(s)⇌Pb(aq)2++2Cl(aq)−{PbCl}_{2\left(s\right)} ⇌ P{b}_{\left(aq\right)}^{2 +} + 2C{l}_{\left(aq\right)}^{-}
Which of the following choices is correct for a mixture of 300mL0.134MPb(NO3)2300mL0.134MPb{\left(N{O}_{3}\right)}_{2} and 100mL100mL 0.4MNaCl?0.4MNaCl ?

Options

  1. ANot enough data provided
  2. BQ<KspQ < {K}_{sp}
  3. CQ>KspQ > {K}_{sp}
  4. DQ=KspQ = {K}_{sp}

Correct answer

C. Q>KspQ > {K}_{sp}

Step-by-step solution

Q=[Pb2+][Cl−]2Q = \left[P{b}^{2 +}\right]{\left[C{l}^{-}\right]}^{2}
=300×0.134400×[100×0.4400]2= \dfrac{300 \times 0.134}{400} \times {\left[\dfrac{100 \times 0.4}{400}\right]}^{2}
=3×0.1344×(0.1)2= \dfrac{3 \times 0.134}{4} \times {\left(0.1\right)}^{2}
=0.105×10−2= 0.105 \times {10}^{- 2}
=1.005×10−3= 1.005 \times {10}^{- 3}
Q>KspQ > {K}_{sp}
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