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JEE Main 201226 MayOnline
Exam
JEE Main
Year
2012
Date
26 May
Mode
Online
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2012 · 26 May · Online · Chemistry · Ionic Equilibrium

The solubility of PbI_2 at 25^ C is 0.7 ~g ~L^ -1 . The solubility...

The solubility of PbI2\mathrm{PbI}_2 at 25∘C25^{\circ} \mathrm{C} is 0.7 g L−10.7 \mathrm{~g} \mathrm{~L}^{-1}. The solubility product of PbI2\mathrm{PbI}_2 at this temperature is (molar mass of PbI2=461.2 g mol−1\mathrm{PbI}_2=461.2 \mathrm{~g} \mathrm{~mol}^{-1} )

Options

  1. A1.40×10−91.40 \times 10^{-9}
  2. B0.14×10−90.14 \times 10^{-9}
  3. C140×10−9140 \times 10^{-9}
  4. D14.0×10−914.0 \times 10^{-9}

Correct answer

D. 14.0×10−914.0 \times 10^{-9}

Step-by-step solution

PbI2⇋Pb+++2\mathrm{PbI}_2 \leftrightharpoons \mathrm{Pb}^{++}+2\mathrm{I}−^{-}Ksp =s×(2s)2=4s3=4×(0.7461.2)3=14.0×10−9\begin{aligned} K_{\text{sp }}= & s \times(2 s)^2=4 s^3 \\ & =4 \times\left(\frac{0.7}{461.2}\right)^3=14.0 \times 10^{-9} \end{aligned}
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