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JIPMER 2008
Exam
JIPMER
Year
2008
Subject
Chemistry
Chapter
Chemical Equilibrium

JIPMER 2008 · Chemistry · Chemical Equilibrium

Solubility product of (PbCl_2) at 298 K is (1 × 10^ -6 ). At this...

Solubility product of (\mathrm{PbCl}_2) at 298 K is (1 \times 10^{-6}). At this temperature solubility of (\mathrm{PbCl}_2) in (\mathrm{mol} / \mathrm{L}) is

Options

  1. A(\left(1 \times 10^{-6}\right)^{1 / 2})
  2. B(\left(1 \times 10^{-6}\right)^{1 / 3})
  3. C(\left(0.25 \times 10^{-6}\right)^{1 / 3})
  4. D(\left(2.5 \times 10^{-6}\right)^{1 / 2})

Correct answer

C. (\left(0.25 \times 10^{-6}\right)^{1 / 3})

Step-by-step solution

(\mathrm{PbCl}_2 \underset{s}{\rightleftharpoons \mathrm{Pb}^{2+}}+\underset{2 s}{2 \mathrm{Cl}^{-}}) KSP=(S)(2S)21×10−6=4S3S=(14×10−6)1/3=(0.25×10−6)1/3\begin{aligned} K_{S P} & =(S)(2 S)^2 \\ 1 \times 10^{-6} & =4 S^3 \\ S & =\left(\frac{1}{4} \times 10^{-6}\right)^{1 / 3} \\ & =\left(0.25 \times 10^{-6}\right)^{1 / 3} \end{aligned})$
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