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JIPMER 2010
Exam
JIPMER
Year
2010
Subject
Chemistry
Chapter
Ionic Equilibrium

JIPMER 2010 · Chemistry · Ionic Equilibrium

The solubility of saturated solution of Ag_2 CrO_4 is s mol L^ -1 ....

The solubility of saturated solution of Ag2CrO4\mathrm{Ag}_2 \mathrm{CrO}_4 is ss mol L−1\mathrm{L}^{-1}. What is its solubility product?

Options

  1. A4s34 s^3
  2. Bs3s^3
  3. C2s32 s^3
  4. D16s216 s^2

Correct answer

A. 4s34 s^3

Step-by-step solution

Ag2CrO4\mathrm{Ag}_2 \mathrm{CrO}_4 ionises completely in the solution as Ag2CrO4⟶s mol/L2Ag+2s mol/L+CrO42−s mol/L\underset{s \mathrm{~mol} / \mathrm{L}}{\mathrm{Ag}_2 \mathrm{CrO}_4 \longrightarrow} \underset{2 s \mathrm{~mol} / \mathrm{L}}{2 \mathrm{Ag}^{+}}+\underset{s \mathrm{~mol} / \mathrm{L}}{\mathrm{CrO}_4^{2-}} Hence, the solubility product, Ksp  of Ag2CrO4=[Ag+]2[CrO42−]=(2s)2(s)=4s3\begin{aligned} K_{\text{sp }} \text{ of } \mathrm{Ag}_2 \mathrm{CrO}_4 & =\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CrO}_4^{2-}\right] \\ & =(2 s)^2(s) \\ & =4 s^3 \end{aligned}
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