Q59 / 160 · JIPMER 2010 · Chemistry

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

  • Correct answer
    A
    $4 s^3$
  • B
    $s^3$
  • C
    $2 s^3$
  • D
    $16 s^2$

Solution

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

Hence, the correct answer is option: 'A'.

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