Q49 / 150 · MHT CET 2021 (23 Sep Shift 2) · Chemistry

The solubility of $\mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_4$ is $2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$ at $298 \mathrm{~K}$. What is it's solubility product?

  • A
    $1.6 \times 10^{-6}$
  • Correct answer
    B
    $3.2 \times 10^{-11}$
  • C
    $1.6 \times 10^{-11}$
  • D
    $3.2 \times 10^{-6}$

Solution

$\begin{aligned} & \mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_{4(\mathrm{~s})} \rightleftharpoons \underset{2 \mathrm{~S}}{2 \mathrm{Ag}_{(\mathrm{aq})}^{+}}+\mathrm{C}_2 \mathrm{O}_{4(\mathrm{aq})}^{2-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{C}_2 \mathrm{O}_4^{2-}\right] \\ & =(2 \mathrm{~S})^2 \mathrm{~S} \\ & =4 \mathrm{~S}^3=4 \times\left(2 \times 10^{-4}\right)^3 \\ & \mathrm{~K}_{\mathrm{sp}}=3.2 \times 10^{-11}\end{aligned}$

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

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