The solubility of sparingly soluble salt $\mathrm{AB}_2$ is $1.0 \times 10^{-4} \mathrm{~mol}$…

The solubility of sparingly soluble salt $\mathrm{AB}_2$ is $1.0 \times 10^{-4} \mathrm{~mol}$ $\mathrm{dm}^{-3}$. What is its solubility product?
  1. $2 \times 10^{-12}$
  2. $4 \times 10^{-8}$
  3. $4 \times 10^{-12}$
  4. $2 \times 10^{-8}$

Solution

$\mathrm{AB}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{A}_{(\mathrm{aq})}^{2+}+2 \mathrm{~B}_{(\mathrm{aq})}^{-}$ At equilibrium, $-\mathrm{S} \quad 2 \mathrm{~S}$ $\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{2+}\right]\left[\mathrm{B}^{-}\right]^2 \\ & =\mathrm{S} .(2 \mathrm{~S})^2=4 \mathrm{~S}^3 \\ & =4 \times\left(1 \times 10^{-4}\right)^3 \\ & \mathrm{~K}_{\mathrm{sp}}=4 \times 10^{-12} \end{aligned}$

Asked in: MHT CET 2021 (24 Sep Shift 2)

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