Q36 / 150 · MHT CET 2008 · Chemistry

Solubility of $\mathrm{Ca}(\mathrm{OH})_{2}$ is $s \mathrm{~mol} \mathrm{~L}^{-1}$. The solubility product $\left(K_{s p}\right)$ under the same condition is

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

Solution

$\begin{aligned} \mathrm{Ca}(\mathrm{OH}) & \rightleftharpoons \mathrm{Ca}^{2+}+2 \mathrm{OH}^{-} \\ K_{s p} &=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2} \\ &=(s)(2 s)^{2}=4 s^{3} \end{aligned}$

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

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