Q1 / 90 · JEE Main 2021 (25 Feb Shift 2) · Chemistry

The solubility of $Ca {(OH)}_{2}$ in water is :
[Given : The solubility product of $Ca {(OH)}_{2}$ in water $= 5.5 \times 10^{- 6}$ ]

Solution

Given that,

The solubility product of $Ca {(OH)}_{2}$ in water $= 5.5 \times 10^{- 6}$

$Ca {(OH)}_{2} ⇌ \underset{s}{{Ca}^{2 +} (aq)} + \underset{2 s}{2 {OH}^{-} (aq)}$

$k_{sp} = s {(2 s)}^{2} \Rightarrow 5.5 \times 10^{- 6} = 4 s^{3}$

$4 s^{3} = 5.5 \times 10^{- 6}$

$s = {(\frac{5.5}{4})}^{1 / 3} \times 10^{- 2}$
$s = 1.11 \times 10^{- 2}$

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