JEE Main 2021 — Ionic Equilibrium Question with Solution

Let, solubility of Ca(OH)₂ in pure water = S mol/L

$$Ca(OH{)}_2$$ $$\rightleftharpoons$$ $$\underset{Smol/L}{C{a}^{2+}}+\underset{2\times S(mol/L)}{2O{H}^{-}}$$

K_sp = [Ca²⁺] [OH^$$-$$]² = S $$\times$$ (2S)² = 4 S³ (mol/L)

The expression of K_sp can also be written as,

K_sp = x^x . y^y . S^{x + y}

= 1¹ . 2² . S^{1 + 2}

= 4 S³ [$$\because$$ For Ca(OH)₂ : x = 1, y = 2]

x and y are the coefficients of cations and anions respectively

$$S={\left(\frac{K_{sp}}{4}\right)}^{1/3}={\left(\frac{5.5\times {10}^{-6}}{4}\right)}^{1/3}$$

$$=1.11\times 10^{-2}$$ ml/L