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

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

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

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

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