The solubility of Ca(OH)2 in water is :
[Given : The solubility product of Ca(OH)2 in water = 5.5 $\times$ 10$-$6]
Solution
<p>Let, solubility of Ca(OH)<sub>2</sub> in pure water = S mol/L</p>
<p>$Ca{(OH)_2}$ $\rightleftharpoons$ $$\mathop {C{a^{2 + }}}\limits_{S\,mol/L} + \mathop {2O{H^ - }}\limits_{2 \times S\,(mol/L)} $$</p>
<p>K<sub>sp</sub> = [Ca<sup>2+</sup>] [OH<sup>$-$</sup>]<sup>2</sup> = S $\times$ (2S)<sup>2</sup> = 4 S<sup>3</sup> (mol/L)</p>
<p>The expression of K<sub>sp</sub> can also be written as,</p>
<p>K<sub>sp</sub> = x<sup>x</sup> . y<sup>y</sup> . S<sup>x + y</sup></p>
<p>= 1<sup>1</sup> . 2<sup>2</sup> . S<sup>1 + 2</sup></p>
<p>= 4 S<sup>3</sup> [$\because$ For Ca(OH)<sub>2</sub> : x = 1, y = 2]</p>
<p>x and y are the coefficients of cations and anions respectively</p>
<p>$$S = {\left( {{{{K_{sp}}} \over 4}} \right)^{1/3}} = {\left( {{{5.5 \times {{10}^{ - 6}}} \over 4}} \right)^{1/3}}$$</p>
<p>$= 1.11 \times {10^{ - 2}}$ ml/L</p>
About this question
Subject: Chemistry · Chapter: Equilibrium · Topic: Chemical Equilibrium and Kc, Kp
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