Question

Calcium hydroxide, Ca(OH)2, is an ionic compound with a solubility product constant, Ksp, of 6.5×10–6. Calculate the solubility of this compound in pure water.

Answer 1

Answer: The solubility of this compound in pure water is 0.012 M

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as $K_{sp}$

The equation for the ionization of the  is given as:

$Ca(OH)_2\rightarrow Ca^{2+}+2OH^-$  

By stoichiometry of the reaction:

1 mole of  $Ca(OH)_2$ gives 1 mole of $Ca^{2+}$ and 2 mole of $OH^-$

When the solubility of  $Ca(OH)_2$ is S moles/liter, then the solubility of $Ca^{2+}$  will be S moles\liter and solubility of $OH^-$ will be 2S moles/liter.

$K_{sp}=[Ca^{2+}][OH^{-}]^2$

$6.5\times 10^{-6}=[S][2S]^2$

$S=0.012M$

Thus solubility of this compound in pure water is 0.012 M