The solubility of an ionic compound in water can be expressed in terms of the mass that will dissolve in a given volume of water, or in terms of the solubility product, Ksp.If only 0.238 g of Ca(OH)2 dissolves in enough water to give 0.230 L of aqueous solution at a given temperature, what is the Ksp value for calcium hydroxide at this temperature?



Answer :

Answer

Kps=1.09 x 10⁻⁵

Procedure

First, we will need to consider the solution equilibrium. In this case, we have the following equation that depicts the equilibrium.

[tex]Ca(OH)_2\leftrightarrows Ca^{2+}+2OH^-[/tex]

The molar solubility will be given by the following equation If only 0.238 g of Ca(OH)₂ is dissolved in enough water to give 0.230 L

[tex]S=\frac{\frac{0.238\text{ g Ca\lparen OH\rparen}_2}{74.09\text{ g mol}^{-1}}}{0.230\text{ L}}=0.01397\text{ M}[/tex]

Therefore the Kps value will be determined as :

[tex]kps=s(2s)^2=(0.01397)(2\times0.01397)^2=1.09\times10^{-5}[/tex]