Question

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 1

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.

`Ca(OH)_2\leftrightarrows Ca^(2+)+2OH^-`

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

`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}`

Therefore the Kps value will be determined as :

`kps=s(2s)^2=(0.01397)(2*0.01397)^2=1.09*10^(-5)`