Question

What is the molar solubility of calcium sulfate in pure water?
${{K}_{sp}}$ = 2.4 $\times {{10}^{-5}}$ for calcium sulfate.
A. What is the molar solubility of calcium sulfate in pure water?
B. What is the mass solubility of calcium sulfate in pure water, expressed in g/L?

Answer 1

Hint: The solubility of sparingly soluble salts in aqueous solution is determined by the ${{K}_{sp}}$ which is the solubility product. In ${{K}_{sp}}$ the concentration is raised to the power number which is the number of ions produced.

Complete answer:
A. We are given calcium sulfate $CaS{{O}_{4}}$ which has ${{K}_{sp}}$ given as 2.4 $\times {{10}^{-5}}$, so it is a sparingly soluble salt, that dissociate as:
$CaS{{O}_{4}}(s)\rightleftharpoons C{{a}^{2+}}(aq)+S{{O}_{4}}^{2-}(aq)$
The solubility product or ${{K}_{sp}}$ is given for both of the products so,
${{K}_{sp}}$= $[C{{a}^{2+}}(aq)]\,\,[S{{O}_{4}}^{2-}(aq)]\,$= 2.4$\times {{10}^{-5}}mo{{l}^{2}}\,{{L}^{-2}}$
The solubility can also be written as $[C{{a}^{2+}}(aq)]\,\,$= s =$\,\,[S{{O}_{4}}^{2-}(aq)]\,$
So, ${{K}_{sp}}$= ${{s}^{2}}$ = 2.4$\times {{10}^{-5}}mo{{l}^{2}}\,{{L}^{-2}}$
So, the molar solubility (s) can be calculated as,
s = $\sqrt{2.4\times {{10}^{-5}}}$ = 4.9$\times {{10}^{-3}}$ mol/L
Therefore, the molar solubility is 4.9$\times {{10}^{-3}}$mol/L
B. Mass solubility is the solubility of the salt by mass. For this, the mass of one mole of calcium sulfate is multiplied by the molar solubility of the salt.
The molar mass of 1 mole of $CaS{{O}_{4}}$= 136.1 g/mol
So, mass solubility = mass of 1 mole of $CaS{{O}_{4}}$$\times $ molar solubility
Mass solubility = 136.1 g/mol$\times $4.9$\times {{10}^{-3}}$mol/L
Mass solubility = 0.067 g/L
Hence, the mass solubility is 0.067 g/L

Note:
We are assuming the salt of calcium sulfate to be anhydrous, as it has to be dissolved in a solution, so the mass of calcium sulfate is taken in anhydrous form as 136.1 g/mol.