Question

What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? ( For calcium sulphate, ${{K}_{sp}}=\text{ }9.1\times {{10}^{-6}}$ ).

Answer 1

Hint: The solubility product of an electrolyte is equal to the product of ionic concentrations of the electrolyte and for calcium sulphate the solubility product is as follows.
\[{{K}_{sp}}=\left[ C{{a}^{2+}} \right]\left[ SO_{4}^{2-} \right]\]
${{K}_{sp}}$ = Solubility product
$\left[ C{{a}^{2+}} \right]$ = Concentration of the calcium
$\left[ SO_{4}^{2-} \right]$ = Concentration of the sulphate

Complete step by step solution:
- In the question it is asked to find the minimum volume of water required to dissolve 1 g of calcium sulphate.
- The given solubility product of calcium sulphate is ${{K}_{sp}}=\text{ }9.1\times {{10}^{-6}}$
- The given mass of the calcium sulphate is 1 g.
- When calcium is added to water it dissolves and the ionic parts are in equilibrium with the calcium sulphate and it is represented as follows.
\[CaS{{O}_{4}}\rightleftarrows C{{a}^{2+}}+SO_{4}^{2-}\]
- The solubility product of the above chemical equation is as follows.
\[{{K}_{sp}} = \left[ C{{a}^{2+}} \right]\left[ SO_{4}^{2-} \right]\]
${{K}_{sp}}$ = Solubility product
$\left[ C{{a}^{2+}} \right]$ = Concentration of the calcium
$\left[ SO_{4}^{2-} \right]$ = Concentration of the sulphate

- Assume the solubility of the calcium sulphate is S
- Then
\[\begin{align}
  & {{K}_{sp}}={{S}^{2}} \\
 & 9.1\times {{10}^{-6}}={{S}^{2}} \\
 & S = 3.02\times {{10}^{-3}}mol/l \\
\end{align}\]

- Molecular weight of calcium sulphate is 136 g/mol.
- Therefore the solubility of calcium sulphate = $3.02\times {{10}^{-3}}\times 136 = 0.41g/l$
- Means we need one liter of water to dissolve 0.41 g of calcium sulphate.
- To dissolve 1 g calcium sulphate we need
\[\begin{align}
 & =\frac{1}{0.41} \\
 & = 2.44l \\
\end{align}\]
- We need 2.44 liter of water to dissolve 1 g of calcium sulphate.

Note: Initially we have to calculate in one liter of water how much calcium sulphate is going to get soluble. Then only we will get how much water is needed to dissolve 1 g of calcium sulphate.