Question

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

Answer 1

$CaS{O}_{4\left(S\right)}\leftrightarrow C{a}_{\left(aq\right)}^{2+}+S{O}_{4\left(aq\right)}^{2-}{K}_{sp}=\left[C{a}^{2+}\right]\left[S{O}_4^{2-}\right]$
Let the solubility of $CaS{O}_4$ be s.
Then, $k_{SP}=s^{2}9.1\times {10}^{-6}=s^{2}s=3.02\times {10}^{-3}mol/L$
Molecular mass of $CaS{O}_4=136$ g/mol
Solubility of in gram/L
$=3.02\times {10}^{-3}\times 136=0.41g/L$
This means that we need 1L of water to dissolve 0.41g of $CaS{O}_4$
Therefore, to dissolve 1g of $CaS{O}_4$ we require $=\frac{1}{0.41}L=2.44L$ of water.