The solubility of calcium sulphate at ${\mathbf{\text{30}}}^{\mathbf{\text{o}}}\mathbf{\text{C}}$is $\mathbf{\text{0.209 g/100 mL}}$solution. Calculate its ${\mathbf{K}}_{\mathbf{sp}}$.

Q_sp- ion-product expression; Q_sp value is obtained when the concentrations of ions formed by dissolution of some compound are multiplied. When the solution is saturated Q_sp value is called K_sp value (solubility-product constant).

$\mathit{M}{\mathit{X}}_{\mathbf{2}}\mathbf{\to }{\mathit{M}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{2}{\mathit{X}}^{\mathbf{-}}$

Solid and liquid state is not included in theK_spequations.

${\mathbf{\text{K}}}_{\mathbf{\text{sp}}}{\mathbf{\text{= [M}}}^{\mathbf{\text{2 +}}}{\mathbf{\text{][X}}}^{\mathbf{\text{-}}}{\mathbf{\text{]}}}^{\mathbf{\text{2}}}$

S (Molar solubility) is equal to the concentration of one mol of the ion formed by dissolution of some compound.

Write the dissolution equation for${\mathbf{\text{CaSO}}}_{\mathbf{\text{4}}}$–

${\text{CaSO}}_{\text{4}}\text{(s)}\iff {\text{Ca}}^{\text{2 +}}{\text{(aq) + SO}}_{\text{4}}^{\text{2 -}}\text{(aq)}$

The solubility is obtained as –

$$\begin{gathered} \mathrm{So}\mathrm{lubility}=0.209\mathrm{g}/100\mathrm{mL} \\ =0.00209\mathrm{g}/\mathrm{mL} \end{gathered}$$

Use molar mass to convert solubility into molar solubility –

$$\begin{gathered} S=0.00209\mathrm{g}/\mathrm{mL}\times \frac{1\mathrm{mol}}{136.14\mathrm{g}}\times \frac{1000\mathrm{mL}}{1\mathrm{L}} \\ =0.0154\mathrm{M} \end{gathered}$$

Now, calculate the value for K_sp as –

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ca}}^{2+}\right]\left[{\mathrm{SO}}_4^{2-}\right] \\ {\mathrm{S}}_{\left[{\mathrm{Ca}}^{2+}\right]}=0.0154\mathrm{M} \\ {\mathrm{S}}_{\left[{\mathrm{SO}}_4^{2-}\right]}=0.0154\mathrm{M} \\ {\mathrm{K}}_{\mathrm{sp}}={\mathrm{S}}^2 \\ {\mathrm{K}}_{\mathrm{sp}}=2.36\times {10}^{-4} \end{gathered}$$

Therefore, the value for K_sp is obtained as $2.36\times {10}^{-4}$.