91Ó°ÊÓ

The solubility product is directly related to the maximum number of moles of the solute that can be dissolved in a litre of solution before the solution becomes saturated. Any additional solute precipitates out of a solution once it is saturated.

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 saturatedQ_sp value is called K_sp value (solubility-product constant).

${\mathbf{\text{MX}}}_{\mathbf{\text{2}}}\mathbf{→}{\mathbf{\text{M}}}^{\mathbf{\text{2 +}}}{\mathbf{\text{+ 2X}}}^{\mathbf{\text{-}}}$

Solid and liquid state is not included in the K_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 the dissolution of some compound.

(a)

Write the equation for dissolution of${\text{SrCO}}_{\text{3}}$–

${\text{SrCO}}_{\text{3}}\text{(s)}⇔{\text{Sr}}^{\text{2 +}}{\text{(aq) + CO}}_{\text{3}}^{\text{2 -}}\text{(aq)}$

The equilibrium concentration is represented as –

$$\begin{gathered} {\text{[Sr}}^{\text{2 +}}\text{]=S} \\ {\text{[CO}}_{\text{3}}^{\text{2 -}}\text{]=S} \end{gathered}$$

Rearranging the equation of and solving –

role="math" localid="1663326003447" $$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Sr}}^{2+}\right]\left[{\mathrm{CO}}_3^{2-}\right] \\ {\mathrm{K}}_{\mathrm{sp}}={\mathrm{S}}^2 \\ {\mathrm{K}}_{\mathrm{sp}}=5.4×{10}^{-10} \\ \mathrm{S}=\sqrt{{\mathrm{K}}_{\mathrm{sp}}} \\ \mathrm{S}=2.3×{10}^{-5}\mathrm{M} \end{gathered}$$

Therefore, the value for solubility is obtained as $2.3×{10}^{-5}\mathrm{M}$.

(b)

In this case, the initial concentration of strontium is $\text{0.13 M}$, so the equilibrium concentrations will be –

role="math" localid="1663326716908" $$\begin{gathered} {\text{[Sr}}^{\text{2 +}}\text{]=0.13 M + S} \\ {\text{[CO}}_{\text{3}}^{\text{2 -}}\text{]=S} \end{gathered}$$

Rearranging the equation of K_sp and solving –

role="math" localid="1663326773418" $$\begin{gathered} K_{\mathrm{sp}}=\left[{\mathrm{Sr}}^{2+}\right]\left[{\mathrm{CO}}_3^{2-}\right] \\ {\mathrm{K}}_{\mathrm{sp}}=(0.13\mathrm{M}+\mathrm{S})·\mathrm{S} \\ {\mathrm{K}}_{\mathrm{sp}}=5.4×{10}^{-10} \\ \mathrm{S}=4.1×{10}^{-9}\mathrm{M} \end{gathered}$$

Therefore, the value for solubility is obtained as role="math" localid="1663326829791" $S=4.1×{10}^{-9}M.$.