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_spvalue is obtained when the concentrations of ions formed by dissolution of some compound are multiplied. When the solution is saturated Q_spvalue is called K_spvalue (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 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.

(a)

Write the equation of dissolution for${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$–

${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(s)}⇔2{\text{Ag}}^{\text{+}}{\text{(aq) + SO}}_4^{\text{2 -}}\text{(aq)}$

In this case, the initial concentration of ${\text{Ag}}^{\text{+}}$ion is $\text{0.22 M + S}$, as there is ${\text{0.22 M AgNO}}_{\text{3}}$so the equilibrium concentrations will be –

role="math" localid="1663351097110" $$\begin{gathered} \left[{\mathrm{Ag}}^{+}\right]=0.22\mathrm{M}+2\mathrm{S} \\ \left[{\mathrm{SO}}_4^{2-}\right]=\mathrm{S} \end{gathered}$$

Rearranging the equation of K_spand solving –

Square ${\text{[Ag}}^{\text{+}}\text{]}$as there are $2$moles of ${\text{Ag}}^{\text{+}}$in the equation.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{SO}}_4^{2-}\right] \\ {\mathrm{K}}_{\mathrm{sp}}=(0.22\mathrm{M}+2\mathrm{S}{)}^2·\mathrm{S} \\ {\mathrm{K}}_{\mathrm{sp}}=1.5×{10}^{-5} \\ \mathrm{S}=3.1×{10}^{-4}\mathrm{M} \end{gathered}$$

Therefore, the value for solubility is obtained as $\mathrm{S}=3.1×{10}^{-4}\mathrm{M}$.

(b)

In this case, the initial concentration of ${\text{SO}}_4^{\text{2 -}}$ion is $\text{0.22 M + S}$, so the equilibrium concentrations will be –

$$\begin{gathered} \left[{\mathrm{Ag}}^{+}\right]=2\mathrm{S} \\ \left[{\mathrm{SO}}_4^{2-}\right]=0.22\mathrm{M}+\mathrm{S} \end{gathered}$$

Rearranging the equation of K_spand solving –

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{SO}}_4^{2-}\right] \\ {\mathrm{K}}_{\mathrm{sp}}=(2\mathrm{S}{)}^2·(0.22\mathrm{M}+\mathrm{S}) \\ {\mathrm{K}}_{\mathrm{sp}}=1.5×{10}^{-5} \\ \mathrm{S}=4.1×{10}^{-3}\mathrm{M} \end{gathered}$$

Therefore, the value for solubility is obtained as $\mathrm{S}=4.1×{10}^{-3}\mathrm{M}$.