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).

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

The 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 of dissolution for ${\text{Ca(IO}}_{\text{3}}{\text{)}}_{\text{2}}$–

${\text{Ca(IO}}_{\text{3}}{\text{)}}_{\text{2}}\text{(s)}⇔{\text{Ca}}^{\text{2 +}}{\text{(aq) + 2IO}}_3^{\text{-}}\text{(aq)}$

In this case, the initial concentration of ${\text{Ca}}^{\text{2 +}}$ion is $\text{0.06 M + S}$, as there is${\text{0.06 M Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ so the equilibrium concentrations will be –

$$\begin{gathered} {\text{[Ca}}^{\text{2 +}}\text{]=0.06M + S} \\ {\text{[IO}}_3^{\text{-}}\text{]=2S} \end{gathered}$$

Rearranging the equation of ${\text{K}}_{\text{sp}}$and solving –

Square ${\text{[IO}}_3^{\text{-}}\text{]}$as there are $2$moles of ${\text{IO}}_3^{\text{-}}$in the equation.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ca}}^{2+}\right]\left[{\mathrm{IO}}_3^{-}\right] \\ {\mathrm{K}}_{\mathrm{sp}}=(0.06\mathrm{M}+\mathrm{S})·(2\mathrm{S}{)}^2 \\ {\mathrm{K}}_{\mathrm{sp}}=7.1×{10}^{-7} \\ \mathrm{S}=1.7×{10}^{-3}\mathrm{M} \end{gathered}$$

Therefore, the value for solubility is obtained as $\mathrm{S}=1.7×{10}^{-3}\mathrm{M}$.

(b)

In this case, the initial concentration of ${\text{IO}}_3^{\text{2 -}}$ion is $\text{0.06 M + 2S}$, as there is ${\text{0.06 M NaIO}}_{\text{3}}$so the equilibrium concentrations will be –

role="math" localid="1663349633177" $$\begin{gathered} \left[{\mathrm{Ca}}^{2+}\right]=\mathrm{S} \\ \left[{\mathrm{IO}}_3^{2-}\right]=0.06\mathrm{M}+2\mathrm{S} \end{gathered}$$

Rearranging the equation of K_spand solving –

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ca}}^{2+}\right]\left[{\mathrm{IO}}_3^{2-}\right] \\ {\mathrm{K}}_{\mathrm{sp}}=\mathrm{S}·(0.06\mathrm{M}+2\mathrm{S}{)}^2 \\ {\mathrm{K}}_{\mathrm{sp}}=7.1×{10}^{-7} \\ \mathrm{S}=1.97×{10}^{-4}\mathrm{M} \end{gathered}$$

Therefore, the value for solubility is obtained as $\mathrm{S}=1.97×{10}^{-4}\mathrm{M}$.