Find the molar solubility of ${\mathbf{BaCrO}}_{\mathbf{4}}\mathbf{(}{\mathbf{K}}_{\mathbf{sp}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{)}$in

(a) pure water and

(b)$\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{M}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{CrO}}_{\mathbf{4}}$ .

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_sp value (solubility-product constant).

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

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 dissolution equation for ${\mathbf{\text{SrCO}}}_{\mathbf{\text{3}}}$ –

${\text{BaSrO}}_{\text{4}}\text{(s)}\iff {\text{Ba}}^{\text{2 +}}{\text{(aq) + CrO}}_4^{\text{2 -}}\text{(aq)}$

The equilibrium concentration is represented as –

$$\begin{gathered} \left[B{a}^{2+}\right]=S \\ \left[{\mathrm{CrO}}_4^{2-}\right]=S \end{gathered}$$

Rearranging the equation of K_sp and solving –

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ba}}^{2+}\right]\left[{\mathrm{CrO}}_4^{2-}\right] \\ {\mathrm{K}}_{\mathrm{sp}}={\mathrm{S}}^2=2.1\times {10}^{-10} \\ \mathrm{S}=\sqrt{{\mathrm{K}}_{\mathrm{sp}}} \\ \mathrm{S}=1.4\times {10}^{-5}\mathrm{M} \end{gathered}$$

Therefore, the value for solubility is obtained as $\mathrm{S}=1.4\times {10}^{-5}\mathrm{M}$.

(b)

In this case, the initial concentration of chromate ion is $1.5\times {10}^{-3}\mathrm{M}$, so the equilibrium concentrations will be –

$$\begin{gathered} \left[{\mathrm{Ba}}^{2+}\right]=\mathrm{S} \\ \left[{\mathrm{CrO}}_4^{2-}\right]=1.5\times {10}^{-3}\mathrm{M}+\mathrm{S} \end{gathered}$$

Rearranging the equation of and solving –

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ba}}^{2+}\right]\left[{\mathrm{CO}}_4^{2-}\right] \\ {\mathrm{K}}_{\mathrm{sp}}=\mathrm{S}·(1.5\times {10}^{-3}\mathrm{M}+\mathrm{S})=2.1\times {10}^{-10} \\ \mathrm{S}=1.4\times {10}^{-7}\mathrm{M} \end{gathered}$$

Therefore, the value for solubility is obtained as localid="1663344663469" $\mathrm{S}=1.4\times {10}^{-7}\mathrm{M}$.