Question

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

Short answer

(a) The molar solubility of ${\text{BaSrO}}_{\text{4}}$in pure water is $\mathrm{S}=1.4\times {10}^{-5}\mathrm{M}$.

(b) The molar solubility of localid="1663344591325" ${\text{BaSrO}}_{\text{4}}$in $1.5\times {10}^{-3}\mathrm{M}{\mathrm{Na}}_2{\mathrm{CrO}}_4$is $\mathrm{S}=1.4\times {10}^{-7}\mathrm{M}$.

Step-by-step solution

Concept Introduction

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.

Solubility of  BaSrO4 in Pure water

(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 $\mathrm{S}=1.4\times {10}^{-5}\mathrm{M}$.

 Solubility of BaSrO4  in Na2CrO4

(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" $\mathrm{S}=1.4\times {10}^{-7}\mathrm{M}$.