Q19.75P - Chemistry: Molecular Nature Of Matter And Change

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

Step-by-Step Solution

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  ${\text{BaSrO}}_{\text{4}}$ in $1.5\times {10}^{-3}\mathrm{M}{\mathrm{Na}}_2{\mathrm{CrO}}_4$ is  $\mathrm{S}=1.4\times {10}^{-7}\mathrm{M}$.

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_sp value is obtained when the concentrations of ions formed by dissolution of some compound are multiplied. When the solution is saturated Q_sp value 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_sp equations.

${\mathbf{\text{K}}}_{\mathbf{\text{sp}}}{\mathbf{\text{ = [M}}}^{\mathbf{\text{2 + }}}{\mathbf{\text{][X}}}^{\mathbf{\text{ - }}}{\mathbf{\text{]}}}^{\mathbf{\text{2}}}$ 

 S (Molar solubility) is equal to the concentration of one mol of the ion formed by the dissolution of some compound.

Solubility of BaSrO 4 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 BaSrO 4 in Na 2 CrO 4

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