Q19.74P - Chemistry: Molecular Nature Of Matter And Change

Question

Find the molar solubility of ${\mathbf{SrCO}}_{\mathbf{3}}\mathbf{(}{\mathbf{K}}_{\mathbf{sp}}\mathbf{=}\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{)}$  in 

(a) pure water and 

(b)  $\mathbf{0}\mathbf{.}\mathbf{13}\mathbf{MSr}\mathbf{(}{\mathbf{NO}}_{\mathbf{3}}{\mathbf{)}}_{\mathbf{2}}$.

Step-by-Step Solution

Answer

(a) The molar solubility of ${\text{SrCO}}_{\text{3}}$  in pure water is $2.3\times {10}^{-5}\mathrm{M}$ .

(b) The molar solubility of  ${\text{SrCO}}_{\text{3}}$ in ${\text{0.13 M Sr(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is  $4.1\times {10}^{-9}\mathrm{M}$.

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).

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

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 SrCO 3 in Pure water

(a)

Write the equation for dissolution of ${\text{SrCO}}_{\text{3}}$–

 ${\text{SrCO}}_{\text{3}}\text{(s)}\iff {\text{Sr}}^{\text{2 + }}{\text{(aq) + CO}}_{\text{3}}^{\text{2 - }}\text{(aq)}$

The equilibrium concentration is represented as –

 $$\begin{gathered} {\text{[Sr}}^{\text{2 + }}\text{]=S} \\ {\text{[CO}}_{\text{3}}^{\text{2 - }}\text{]=S} \end{gathered}$$

Rearranging the equation of   and solving –

  $$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Sr}}^{2+}\right]\left[{\mathrm{CO}}_3^{2-}\right] \\ {\mathrm{K}}_{\mathrm{sp}}={\mathrm{S}}^2 \\ {\mathrm{K}}_{\mathrm{sp}}=5.4\times {10}^{-10} \\ \mathrm{S}=\sqrt{{\mathrm{K}}_{\mathrm{sp}}} \\ \mathrm{S}=2.3\times {10}^{-5}\mathrm{M} \end{gathered}$$

Therefore, the value for solubility is obtained as $2.3\times {10}^{-5}\mathrm{M}$ .

Solubility of SrCO 3 in Sr(NO 3 ) 2

(b)

In this case, the initial concentration of strontium is  $\text{0.13 M}$, so the equilibrium concentrations will be –

$$\begin{gathered} {\text{[Sr}}^{\text{2 + }}\text{]=0.13 M + S} \\ {\text{[CO}}_{\text{3}}^{\text{2 - }}\text{]=S} \end{gathered}$$ 

Rearranging the equation of  K_sp and solving –

  $$\begin{gathered} K_{\mathrm{sp}}=\left[{\mathrm{Sr}}^{2+}\right]\left[{\mathrm{CO}}_3^{2-}\right] \\ {\mathrm{K}}_{\mathrm{sp}}=(0.13\mathrm{M}+\mathrm{S})·\mathrm{S} \\ {\mathrm{K}}_{\mathrm{sp}}=5.4\times {10}^{-10} \\ \mathrm{S}=4.1\times {10}^{-9}\mathrm{M} \end{gathered}$$

 

Therefore, the value for solubility is obtained as $S=4.1\times {10}^{-9}M.$ .