Question

Is it possible to precipitate $\mathbf{99}\mathbf{.}\mathbf{0}\mathbf{\%}$of role="math" localid="1663573455203" $\mathbf{0}\mathbf{.}\mathbf{010}{\mathbf{MCe}}^{\mathbf{3}\mathbf{+}}$by adding oxalate $\mathbf{\left(}{\mathbf{C}}_{\mathbf{2}}{{\mathbf{O}}^{\mathbf{2}}}_{\mathbf{4}}\mathbf{\right)}$without precipitating $\mathbf{0}\mathbf{.}\mathbf{010}{\mathbf{MCe}}^{\mathbf{2}\mathbf{+}}$?

role="math" localid="1663574065069" $$\begin{gathered} {\mathbf{CaC}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{K}}_{\mathbf{sp}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{{}_{\mathbf{-}\mathbf{8}}} \\ {\mathbf{Ce}}_{\mathbf{2}}\mathbf{}\mathbf{(}\mathbf{C}\mathbf{2}\mathbf{}\mathbf{O}\mathbf{4}\mathbf{}\mathbf{)}\mathbf{3}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{K}}_{\mathbf{sp}}\mathbf{}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{30}} \end{gathered}$$

Short answer

There is no such thing as a participant formation.

Step-by-step solution

Step 1:

The possibility to precipitate $\mathbf{99}\mathbf{\%}$of $\mathbf{0}\mathbf{.}\mathbf{010}{\mathbf{MCe}}^{\mathbf{3}\mathbf{+}}$by adding oxalate without precipitating $\mathbf{0}\mathbf{.}\mathbf{010}{\mathbf{MCe}}^{\mathbf{2}\mathbf{+}}$must be predicted.

check for precipitate

Given

$$\begin{gathered} {\mathrm{CaC}}_2{\mathrm{O}}_4{\mathrm{K}}_{\mathrm{sp}}=1.3\times {10}^{{}_{-8}} \\ {\mathrm{Ce}}_2(\mathrm{C}2\mathrm{O}4)3{\mathrm{K}}_{\mathrm{sp}}=5.9\times {10}^{-30} \end{gathered}$$

There is a need to reduce concentration of ${\mathrm{Ce}}^{3+}$to$1.0\%\mathrm{of}$ $0.010\mathrm{M}(=0.00010\mathrm{M})$. The concentration of oxalate in equilibrium WITH $0.00010{\mathrm{MCe}}^{3+}$is calculated as follows,

${\left[{\mathrm{Ce}}^{3+}\right]}^2[{\mathrm{C}}_2{\mathrm{O}}_4^2{]}^3=\mathrm{Ksp}"=5.9\times {10}^{-30}$

${(0.00010)}^2[{\mathrm{C}}_2{\mathrm{O}}_4^2{]}^3=5.9\times {10}^{-30}$

$\left[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right]=\left(\frac{5.9\times \times {10}^{-30}}{{\left(0.0001\right)}^2}\right)=8.4\times {10}^{-8}\mathrm{M}$

If $8.4\times {10}^{-8}{{\mathrm{MC}}_4}^{2-}$will precipitate $0.010{\mathrm{MCa}}^{2+}$, calculate $\mathrm{Q}$for ${\mathrm{CaC}}_2{\mathrm{O}}_4$:

$$\begin{gathered} \left[{\mathrm{Ca}}^{2+}]=\right[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}]= \\ (0.010\left)\right(8.4\times {10}^{-8}) \\ =8.4\times {10}^{-8} \end{gathered}$$

Because$\mathrm{Q}<{\mathrm{K}}_{\mathrm{sp}}$for${\mathrm{CaC}}_2{\mathrm{O}}_4=1.3\times {10}^{-8}{\mathrm{Ca}}^{2+}$, so will not precipitate.