Question

An aqueous solution at 25°C is 0.10 M in both Mg²⁺ and Pb²⁺ ions. We wish to separate the two kinds of metal ions by taking advantage of the different solubilities of their oxalates, MgC₂O₄ and PbC₂O₄ .

(a) What is the highest possible oxalate ion concentration that allows only one solid oxalate salt to be present at equilibrium? Which ion is present in the solid— Mg²⁺ or Pb²⁺ ?

(b) What fraction of the less soluble ion still remains in solution under the conditions of part (a)?

Short answer

1. The possible oxalate ion concentration is $\boxed{8.6X{10}^{-4}M}$there is no magnesium oxalate precipitated, and here $\boxed{P{b}^{2+}}$ion present as solid.
2. The fraction of ion remains in solution is .

Step-by-step solution

Step(1):-Finding Oxalate ion Concentration

(a) The equation for solubility products of MgC₂O₄ and PbC₂O₄ are:-

$$\begin{gathered} Mg{C}_2{O}_4\left(s\right)\to M{g}^{2+}\left(aq\right)+C_2{O}_2^{2-}\left(aq\right);K_{SP}=8.6X{10}^{-5} \\ Pb{C}_2{O}_4\left(s\right)\to P{b}^{2+}\left(aq\right)+C_2{O}_2^{2-}\left(aq\right);K_{SP}=2.7X{10}^{-11} \end{gathered}$$

Magnesium oxalate MgC₂O₄ is for more soluble than lead oxalate PbC₂O₄. It is possible to add enough oxalate ions (C₂O₄^2-) to precipitate almost all the Pb²⁺ ions but leave the Mg²⁺ions solution.

For Mg²⁺ ion to remain in solution, its reaction quotient must smaller than K_SP , that is

$Q=\left[M{g}^{2+}\right]\left[C_2{O}_4^{2-}\right]<K_{SP}$

Inserting the value of $K_{SP}$ and the concentration of Mg²⁺ gives:

$$\begin{gathered} \left[C_2{O}_4\right]<\frac{K_{SP}}{\left[M{g}^{2+}\right]} \\ \left[C_2{O}_4\right]<\frac{8.6X{10}^{-5}}{0.10} \\ \left[C_2{O}_4\right]<8.6X{10}^{-4} \end{gathered}$$

Therefore, the possible oxalate ion concentration is role="math" localid="1663410799251" $\boxed{8.6X{10}^{-4}M}$there is no magnesium oxalate precipitated, and here $\boxed{P{b}^{2+}}$ion present as solid.

Step(2):-Finding the fraction of Pb2+ ion

(b) To reduce the lead ion concentration in solution as for as possible (that’s, to precipitate out as much lead oxalate as possible), the oxalate ion concentration should keep as high as possible without exceeding ,$\boxed{8.6X{10}^{-4}M}$ ,if exactly this concentration (C₂O₄^2-) is chosen, then at equilibrium.

$\begin{array}{rcl}\left[P{b}^{2+}\right]\left[C_2{O_4}^{2-}\right]& =& K_{SP}\\ \left[P{b}^{2+}\right]& =& \frac{K}{\left[C_2{O_4}^{2-}\right]}\\ & =& \frac{2.7X{10}^{-11}}{8.6X{10}^{-4}}\\ & =& 3.1X{10}^{-8}M\end{array}$

At that concentration of (C₂O₄^2-) the concentration of Pb²⁺ ion has been reduced to 3.1X10^-8 from the original concentration of 0.10M.

The fraction of Pb²⁺ ion remains in solution is

$$\begin{gathered} =\frac{3.1X{10}^{-8}M}{0.10M} \\ =3.1X{10}^{-7} \end{gathered}$$

Therefore, the fraction of Pb²⁺ ion remains in solution is $\boxed{3.1X{10}^{-7}}$.