Question

The cations in an aqueous solution that contains 0.150 M Ba(NO₃)₂ and 0.0800M Ca(NO₃)₂ are to be separated by taking advantage of the difference in the solubilities of their sulfates. K_sp (BaSO₄) = 1.1x10^-10 and K_sp(CaSO₄) = 2.4x10^-5. What should be the concentration of sulfate ion for the best separation?

Short answer

The concentration of sulfate ion for separation is $\boxed{3.0X{10}^{-4}M}$, and therefore $\boxed{B{a}^{2+}}$ ion is present in the solution.

Step-by-step solution

Step(1):- we have the concentration of Ba2+ , Ca2+ are 0.150M and 0.800M respectively.

The equation for the solubility of BaSO₄ and CaSO₄ are

Calcium sulfate CaSO₄ is for more soluble than barium sulfate BaSO₄ It is possible to add enough sulfate ions $\left(S{O}_4^{2-}\right)$to precipitate almost at the Ba²⁺ ions but leaves Ca²⁺ in the solution.

Step(2):- Finding Concentration

For Ca²⁺ ion to remain in solution, its reaction quotient must smaller K_SP, that is

$Q=\left[C{a}^{2+}\right]\left[S{O}_4^{2-}\right]<K_{SP}$

Inserting the value of K_SP and the concentration of Ca²⁺ gives, that is

$$\begin{gathered} \left[S{O}_4^{2-}\right]<\frac{K_{SP}}{\left[C{a}^{2+}\right]} \\ \left[S{O}_4^{2-}\right]<\frac{2.4X{10}^{-5}}{0.0800} \\ \left[S{O}_4^{2-}\right]<3.0X{10}^{-4}M \end{gathered}$$

Thus, the concentration of sulfate ion for separation is $\boxed{3.0X{10}^{-4}M}$, and therefore$\boxed{B{a}^{2+}}$ion is present in the solution.