Question

An aqueous solution at 25°C is 0.10 M in Ba²⁺ and 0.50 M in Ca²⁺ ions. We wish to separate the two by taking advantage of the different solubilities of their fluorides, BaF₂ and CaF₂ .

(a) What is the highest possible fluoride ion concentration that allows only one solid fluoride salt to be present at equilibrium? Which ion is present in the solid— Ba²⁺ or Ca²⁺ ?

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

Short answer

The highest possible fluoride ion concentration is $\boxed{4.1x{10}^{-7}M}$

Step-by-step solution

Step(1): Finding highest possible fluoride ion concentration.

(a) The equation for solubility products of BaF₂ and CaF₂ . It is possible to add enough fluoride ion [F^-] to precipitate almost all the Ca²⁺ ions but leave the Ba²⁺ in the solution.

For Ba²⁺ion to remain in solution, its reaction quotient must be smaller than K_SP , that is

$Q=\left[B{a}^{2+}\right]{\left[F^{-}\right]}^2<K_{SP}$

Substitute the value of K_SP and the concentration of Ba²⁺ gives the concentration of [F^-] as follows:

$$\begin{gathered} {\left[F^{-}\right]}^2<\frac{K_{SP}}{\left[B{a}^{2+}\right]} \\ {\left[F^{-}\right]}^2<1.7X{10}^{-6} \\ {\left[F^{-}\right]}^2<4.1X{10}^{-3}M \end{gathered}$$

Therefore , the highest possible fluoride ion concentration is $\boxed{4.1X{10}^{-7}M}$and here Ca²⁺ ion present in the solid.

Step(2): Finding the fraction of Ca2+ ion.

(b) The less soluble ions is Ca²⁺ , under the conditions of part(a), that is at this fluoride ion concentration $\boxed{4.1X{10}^{-7}M}$ , calculate the concentration of Ca²⁺ in solution as follows:

$\begin{array}{rcl}\left[C{a}^{2+}\right]{\left[F^{-}\right]}^2& =& K_{SP}\\ \left[C{a}^{2+}\right]& =& \frac{K_{SP}}{{\left[F^{-}\right]}^2}\\ & =& \frac{3.9X{10}^{-11}}{{\left(4.2X{10}^{-6}\right)}^2}\\ & =& 2.3X{10}^{-6}M\end{array}$

The concentration Ca²⁺ ion is 2.3X10^-6 . The original concentration of Ca²⁺ is given as 0.50M.

Calculate the fraction of Ca²⁺ remains in solution as follows:

$\begin{array}{rcl}ThefractionofC{a}^{2+}remainsinsolution& =& \frac{\left[C{a}^{2+}\right]insolution}{total\left[C{a}^{2+}\right]}\\ & =& \frac{2.3X{10}^{-6}M}{0.50M}\\ & =& 4.6X{10}^{-6}\end{array}$

Therefore, the fraction of Ca²⁺ ion remains in the solution is $\boxed{4.6X{10}^{-6}}$.