Question

A saturated solution of barium fluoride, BaF2BaF2, was prepared by dissolving solid BaF2BaF2 in water. The concentration of Ba2+Ba2+ ion in the solution was found to be 7.52×10−3 MM . Calculate KspKspK_sp for BaF2BaF2.

Answer 1

Answer: $1.70\times 10^{-6}$

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as

The equation for the ionization of the $BaF_2$ is given as:

$BaF_2\rightarrow Ba^{2+}+2F^-$

When the solubility of $BaF_2$ is S moles/liter, then the solubility of $Ba^{2+}$  will be S moles/liter and solubility of $F^-$ will be 2S moles/liter.

By stoichiometry of the reaction:

1 mole of $BaF_2$ gives 1 mole of $Ba^{2+$ and 2 moles of $F^-$

$K_{sp}=[Ba^{2+}][F^{-}]^2$

$K_{sp}=s\times (2s)^2$

$K_{sp}=4\times (7.52\times 10^{-3})^3$

$K_{sp}=1.70\times 10^{-6}$

Thus $K_{sp}$ for $BaF_2$ is $1.70\times 10^{-6}$