Answer: [tex]1.70\times 10^{-6}[/tex]
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 [tex]BaF_2[/tex] is given as:
[tex]BaF_2\rightarrow Ba^{2+}+2F^-[/tex]
When the solubility of [tex]BaF_2[/tex] is S moles/liter, then the solubility of [tex]Ba^{2+}[/tex] will be S moles/liter and solubility of [tex]F^-[/tex] will be 2S moles/liter.
By stoichiometry of the reaction:
1 mole of [tex]BaF_2[/tex] gives 1 mole of [tex]Ba^{2+[/tex] and 2 moles of [tex]F^-[/tex]
[tex]K_{sp}=[Ba^{2+}][F^{-}]^2[/tex]
[tex]K_{sp}=s\times (2s)^2[/tex]
[tex]K_{sp}=4\times (7.52\times 10^{-3})^3[/tex]
[tex]K_{sp}=1.70\times 10^{-6}[/tex]
Thus [tex]K_{sp}[/tex] for [tex]BaF_2[/tex] is [tex]1.70\times 10^{-6}[/tex]
