Explanation:
Suppose the solubility of [tex]Hg_{2}Br_{2}[/tex] is x. Hence, upon dissociation equilibrium reaction for [tex]Hg_{2}Br_{2}[/tex] will be as follows.
[tex]Hg_{2}Br_{2}(s) \rightleftharpoons Hg^{2+}_{2}(aq) + 2Br^{-}(aq)[/tex]
At equilibrium: x 2x
Therefore, equation for [tex]K_{sp}[/tex] will be as follows.
[tex]K_{sp}[/tex] = [tex][Hg^{2+}_{2}][Br^{-}]^{2}[/tex]
= [tex](x) \times (2x)^{2}[/tex]
= [tex]4x^{3}[/tex]
x = [tex](\frac{K_{sp}}{4})^{1/3}[/tex]
x = [tex](\frac{6 \times 10^{-23}}{4})^{1/3}[/tex]
= [tex]2 \times 10^{-8}[/tex]
Thus, we can conclude that the water solubility of [tex]Hg_{2}Br_{2}[/tex] is [tex]2 \times 10^{-8}[/tex].
