Answer:
All three hydroxides will precipitate
Explanation:
General solubility equilibrium of bivalent metal hydroxides is represented as-
[tex]M(OH)_{2}\rightleftharpoons M^{2+}+2OH^{-}[/tex]
solubility product, [tex]K_{sp}=[M^{2+}][OH^{-}]^{2}[/tex]
If , at pH 8.50, product of [tex][M^{2+}][/tex] and [tex][OH^{-}]^{2}[/tex] exceeds [tex]K_{sp}[/tex] values of given hydroxides then hydroxides of given metal ions will precipitate.
pH = 8.50
or, pOH = 14-8.50 = 5.50
or, [tex]-log[OH^{-}][/tex] = 5.50
or, [tex][OH^{-}][/tex] = [tex]10^{-5.50}[/tex]
or, [tex][OH^{-}][/tex] = [tex]3.16\times 10^{-6}[/tex]
As each salt solutions have concentration of 0.010 M therefore concentrations of each metal ions ([tex]M^{2+}[/tex]) are 0.010 M
So, [tex][M^{2+}][OH^{-}]^{2}=0.010\times (3.16\times 10^{-6})^{2}=3.16\times 10^{-8}[/tex]
As product of [tex][M^{2+}][/tex] and [tex][OH^{-}]^{2}[/tex] exceeds [tex]K_{sp}[/tex] values of every hydroxides therefore all three hydroxides will precipitate.
