Answer
The necessary concentration of hydroxide ion for precipitation of magnesium hydroxide must be greater than $1.0 \times 10^{-5}M$
Work Step by Step
1. Calculate the molar mass $(Mg)$:
24.31* 1 = 24.31g/mol
2. Calculate the number of moles $(Mg)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 1.35}{ 24.31}$
$n(moles) = 0.0555$
3. Find the concentration in mol/L $(Mg)$:
$0.0555$ mol in 1L: $0.0555 M (Mg^{2+})$
4. Write the $K_{sp}$ expression:
$ Mg(OH)_2(s) \lt -- \gt 1Mg^{2+}(aq) + 2OH^{-}(aq)$
$ K_{sp} = [Mg^{2+}]^ 1[OH^{-}]^ 2$
$ 5.6 \times 10^{-12} = 0.0555 \times [OH^-]^2$
$1.0 \times 10^{-10} = [OH^-]^2$
$1.0 \times 10^{-5} = [OH^-]$
