# Chapter 17 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria - Study Questions - Page 677c: 69

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: $Mg(OH)_2(s) \lt -- \gt 1Mg^{2+}(aq) + 2OH^{-}(aq)$ $K_{sp} = [Mg^{2+}]^ 1[OH^{-}]^ 2$ $5.6 \times 10^{-12} = 0.0555 \times [OH^-]^2$ $1.0 \times 10^{-10} = [OH^-]^2$ $1.0 \times 10^{-5} = [OH^-]$

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