# Chapter 15 - Additional Aqueous Equilibria - Questions for Review and Thought - Topical Questions - Page 693c: 68a

The $K_{sp}$ for $Mg(OH)_2$ at that temperature is equal to $1.8 \times 10^{-11}$

#### Work Step by Step

1. Calculate the molar mass: 24.31* 1 + 2 * ( 16* 1 + 1.01* 1 ) = 58.33g/mol 2. Calculate the number of moles 9.6 mg = 0.0096 g $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 9.6\times 10^{- 3}}{ 58.33}$ $n(moles) = 1.646\times 10^{- 4}$ 3. Find the concentration in mol/L: $1.646 \times 10^{-4}$ mol in 1L: $1.646 \times 10^{-4} M$ 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. Determine the ion concentrations: $[Mg^{2+}] = [Mg(OH)_2] * 1 = [1.646 \times 10^{-4}] * 1 = 1.646 \times 10^{-4}$ $[OH^{-}] = [Mg(OH)_2] * 2 = 3.292 \times 10^{-4}$ 6. Calculate the $K_{sp}$: $K_{sp} = (1.646 \times 10^{-4})^ 1 \times (3.292 \times 10^{-4})^ 2$ $K_{sp} = (1.646 \times 10^{-4}) \times (1.083 \times 10^{-7})$ $K_{sp} = (1.782 \times 10^{-11})$

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