## Chemistry: The Molecular Science (5th Edition)

Hydroxide ion concentration: $9.4 \times 10^{-3}M$.
1. Calculate the molar mass: $Mg$ = 24.31g/mol 2. Calculate the number of moles $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 5\times 10^{- 6}}{ 24.31}$ $n(moles) = 2.057\times 10^{- 7}$ 3. Find the concentration in mol/L: $2.057 \times 10^{-7}$ mol in 1L: $2.057 \times 10^{-7} M$ 4. Write the $K_{sp}$ expression: $Mg(OH)_2(s) \lt -- \gt 1Mg^{2+}(aq) + 2OH^{-}(aq)$ $1.8 \times 10^{-11} = [Mg^{2+}]^ 1[OH^{-}]^ 2$ 5. Find the $OH^{-}$ concentration. $1.8 \times 10^{-11}= ( 2.057 \times 10^{-7})^ 1 \times ( [OH^{-}])^ 2$ $1.8 \times 10^{-11}= (2.057 \times 10^{-7})^ 1 \times ([OH^{-}])^ 2$ $1.8 \times 10^{-11}= 2.057 \times 10^{-7} \times ([OH^{-}])^ 2$ $\frac{1.8 \times 10^{-11}}{2.057 \times 10^{-7}} = ([OH^{-}])^ 2$ $8.752 \times 10^{-5} = ([OH^{-}])^ 2$ $\sqrt [ 2] {8.752 \times 10^{-5}} = [OH^{-}]$ $9.355 \times 10^{-3} = [OH^{-}]$