Answer
Hydroxide ion concentration: $9.4 \times 10^{-3}M$.
Work Step by Step
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^{-}]$
$9.355 \times 10^{-3} = [OH^{-}]$
