Answer
For a saturated zinc hydroxide solution:
$pH = 9.53$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ Zn(OH)_2(s) \lt -- \gt 1Zn^{2+}(aq) + 2OH^{-}(aq)$
$1.8 \times 10^{-14} = [Zn^{2+}]^ 1[OH^{-}]^ 2$
2. Considering a pure solution: $[Zn^{2+}] = 1S$ and $[OH^{-}] = 2S$
$1.8 \times 10^{-14}= ( 1S)^ 1 \times ( 2S)^ 2$
$1.8 \times 10^{-14} = 4S^ 3$
$4.5 \times 10^{-15} = S^ 3$
$ \sqrt [ 3] {4.5 \times 10^{-15}} = S$
$1.7 \times 10^{-5} = S$
- This is the molar solubility value for this salt.
Since each molecule has 2 hydroxide ions:
$[OH^-] = 2 * 1.7 \times 10^{-5} = 3.4 \times 10^{-5}$
3. Calculate the pOH:
$pOH = -log[OH^-]$
$pOH = -log( 3.4 \times 10^{- 5})$
$pOH = 4.47$
4. Find the pH:
$pH + pOH = 14$
$pH + 4.47 = 14$
$pH = 9.53$
