Answer :
To determine the pH of a saturated solution of magnesium hydroxide (Mg(OH)2), we need to consider the dissociation of the compound in water.
Magnesium hydroxide dissociates in water as follows:
\[Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^-\]
The dissociation constant for magnesium hydroxide (\(K_{\text{sp}}\)) is \(5.6 \times 10^{-12}\). In a saturated solution of magnesium hydroxide, the concentrations of \(Mg^{2+}\) and \(OH^-\) ions will be determined by the solubility product.
Since magnesium hydroxide dissociates to form twice as many hydroxide ions (\(OH^-\)) as magnesium ions (\(Mg^{2+}\)), we can represent the equilibrium concentrations as follows:
\[Mg(OH)_2 \rightarrow Mg^{2+} + 2OH^-\]
Let's denote the concentration of Mg^{2+} ions as \(x\) and the concentration of OH^- ions as \(2x\) (due to the stoichiometry of the reaction).
Now, we can set up the equilibrium expression and solve for \(x\):
\[K_{\text{sp}} = [Mg^{2+}][OH^-]^2\]
Substituting the values:
\[5.6 \times 10^{-12} = x \times (2x)^2\]
\[5.6 \times 10^{-12} = 4x^3\]
Solving for \(x\):
\[x = \sqrt[3]{\frac{5.6 \times 10^{-12}}{4}}\]
\[x \approx 5.3 \times 10^{-5} \, \text{M}\]
Since the concentration of hydroxide ions is \(2x\), we have:
\[2x \approx 2 \times 5.3 \times 10^{-5} \, \text{M}\]
\[2x \approx 1.06 \times 10^{-4} \, \text{M}\]
Now, we can calculate the pOH (negative logarithm of the hydroxide ion concentration):
\[pOH = -\log(1.06 \times 10^{-4})\]
\[pOH \approx 3.974\]
Finally, we can find the pH using the relation:
\[pH + pOH = 14\]
\[pH = 14 - pOH\]
\[pH \approx 14 - 3.974\]
\[pH \approx 10.026\]
Therefore, the pH of a saturated solution of magnesium hydroxide is approximately 10.03.
