Question

The pH at which Mg(OH)\(_2\) \([K_{sp} = 1 \times 10^{-11}]\) begins to precipitate from a solution containing 0.10 M Mg\(^{2+}\) ions is \( \_\_\_\_ \).

Answer 1

Correct Answer: 9

This problem requires determining the pH at which magnesium hydroxide, Mg(OH)₂, initiates precipitation from a solution containing a specified concentration of magnesium ions, Mg²⁺. The solubility product constant (\(K_{sp}\)) for Mg(OH)₂ is provided.

Concept Utilized:

The core principle applied is the solubility product constant (\(K_{sp}\)). A sparingly soluble salt commences precipitation when the ionic product of its constituent ions in the solution attains a value equal to its \(K_{sp}\).

The fundamental steps and equations are:

1. Dissociation Equilibrium: The equilibrium reaction for Mg(OH)₂ is represented as: \[ \text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\text{OH}^-(aq) \]
2. Solubility Product Expression: The \(K_{sp}\) is defined by the equation: \[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \]
3. Precipitation Criterion: Precipitation occurs when the ionic product, \(Q_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2\), equals \(K_{sp}\).
4. pH and pOH Relationship: At 25°C, the pH and pOH of an aqueous solution are linked by: \[ \text{pOH} = -\log_{10}[\text{OH}^-] \] \[ \text{pH} + \text{pOH} = 14 \]

Detailed Solution Procedure:

Step 1: Document the given parameters and the condition for precipitation initiation.

• \(K_{sp}\) of Mg(OH)₂ = \(1 \times 10^{-11}\)
• Concentration of Mg²⁺ ions, \([\text{Mg}^{2+}] = 0.10 \, \text{M}\)

Precipitation commences when the following condition is satisfied:

\[[\text{Mg}^{2+}][\text{OH}^-]^2 = K_{sp}\]

Step 2: Calculate the required hydroxide ion concentration, \([\text{OH}^-]\), for precipitation to begin.

Substitute the known values into the equation:

\[(0.10) \times [\text{OH}^-]^2 = 1 \times 10^{-11}\]

Solve for \([\text{OH}^-]\):

\[[\text{OH}^-]^2 = \frac{1 \times 10^{-11}}{0.10} = \frac{1 \times 10^{-11}}{10^{-1}} = 1 \times 10^{-10}\]\[[\text{OH}^-] = \sqrt{1 \times 10^{-10}} = 1 \times 10^{-5} \, \text{M}\]

This represents the minimum hydroxide ion concentration necessary for Mg(OH)₂ precipitation to start.

Step 3: Determine the pOH of the solution.

The pOH is calculated as the negative base-10 logarithm of the hydroxide ion concentration:

\[\text{pOH} = -\log_{10}[\text{OH}^-] = -\log_{10}(1 \times 10^{-5})\]\[\text{pOH} = -(-5) = 5\]

Final Calculation and Outcome:

Step 4: Derive the solution's pH from the calculated pOH.

Utilizing the relationship \( \text{pH} + \text{pOH} = 14 \):

\[\text{pH} = 14 - \text{pOH}\]\[\text{pH} = 14 - 5 = 9\]

Therefore, Mg(OH)₂ initiates precipitation in the solution when the pH reaches 9.