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

The problem asks for the pH at which magnesium hydroxide, Mg(OH)₂, will start to precipitate from a solution containing a known concentration of magnesium ions, Mg²⁺. We are given the solubility product constant (\(K_{sp}\)) for Mg(OH)₂.

Concept Used:

The solution is based on the concept of the solubility product constant (\(K_{sp}\)). Precipitation of a sparingly soluble salt begins when the ionic product of its constituent ions in the solution just equals its \(K_{sp}\) value. The key steps and formulas are:

1. Dissociation Equilibrium: For Mg(OH)₂, the equilibrium is: \[ \text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\text{OH}^-(aq) \]
2. Solubility Product Expression: The \(K_{sp}\) expression is: \[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \]
3. Condition for Precipitation: Precipitation starts when the ionic product \(Q_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2\) is equal to \(K_{sp}\).
4. Relationship between pH and pOH: The pH and pOH of an aqueous solution at 25°C are related by: \[ \text{pOH} = -\log_{10}[\text{OH}^-] \] \[ \text{pH} + \text{pOH} = 14 \]

Step-by-Step Solution:

Step 1: Write down the given values and the condition for precipitation.

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

Precipitation will begin when the following condition is met:

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

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

Substitute the known values into the equation:

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

Now, 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 is the minimum concentration of hydroxide ions needed in the solution to start the precipitation of Mg(OH)₂.

Step 3: Calculate the pOH of the solution.

The pOH is the negative logarithm of the hydroxide ion concentration:

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

Final Computation & Result:

Step 4: Calculate the pH of the solution from the pOH.

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

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

Thus, Mg(OH)₂ begins to precipitate from the solution when the pH reaches 9.

Answer 2

Precipitation occurs when \( Q_p = K_{sp} \). The solubility product expression is:

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

Given:

\( 0.1 \times [\text{OH}^-]^2 = 10^{-11} \)

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

\( [\text{OH}^-] = 10^{-5} \)

Then,

\( \text{pOH} = 5 \Rightarrow \text{pH} = 9 \)