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 \( \_\_\_\_ \).
Correct Answer: 9
Approach Solution - 1
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:
- Dissociation Equilibrium: For Mg(OH)₂, the equilibrium is: \[ \text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\text{OH}^-(aq) \]
- Solubility Product Expression: The \(K_{sp}\) expression is: \[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \]
- Condition for Precipitation: Precipitation starts when the ionic product \(Q_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2\) is equal to \(K_{sp}\).
- 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.
Approach Solution -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,
\( \text{pOH} = 5 \Rightarrow \text{pH} = 9 \)
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