`K_(sp)` of `Mg(OH)_(2)` is `4.0 xx 10^(-6)`. At what minimum `pH, Mg^(2+)` ions starts precipitating `0.01MgCl_(2)`

Options

A. `2+log 2`
B. `2 - log 2`
C. `12 + log 2`
D. `12 - log 2`

Explanation

To determine the minimum pH at which magnesium ions start precipitating from a solution of 0.01 M MgCl₂, we can follow these steps:

### Step 1: Write the dissociation equation for magnesium hydroxide
Magnesium hydroxide (Mg(OH)₂) dissociates in water as follows:
\[ \text{Mg(OH)}_2 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + 2 \text{OH}^- (aq) \]

### Step 2: Write the expression for the solubility product constant \( K_{sp} \)
The solubility product constant \( K_{sp} \) for Mg(OH)₂ is given by:
\[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \]
Given \( K_{sp} = 4.0 \times 10^{-6} \).

### Step 3: Determine the concentration of magnesium ions
The concentration of magnesium ions in the solution from MgCl₂ is:
\[ [\text{Mg}^{2+}] = 0.01 \, \text{M} = 1.0 \times 10^{-2} \, \text{M} \]

### Step 4: Substitute into the \( K_{sp} \) expression
Substituting the concentration of magnesium ions into the \( K_{sp} \) expression, we have:
\[ 4.0 \times 10^{-6} = (1.0 \times 10^{-2})[\text{OH}^-]^2 \]

### Step 5: Solve for the hydroxide ion concentration \([\text{OH}^-]\)
Rearranging the equation to solve for \([\text{OH}^-]^2\):
\[ [\text{OH}^-]^2 = \frac{4.0 \times 10^{-6}}{1.0 \times 10^{-2}} = 4.0 \times 10^{-4} \]
Taking the square root gives:
\[ [\text{OH}^-] = \sqrt{4.0 \times 10^{-4}} = 2.0 \times 10^{-2} \, \text{M} \]

### Step 6: Calculate pOH
Now, we calculate pOH using the hydroxide ion concentration:
\[ \text{pOH} = -\log[\text{OH}^-] = -\log(2.0 \times 10^{-2}) \]
Using the properties of logarithms:
\[ \text{pOH} = -\log(2.0) - \log(10^{-2}) = -\log(2.0) + 2 \]
Using \(\log(2) \approx 0.301\):
\[ \text{pOH} \approx -0.301 + 2 = 1.699 \approx 1.7 \]

### Step 7: Calculate pH
Using the relation \( \text{pH} + \text{pOH} = 14 \):
\[ \text{pH} = 14 - \text{pOH} = 14 - 1.7 = 12.3 \]

### Step 8: Express pH in terms of log
We can express pH as:
\[ \text{pH} = 12 + \log(2) \]

### Final Answer
Thus, the minimum pH at which magnesium ions start precipitating is:
\[ \text{pH} = 12 + \log(2) \]