What is the minimum pH necessary to cause a precipitate of `Pb(OH)_(2)` `(K_(sp)=1.2xx10^(-5))` to form in a 0.12 M `PbCl_(2)` solution?

Options

A. `12.4`
B. `10.8`
C. `12.0`
D. `11.1`

Answer

To find the minimum pH necessary to cause a precipitate of `Pb(OH)₂` to form in a 0.12 M `PbCl₂` solution, we can follow these steps:

### Step 1: Write the dissociation equation for `Pb(OH)₂`
The dissociation of lead(II) hydroxide can be represented as:
\[ \text{Pb(OH)}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{OH}^- (aq) \]

### Step 2: Write the expression for the solubility product constant (Ksp)
The solubility product constant expression for this equilibrium is:
\[ K_{sp} = [\text{Pb}^{2+}][\text{OH}^-]^2 \]
Given that \( K_{sp} = 1.2 \times 10^{-5} \).

### Step 3: Determine the concentration of `Pb²⁺`
In the 0.12 M `PbCl₂` solution, the concentration of `Pb²⁺` ions is:
\[ [\text{Pb}^{2+}] = 0.12 \, \text{M} \]

### Step 4: Substitute into the Ksp expression
Substituting the known concentration of `Pb²⁺` into the Ksp expression gives:
\[ 1.2 \times 10^{-5} = (0.12)[\text{OH}^-]^2 \]

### Step 5: Solve for the concentration of `OH⁻`
Rearranging the equation to solve for \([\text{OH}^-]^2\):
\[ [\text{OH}^-]^2 = \frac{1.2 \times 10^{-5}}{0.12} \]

Calculating this gives:
\[ [\text{OH}^-]^2 = 1.0 \times 10^{-4} \]

Taking the square root to find \([\text{OH}^-]\):
\[ [\text{OH}^-] = \sqrt{1.0 \times 10^{-4}} = 0.01 \, \text{M} \]

### Step 6: Calculate pOH
Using the concentration of hydroxide ions, we can calculate pOH:
\[ \text{pOH} = -\log[\text{OH}^-] = -\log(0.01) = 2 \]

### Step 7: Calculate pH
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
Thus,
\[ \text{pH} = 14 - \text{pOH} = 14 - 2 = 12 \]

### Conclusion
The minimum pH necessary to cause a precipitate of `Pb(OH)₂` to form in a 0.12 M `PbCl₂` solution is:
\[ \text{pH} = 12.0 \]

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