The solubility of `Pb(OH)_(2)` in water is `6.7xx10^(-6)`M. Calculate the solubility of `Pb(OH)_(2)` in a buffer solution of `pH=8`.

Options

A. `1.2xx10^(-2)`
B. `1.6xx10^(-3)`
C. `1.6xx10^(-2)`
D. `1.2xx10^(-3)`

Explanation

To solve the problem of calculating the solubility of \( \text{Pb(OH)}_2 \) in a buffer solution with a pH of 8, we can follow these steps:

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

### Step 2: Define the solubility
Let the solubility of \( \text{Pb(OH)}_2 \) in pure water be \( S \). From the dissociation, we can say:
- The concentration of \( \text{Pb}^{2+} \) ions will be \( S \).
- The concentration of \( \text{OH}^- \) ions will be \( 2S \).

### Step 3: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) for \( \text{Pb(OH)}_2 \) can be expressed as:
\[ K_{sp} = [\text{Pb}^{2+}][\text{OH}^-]^2 = S \cdot (2S)^2 = S \cdot 4S^2 = 4S^3 \]

### Step 4: Substitute the given solubility into the \( K_{sp} \) expression
Given that the solubility \( S \) in pure water is \( 6.7 \times 10^{-6} \) M, we can calculate \( K_{sp} \):
\[ K_{sp} = 4(6.7 \times 10^{-6})^3 \]
Calculating this gives:
\[ K_{sp} = 4 \cdot 2.99 \times 10^{-17} \approx 1.2 \times 10^{-16} \]

### Step 5: Calculate the \( \text{pOH} \) from the given \( \text{pH} \)
Given the buffer solution has a pH of 8, we can find the pOH:
\[ \text{pOH} = 14 - \text{pH} = 14 - 8 = 6 \]

### Step 6: Calculate the concentration of \( \text{OH}^- \)
Using the pOH, we can find the concentration of hydroxide ions:
\[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-6} \, \text{M} \]

### Step 7: Set up the \( K_{sp} \) expression for the buffer solution
In the buffer solution, the \( K_{sp} \) expression remains the same:
\[ K_{sp} = [\text{Pb}^{2+}][\text{OH}^-]^2 \]
Substituting the known values:
\[ 1.2 \times 10^{-16} = [\text{Pb}^{2+}] \cdot (10^{-6})^2 \]
This simplifies to:
\[ 1.2 \times 10^{-16} = [\text{Pb}^{2+}] \cdot 10^{-12} \]

### Step 8: Solve for the concentration of \( \text{Pb}^{2+} \)
Rearranging the equation gives:
\[ [\text{Pb}^{2+}] = \frac{1.2 \times 10^{-16}}{10^{-12}} = 1.2 \times 10^{-4} \, \text{M} \]

### Step 9: Calculate the solubility in the buffer solution
Since the solubility of \( \text{Pb(OH)}_2 \) in the buffer solution is equal to the concentration of \( \text{Pb}^{2+} \):
\[ \text{Solubility of } \text{Pb(OH)}_2 = 1.2 \times 10^{-4} \, \text{M} \]

### Final Answer
The solubility of \( \text{Pb(OH)}_2 \) in a buffer solution of pH 8 is \( 1.2 \times 10^{-4} \, \text{M} \).

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