Step 1: Understanding the Question:
This question is based on ionic equilibrium, specifically calculating the solubility of a sparingly soluble metal hydroxide in a buffer solution of a known $\text{pH}$.
We first need to determine the solubility product constant ($K_{sp}$) of lead hydroxide from its solubility in pure water, and then find its modified solubility in the buffer.
Step 2: Key Formulas and Approach:
Solubility in Water: For $\text{Pb(OH)}_2$, the dissociation is $\text{Pb(OH)}_2 \rightleftharpoons \text{Pb}^{2+} + 2\text{OH}^-$. The relation is:
\[
K_{sp} = 4s^3
\]
Buffer Solution Properties: In a buffer solution, the $\text{pH}$ remains constant. We can find the concentration of $\text{OH}^-$ ions using:
\[
\text{pH} + \text{pOH} = 14 \quad \text{and} \quad [\text{OH}^-] = 10^{-\text{pOH}}
\]
Solubility in Buffer: Let the new solubility be $s'$. The relation is $K_{sp} = s' [\text{OH}^-]^2$.
Step 3: Detailed Explanation:
Calculate $K_{sp$ of Lead Hydroxide:}
The solubility in water is $s = 6.7 \times 10^{-6}\text{ M}$.
\[
K_{sp} = 4s^3 = 4 \times (6.7 \times 10^{-6})^3
\]
\[
K_{sp} = 4 \times 300.76 \times 10^{-18} \approx 1.2 \times 10^{-15}
\]
Calculate Hydroxyl Ion Concentration in the Buffer:
The buffer solution has a $\text{pH} = 8$.
\[
\text{pOH} = 14 - 8 = 6
\]
\[
[\text{OH}^-] = 10^{-6}\text{ M}
\]
Calculate the Solubility in the Buffer ($s'$):
In the buffer solution, the concentration of $\text{OH}^-$ is maintained strictly at $10^{-6}\text{ M}$.
\[
K_{sp} = [\text{Pb}^{2+}][\text{OH}^-]^2
\]
\[
1.2 \times 10^{-15} = s' \times (10^{-6})^2
\]
\[
1.2 \times 10^{-15} = s' \times 10^{-12}
\]
\[
s' = \frac{1.2 \times 10^{-15}}{10^{-12}} = 1.2 \times 10^{-3}\text{ M}
\]
Step 4: Final Answer:
The solubility of lead hydroxide in the buffer solution is $1.2 \times 10^{-3}\text{ M}$, which corresponds to Option (D).