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`.

Explanation

To solve the problem of calculating the solubility of `Pb(OH)₂` in a buffer solution with a pH of 8, we will follow these steps:

### Step 1: Determine the dissociation of `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: Define the solubility
Let the solubility of `Pb(OH)₂` in water be \( S \). From the dissociation equation:
- The concentration of `Pb²⁺` ions will be \( S \).
- The concentration of `OH⁻` ions will be \( 2S \).

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

### Step 4: Calculate \( K_{sp} \) using the given solubility in water
Given that the solubility of `Pb(OH)₂` in water is \( 6.7 \times 10^{-6} \, M \):
\[ K_{sp} = 4(6.7 \times 10^{-6})^3 \]

Calculating this:
\[ K_{sp} = 4 \times (6.7 \times 10^{-6})^3 \]
\[ K_{sp} = 4 \times 2.998 \times 10^{-16} \]
\[ K_{sp} \approx 1.203 \times 10^{-15} \]

### Step 5: Determine the concentration of `OH⁻` in the buffer solution
For a buffer solution with a pH of 8:
- Calculate \( pOH \):
\[ pOH = 14 - pH = 14 - 8 = 6 \]

- Calculate the concentration of `OH⁻`:
\[ [\text{OH}^-] = 10^{-pOH} = 10^{-6} \, M \]

### Step 6: Set up the equation for solubility in the buffer
Using the \( K_{sp} \) expression in the buffer solution:
\[ K_{sp} = [\text{Pb}^{2+}][\text{OH}^-]^2 \]
Substituting the known values:
\[ 1.203 \times 10^{-15} = [\text{Pb}^{2+}] \cdot (10^{-6})^2 \]
\[ 1.203 \times 10^{-15} = [\text{Pb}^{2+}] \cdot 10^{-12} \]

### Step 7: Solve for the concentration of `Pb²⁺`
Rearranging the equation to find \( [\text{Pb}^{2+}] \):
\[ [\text{Pb}^{2+}] = \frac{1.203 \times 10^{-15}}{10^{-12}} \]
\[ [\text{Pb}^{2+}] = 1.203 \times 10^{-3} \, M \]

### Conclusion
The solubility of `Pb(OH)₂` in a buffer solution of pH 8 is:
\[ \text{Solubility} = 1.203 \times 10^{-3} \, M \]

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