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

### Step 1: Write the Dissociation Equation
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 Solubility and Ksp
Let the solubility of \( \text{Pb(OH)}_2 \) in pure water be \( s \). From the dissociation, we can express the concentrations of the ions:
- \( [\text{Pb}^{2+}] = s \)
- \( [\text{OH}^-] = 2s \)

The solubility product constant \( K_{sp} \) for \( \text{Pb(OH)}_2 \) is given by:
\[ K_{sp} = [\text{Pb}^{2+}][\text{OH}^-]^2 = s \cdot (2s)^2 = 4s^3 \]

### Step 3: Calculate Ksp Using Given Solubility
We know the solubility of \( \text{Pb(OH)}_2 \) in pure water is \( 6.7 \times 10^{-6} \, M \). Therefore, substituting \( s = 6.7 \times 10^{-6} \):
\[ K_{sp} = 4(6.7 \times 10^{-6})^3 \]
Calculating this gives:
\[ K_{sp} = 4 \times (2.99 \times 10^{-17}) = 1.2 \times 10^{-15} \]

### Step 4: Determine the pOH and [OH⁻] in the Buffer
Given the pH of the buffer is 8, we can find the pOH:
\[ \text{pOH} = 14 - \text{pH} = 14 - 8 = 6 \]
Now, calculate the hydroxide ion concentration:
\[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-6} \, M \]

### Step 5: Set Up the Ksp 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:
\[ 1.2 \times 10^{-15} = [\text{Pb}^{2+}](10^{-6})^2 \]
This simplifies to:
\[ 1.2 \times 10^{-15} = [\text{Pb}^{2+}] \cdot 10^{-12} \]

### Step 6: Solve for [Pb²⁺]
Rearranging the equation gives:
\[ [\text{Pb}^{2+}] = \frac{1.2 \times 10^{-15}}{10^{-12}} = 1.2 \times 10^{-3} \, M \]

### Step 7: Conclusion
Thus, the solubility of \( \text{Pb(OH)}_2 \) in the buffer solution at pH 8 is:
\[ \text{Solubility} = 1.2 \times 10^{-3} \, M \]

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