Question

The solubility of \(\text{Pb(OH)}_2\) in water is \(6.7 \times 10^{-6}\text{ M}\). Its solubility in a buffer solution of \(\text{pH} = 8\) would be:

• A. \( 1.2 \times 10^{-2} \)
• B. \( 1.6 \times 10^{-3} \)
• C. \( 1.6 \times 10^{-2} \)
• D. \( 1.2 \times 10^{-3} \)

Hint:

Always remember that a buffer solution acts as an infinite reservoir for \(\text{H}^+\) or \(\text{OH}^-\) ions. Therefore, unlike when dissolving a salt in pure water or standard ionic solutions, you completely ignore any extra \(\text{OH}^-\) ions produced by the salt dissociation (\(2s'\)) because the buffer dynamically absorbs them to lock the final concentration exactly to \(10^{-\text{pOH}}\).

Answer 1

The Correct Option is D

Concept: This problem requires determining the solubility product constant (\(K_{sp}\)) of a sparingly soluble salt in pure water first, then evaluating its modified solubility under a fixed external \(\text{pH}\) condition:
• For a salt like \(\text{Pb(OH)}_2\) with a molar solubility \(s\) in pure water, its dissociation equilibrium yields: \[ K_{sp} = [\text{Pb}^{2+}][\text{OH}^-]^2 = (s)(2s)^2 = 4s^3 \]
• In a buffer solution, the \(\text{pH}\) is maintained at a constant value, fixing the concentration of hydroxyl ions (\([\text{OH}^-]\)). The new solubility \(s'\) is determined by satisfying the constant \(K_{sp}\) relationship.

Step 1: Calculating the solubility product constant ($K_{sp}$) from solubility in pure water.
The given solubility of \(\text{Pb(OH)}_2\) in pure water is \(s = 6.7 \times 10^{-6}\text{ M}\). Using the relation for a \(1:2\) electrolyte system: \[ K_{sp} = 4s^3 \] \[ K_{sp} = 4 \times \left(6.7 \times 10^{-6}\right)^3 \] \[ K_{sp} = 4 \times 300.763 \times 10^{-18} \approx 1.2 \times 10^{-15} \]

Step 2: Determining $[\text{OH}^-]$ from the buffer solution $\text{pH}$.
The buffer solution maintains a constant environment of \(\text{pH} = 8\). Using the relation between \(\text{pH}\) and \(\text{pOH}\) at \(25^\circ\text{C}\): \[ \text{pH} + \text{pOH} = 14 \quad \Rightarrow \quad \text{pOH} = 14 - 8 = 6 \] Now, computing the concentration of hydroxyl ions from the logarithmic definition: \[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-6}\text{ M} \]

Step 3: Calculating the new solubility ($s'$) in the buffer solution.
Let the new molar solubility of \(\text{Pb(OH)}_2\) in the buffer solution be \(s'\). The dissociation equilibrium yields: \[ \text{Pb(OH)}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{OH}^-(aq) \] Here, the equilibrium concentrations are \([\text{Pb}^{2+}] = s'\) and \([\text{OH}^-]\) is fixed entirely by the buffer at \(10^{-6}\text{ M}\). Substitute these conditions into the \(K_{sp}\) expression: \[ K_{sp} = [\text{Pb}^{2+}][\text{OH}^-]^2 \] \[ 1.2 \times 10^{-15} = (s') \times \left(10^{-6}\right)^2 \] \[ 1.2 \times 10^{-15} = s' \times 10^{-12} \] Isolating for the new solubility variable \(s'\): \[ s' = \frac{1.2 \times 10^{-15}}{10^{-12}} = 1.2 \times 10^{-3}\text{ M} \]