1. Understand the Equilibrium
Lead bromide (PbBr₂) is a sparingly soluble ionic compound. When it dissolves in water, it establishes an equilibrium:
```
PbBr₂(s) ⇌ Pb²⁺(aq) + 2Br⁻(aq)
```
2. Set Up the Solubility Product (Ksp) Expression
The solubility product constant (Ksp) for PbBr₂ is the product of the ion concentrations at equilibrium, raised to their stoichiometric coefficients:
```
Ksp = [Pb²⁺] [Br⁻]²
```
3. Define Molar Solubility
Molar solubility (s) is the concentration of the metal cation (Pb²⁺ in this case) that dissolves in a saturated solution. Since the ratio of Pb²⁺ to Br⁻ is 1:2, the bromide concentration at equilibrium is 2s.
4. Substitute into the Ksp Expression
Substitute the molar solubility expressions into the Ksp equation:
```
Ksp = (s) (2s)² = 4s³
```
5. Look Up the Ksp Value
The Ksp for PbBr₂ is 6.6 x 10⁻⁶ at 25°C.
6. Solve for Molar Solubility (s)
Now you can solve for the molar solubility (s):
```
6.6 x 10⁻⁶ = 4s³
s³ = 1.65 x 10⁻⁶
s = ∛(1.65 x 10⁻⁶)
s ≈ 1.18 x 10⁻² M
```
Therefore, the molar solubility of PbBr₂ in water is approximately 1.18 x 10⁻² M.
