1. Understand the Equilibrium
* AgBr is a sparingly soluble salt, meaning it dissolves to a small extent in water. The equilibrium reaction is:
AgBr(s) ⇌ Ag⁺(aq) + Br⁻(aq)
* The solubility product constant (Ksp) for AgBr is 5.0 x 10⁻¹³
2. Account for the Common Ion Effect
* Silver nitrate (AgNO₃) is a soluble salt that provides a common ion (Ag⁺) with the AgBr solution. This common ion will shift the equilibrium of the AgBr dissolution to the left, reducing the solubility of AgBr.
3. Set Up an ICE Table
* We'll use an ICE (Initial, Change, Equilibrium) table to keep track of the concentrations.
| | Ag⁺ (M) | Br⁻ (M) |
|-----|---------|---------|
| I | 3.0 x 10⁻² | 0 | (Initial concentration from AgNO₃)
| C | +s | +s |
| E | 3.0 x 10⁻² + s | s |
* s represents the molar solubility of AgBr
4. Write the Ksp Expression
* Ksp = [Ag⁺][Br⁻] = 5.0 x 10⁻¹³
5. Substitute and Solve
* Substitute the equilibrium concentrations from the ICE table into the Ksp expression:
(3.0 x 10⁻² + s)(s) = 5.0 x 10⁻¹³
* Since the Ksp is very small, we can assume that s is negligible compared to 3.0 x 10⁻²:
(3.0 x 10⁻²)(s) ≈ 5.0 x 10⁻¹³
* Solve for s:
s ≈ 5.0 x 10⁻¹³ / 3.0 x 10⁻²
s ≈ 1.7 x 10⁻¹¹ M
Therefore, the molar solubility of AgBr in a 3.0 x 10⁻² M silver nitrate solution is approximately 1.7 x 10⁻¹¹ M.
