1. Understand Ammonia's Properties
* Ammonia is a weak base, meaning it doesn't fully ionize in water.
* The equilibrium reaction for ammonia in water is:
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)
2. Use the Kb Value
* The base ionization constant (Kb) for ammonia is 1.8 x 10⁻⁵. This value tells us the extent to which ammonia ionizes in water.
3. Set Up an ICE Table
| | NH₃ | NH₄⁺ | OH⁻ |
|-----|--------|--------|--------|
| I | 0.440 | 0 | 0 |
| C | -x | +x | +x |
| E | 0.440-x | x | x |
* I (Initial): We start with 0.440 M ammonia and no products.
* C (Change): The reaction shifts to the right, consuming ammonia and forming products.
* E (Equilibrium): We represent the change in concentration with 'x'.
4. Write the Kb Expression
Kb = [NH₄⁺][OH⁻] / [NH₃]
5. Substitute and Solve for x
1.8 x 10⁻⁵ = (x)(x) / (0.440 - x)
* Since Kb is small, we can assume that 'x' is much smaller than 0.440, so we can simplify the equation:
1.8 x 10⁻⁵ ≈ x²/0.440
* Solve for x:
x² ≈ 7.92 x 10⁻⁶
x ≈ 2.81 x 10⁻³ M
6. Calculate the pOH
pOH = -log[OH⁻] = -log(2.81 x 10⁻³) ≈ 2.55
7. Calculate the pH
pH + pOH = 14
pH = 14 - pOH ≈ 14 - 2.55 ≈ 11.45
Therefore, the pH of a 0.440 M ammonia solution is approximately 11.45.
