1. Understand the Chemistry

* NH₃ (Ammonia): A weak base.

* NH₄Cl (Ammonium chloride): The salt of a weak base (NH₃) and a strong acid (HCl). This salt will hydrolyze in water, producing H⁺ ions and making the solution acidic.

2. Set up the Equilibrium

The relevant equilibrium is the hydrolysis of the ammonium ion (NH₄⁺):

NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq)

3. Calculate the Initial Concentrations

* [NH₄⁺]:

- Molar mass of NH₄Cl = 53.49 g/mol

- Moles of NH₄Cl = 0.535 g / 53.49 g/mol = 0.01 mol

- [NH₄⁺] = 0.01 mol / 0.05 L = 0.2 M

* [NH₃]:

- [NH₃] = 0.1 M (given)

4. Use the ICE Table

| | NH₄⁺ | H₂O | NH₃ | H₃O⁺ |

|-------------|----------|----------|----------|----------|

| Initial (I) | 0.2 M | | 0.1 M | 0 |

| Change (C) | -x | | +x | +x |

| Equilibrium (E) | 0.2-x | | 0.1+x | x |

5. Write the Equilibrium Expression

Kₐ for NH₄⁺ is 5.6 x 10⁻¹⁰ (you can find this value in a table of acid dissociation constants)

Kₐ = [NH₃][H₃O⁺] / [NH₄⁺] = (0.1+x)(x) / (0.2-x)

6. Make Approximations

Since Kₐ is very small, we can assume that x is negligible compared to 0.1 and 0.2. This simplifies the equation:

5.6 x 10⁻¹⁰ ≈ (0.1)(x) / 0.2

7. Solve for x

x = [H₃O⁺] ≈ 1.12 x 10⁻⁹ M

8. Calculate pH

pH = -log[H₃O⁺] = -log(1.12 x 10⁻⁹) ≈ 8.95

Therefore, the pH of the solution is approximately 8.95.