What is the pH of a 9.50 × 10⁻² M Ammonia Solution and What is the Percent Ionization of Ammonia at This Concentration?

Ammonia (NH₃) is a weak base with a K_b value of 1.8 × 10^-5. To find the pH of a 9.50 × 10^-2 M ammonia solution, we need to follow these steps:

Step 1: Write the Ionization Equation

NH₃ + H₂O ⇌ NH₄⁺ + OH^–

Step 2: Set Up the Expression for K_b

K_b = [NH₄⁺][OH^–] / [NH₃]

Given K_b = 1.8 × 10^-5

Step 3: Assume the Change in Concentration

Let x be the concentration of NH₄⁺ and OH^– ions formed.

Initial concentration of NH₃ = 9.50 × 10^-2 M

At equilibrium:

[NH₃] = 9.50 × 10^-2 – x

[NH₄⁺] = x

[OH^–] = x

Step 4: Substitute into the K_b Expression

1.8 × 10^-5 = x² / (9.50 × 10^-2 – x)

Since K_b is small, we can assume x is much smaller than 9.50 × 10^-2, so 9.50 × 10^-2 – x ≈ 9.50 × 10^-2

1.8 × 10^-5 ≈ x² / 9.50 × 10^-2

x² ≈ 1.8 × 10^-5 × 9.50 × 10^-2

x² ≈ 1.71 × 10^-6

x ≈ √(1.71 × 10^-6)

x ≈ 1.31 × 10^-3 M

Step 5: Calculate pOH and pH

[OH^–] = 1.31 × 10^-3 M

pOH = -log[OH^–] = -log(1.31 × 10^-3) ≈ 2.88

pH = 14 – pOH = 14 – 2.88 ≈ 11.12

Step 6: Calculate Percent Ionization

Percent Ionization = (x / initial concentration) × 100

Percent Ionization = (1.31 × 10^-3 / 9.50 × 10^-2) × 100 ≈ 1.38%

Therefore, the pH of a 9.50 × 10^-2 M ammonia solution is approximately 11.12, and the percent ionization of ammonia at this concentration is approximately 1.38%.