Calculate the pH of a 100 mL solution of 0.15 M NH3, given the Ka of NH4+ is 5.8 x 10-10. (1 pt)

Answer 1

To calculate the pH of a solution of NH₃, consider the dissociation of ammonia into NH₄⁺ ions and OH^- ions, as NH₃ is a weak base.

The equilibrium expression for the reaction is:

NH₃ (aq) + H₂O (l) ⇌ NH₄⁺ (aq) + OH⁻ (aq).

Initially, the concentration of NH₃ is 0.15 M and there is no NH₄⁺ or OH^- in the solution.

Define "x" as the change in the concentration of NH₄⁺ and OH^- at equilibrium. After the reaction reaches equilibrium, the concentrations will be:

NH₃: (0.15−x) M

NH₄⁺:​ x

OH⁻: x

The equilibrium expression is: Kw = [NH₄⁺​][OH⁻]​ / [NH₃]

Kw = 1.0 × 10⁻¹⁴ at 298 K

NH₃ is a weak base, so we can use Kw to find the concentration of OH- ions.

Therefore, 1.0 × 10⁻¹⁴ = [x] [x] / 0.15 − x

Approximate 0.15 - x as 0.15. This is okay when we are dealing with a weak base like NH₃.

So: 1.0 × 10⁻¹⁴ = x² / 0.15 and x² = 1.0 × 10⁻¹⁴ × 0.15. That means x² = 1.5 × 10⁻¹⁵ and x = 1.22 x 10^-8​. This is the concentration of OH^- ions.

pOH = - log₁₀(OH^-) = - log₁₀(1.22 x 10^-8) ≈ 7.91

pH = 14 - pOH = 14 − 7.91 ≈ 6.09