Determine the ammonia concentration of an aqueous solution that has a pH of 11.00.

Question

Determine the ammonia concentration of an aqueous solution that has a pH of 11.00. 

The equation for the dissociation of NH₃ (Kb = 1.8 × 10^-5) is

  NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq)

Answer 1

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq)

Kb = [NH₄⁺] [OH^-] / [ NH₃]

From the pH = 11.00, we can find the pOH. From the pOH, we can find the [OH-]

pOH = 14 - pH

pOH = 14 - 11.00

pOH = 3.00

[OH-] = 1x10^-3 This is also the [NH₄⁺]

Now, plug values into the Kb expression and solve for [NH₃]

1.8x10^-5 = (1.00x10^-3)(1.00x10^-3) / x - 1.00x10^-3

1.8x10^-5x - 1.8x10^-8 = 1.00x10^-6

1.8x10^-5x = 1.018x10^-6

x = [NH₃] = 0.0566 M