How to find the Ph

Question

What is the pH of a 0.15 M solution of 100 mL NH₃, when the ka of NH₄+ is 5.8 x 10^-10?

Answer 1

NH₃ + H₂O ==> NH₄⁺ + OH^-

Kb = [NH₄⁺][OH^-] / [NH₃]

KaKb = 1x10^-14

Kb = 1x10^-14 / 5.8x10^-10

Kb = 1.72x10-5

1.72x10^-5 = (x)(x) / 0.15 - x (and assuming x is small we can neglect it in the denominator)

x² = 2.58x10^-6

x = 1.61x10^-3 M = [OH^-] (note: our assumption above was correct as this is only about 1% of 0.15 m)

pOH = -log OH = -log 1.61x10-3 = 2.79

pH = 14 - 2.79

pH = 11.2