What is the pH of 0.35 M NH3 aqueous solution if the Kb for NH3 is 3.5 x 10-5?

Answer 1

NH3 + H2O ==>. NH4⁺ + OH^-

Kb = 3.5x10^-5 = [NH4⁺][OH-]/[NH3]

3.5x10^-5 = (x)(x)/0.35

x² = 3.5x10^-3 M = [OH-]

pOH = - log 3.5x10^-3 = 2.46

pH = 11.54