HNO3 and NH3 titration

Question

Consider the titration of 27.8 mL of 0.447 M NH3 (Kb = 1.79 ×10-5) with 0.267 M HNO3. What is the pH of 100% titration?

Answer 1

HNO₃ + NH₃ ==> NH₄NO₃ ==> NH₄⁺ + NO₃^- ... balanced equation

moles NH₃ = 0.0278 L x 0.447 mol/L = 0.01243 moles

moles HNO₃ = 0.01243 moles

volume HNO₃ needed = (x L)(0.267 mol/L) = 0.01243 and x = 0.0466 L = 46.6 mls

Total volume at neutralization = 27.8 ml + 46.6 mls = 74.4 ml = 0.0744 L

[NH₄⁺] at neutralization = 0.01243 moles/0.0744 L = 0.167 M

NH₄⁺ + H₂O ==> H₃O⁺ + NH₃

Ka for NH₄ acting as an acid = 1x10^-14 / Kb NH₃ = 1x10^-14/1.79x10^-5

Ka = 5.6x10^-10

Ka = 5.6x10^-10 = [H₃O⁺][NH₃]/[NH₄⁺] = (x)(x)/0.167-x and if we assume x is small vs. 0.167 we can ignore

x² = 9.35x10^-11

x = 9.7x10^-6 = [H₃O⁺] (note: this is small vs. 0.167 so our assumption was correct)

pH = -log [H₃O⁺] = -log 9.7x10^-6

pH = 5.01