Answer to Question #146081 in General Chemistry for Nomawethu

Question

What is the PH at the equivalence point in the titration of 100ml of 0.20m ammonia with 0.10M hydrochloric acid?

Given NH3 as 1.75×10^-5

Answer 1

NH₃ + HCl = NH₄⁺ + Cl^-

Moles of NH₃ = 0.2 × 100 ×10^-3 = 0.02 moles

Vol HCl = 0.02moles/ 0.10M = 0.2L

Total volume = 0.2+ 0.1 = 0.3L

Concentration of ammonium ion = 0.02/0.3 = 0.067M

Ka = Kw/Kb = 10^-14/ 1.75×10^-5 = 5.71 × 10^-10

Ka = NH₃ . H₃O / NH₄⁺

= x*x/(0.067-x)

0.067-x ~ 0.067

Ka= x²/ 0.067 = 5.71 × 10^-10

x= (0.067 × 5.71 ×10^-10)^1/2

x= 6.19 × 10^-6 = [H₃O]⁺

PH = -log [6.19 ×10^-6]

= 5.21