Question

What is the pH at the equivalence point of a titration of the weak base NH3(aq) with the strong acid HBr(aq) if 30.00 mL of 0.200 M NH3 solution requires 30.00 mL of 0.200 M HBr to reach the equivalence point? Kb = 1.8 x 10–5 for NH3.

Answer 1

pH in equivalence point

NH3 + HBr = NH4+

NH4+ + H2O = NH3 + H3O+

Ka = [NH3][H3O+]/[NH4+]

then, this will be slightly acidic

[NH4+] = mmol of NH4+ /(total volume )

total V= V1+V2 = 30+30 = 60

then [NH4+] = 30*0.2/60 = 0.10

in equilbirium

Ka = [NH3][H3O+]/[NH4+]

KA = Kw/Kb = (10^-14)/(1.8*10^-5) = 5.55*10^-10

[NH3] = [H3O+] = x

[NH4+] = 0.10 - x

substitute in

Ka = [NH3][H3O+]/[NH4+]

5.55*10^-10 = (x*x)/(0.1-x)

x = [H3O+] = 7.45*10^-6

pH = -log(H) =-log(7.45*10^-6) = 5.1278

pH = 5.1278 acidic, as expected