Consider the titration of 50.0 ml of 0.20 m nh3 (kb=1.8×10−5) with 0.20 m hno3. calculate the ph after addition of 50.0 ml of the titrant at 25 ∘c.

Total moles of NH3 = 50 ml x 0.20 m = 0.01 mol 
Total moles of HNO3 = 50 ml x 0.20 m = 0.01 mol
 NH3 + H+ = NH4+, moles of NH4+ = 0.01 M x l
 H3O+ = NH3 = N
 Total volume of both compounds = 50 ml + 50 ml = 100 ml = 0.1 l
 For ammonium ion NH4+ = 0.01 M x l / 0.1 l = 0.1 M
 Kw = 1.0 x 10^-14; Kb = 1.8 x 10^-5; Ka = Kw/Kb = 1.0 x 10^-14 / 1.8 x 10^-5 
  Ka = 5.55 x 10^-10 = [H3O+ + 1][NH3] / [NH4^+1] = N x N / (0.1 - N)
 Since the value of the N is very less and almost negligible 5.55 x 10^-10 = N^2/0.1 
 N^2 = 55.5 x 10^-12 => N = 7.44 x 10^-6 => H3O+ = 7.44 x 10^-6 Hence the pH value is 5.13

False because saturated fatty acids contain only single bond