Verified answer

anilinium nitrate is a salt of weak base and strong acid ....

so its pH will be given by :

pH = 1/2 [ pKw -pKb -log C ]

pKw is -log Kw where Kw is dissociation constant of water ...as Kw is 10^-14 so pKw = 14

given C = 0.19 M....so log C = -0.72

Kb = 4.2 X 10^-10 .....so pKb = -log (4.2 X 10^-10) = 9.38

putting in pH expression : pH = 1/2 [ 14 - 9.38 - (-0.72)] = 1/2 [5.34] = 2.67

see this link --

http://chemistry.tutorvista.com/inorganic-chemistr...

The salt of a weak base in water will tend to form the weak base and a hydronium ion

C6H5NH3^+1 + H2O <--> C6H5NH2 + H3O^+1

Let X = [H3O+] = [C6H5NH2]

Ka = [C6H5NH2][H3O+] / [C6H5NH3+]

Since Ka x Kb = Kw you can solve for Ka knowing Kb (4.2x10^-10) and Kw (1x10^-14)

Ka = 10^-14 / (4.2x10^-10) = 2.38x10^-5

2.38x10^-5 = X^2 / (0.19-X)

Since X will be small compared to 0.19, 0.19-X is equal to 0.19 (2 s.f.)

Solving for X-

X = 4.52x10^-6 = [H+]

pH = 5.34

I agree with 2.67.