What is the ph of a 0.35 m solution of anilinium nitrate (c6h5nh3no3) ? kb for aniline is 4.2 * 10-10. your answer must be within ± 0.4%?

When the balanced reaction equation is:

and by using ICE table:

C6H5NH3 + (aq) + H2O (l) ↔ C6H5NH2 (aq) + H3O + (aq)

initial 0.35 0 0

change - X + X + X

Equ (0.35-X) X X

so, the Ka expression = [C6H5NH2][H3O]/[C6H5NH3+]

when Ka = Kw / Kb

and we know that Kw = 1 x 10 ^-14 & Kb is given = 4.2 x 10^-10

∴ Ka = (1 x 10^-14) / (4.2 x 10^-10)

= 2.4 x 10^-5

by substitution on Ka expression:

2.4 x 10^-5 = X*X / (0.35-X) by solving for X

∴ X = 0.00289 M

∴[H3O+] = X = 0.00289

when PH = - ㏒[H3O+]

= - ㏒ 0.00289

∴PH = 2.54