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
