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

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