[soupastupid]

i don't know how to do this problem.

[soupastupid]

is it

Cl = .220
C6H5NH2 = 2.18*10^-1
H3O+ = 4.21*10^-12
OH- = 2.38*10^-3
C6H5NH3+ = 2.38*10^-3

[enahs]

You have to know some sort of dissociation constants. Were these given or did you look them up?
Or how did you come up with your numbers?

[soupastupid]

i used

kw = 1*10^-14
kbase of C6H5NH3Cl = 3.9*10^-10

was i suppose to use those values?
and then i set up an ICE BOX

do u kno wat that is?

its hard to explain what i did

nevermind

i think i got it

tanx anyway

[Borek]

H3O+ = 4.21*10^-12
OH- = 2.38*10^-3

Think it over.

[soupastupid]

think it over?

im assuming Cl is counterion right?

and so the equation becomes

C6H5NH2 + H2O = C6H5NH3+ + OH-

.220 for initial C6H5NH2 and Cl

and i put

(kw/ka)*(.220) = x^2

kw = 1*10^-14
kbase of C6H5NH3Cl = 3.9*10^-10

i get x = 2.38*10^-3
x= [OH-] and [H3O+] = kw/[OH-]
[H3O+] = 4.21*10^-12


this is wrong?
wat should i do?

[Borek]

[H3O+] = 4.21*10^-12

Protonated aniline is a weak acid, yet you have calculated that the solution is basic.

What are initial species in the solution?

[soupastupid]

initial species?

i think its suppose to be base

intial species of Aniline

C6H5NH2 and H20
there is .220 M of aniline because Cl is a counterion (how do I know its a counterion? im just guessing it is)

this makes C6H5NH3+ and OH-

right?

[Borek]

No, you start with aniline hydrochloride, that means aniline was fully protonated.

Think: why should aniline (neutral molecule) need counter ion?