Question

Calculate the concentrations of all species present in a 0.33 M solution of ethylammonium chloride (C2H5NH3Cl).



Okay so i tried doing

HA = H + A
ice table regular it doesnt work

idk what Ka to use. b/c of the chloride.


i have to find [C2H5NH2][H+][OH -][C2H5NH3+][Cl -]

Answer 1

This is a salt so it dissolves (and hydrolyzes in H2O).
C2H5HN3Cl(aq)==> C2H4NH3^+ + Cl^-
That gives you the Cl^- immediately. It's 0.33 M.
Now the salt dissolves in water (use NH4^+ as an example).
C2H5NH3^ + HOH ==> C2H4NH2 + H3O^+
Does that help? Use Ka = Kw/Kb and Kb is for C2H4NH2 (ethylamine).

Answer 2

no because i keep getting .01359 for C2H4NH3^+

but obviously that's wrong

i did x^2/.33-x=5.6e-4

Answer 3

No, I think if you will read my former response closely that isn't what I said to do.
C2H5NH3^+ + HOH ==> C2H5NH2 + H3O^+

Ka = Kw/Kb = (C2H5NH2)(H3O^+)/(C2H5NH3^+)
Now plug in Kw for water, Kb for ethylamine, 0.33 for the salt and calculate the y and y in the numerator. I get approximately 3 x 10^-6 M for H3O^+ so it should be close to that. I don't know what your value is for Kb but I looked up ethylamine on the internet and found 4.7 x 10^-4. Obviously you should use your book's value.

Answer 4

I don't understand why u did the Ka=Kw/Kb
why cant u just use Kb? isnt it a base?

Answer 5

C2H5NH2 is a base but you don't have the base. You have the salt, C2H5NH3Cl which contains the conjugate acid of the base or C2H5NH3^+. It's the acid that is hydrolyzing (reacting) with the water to give the base. You can look up Kb but no tables (at least none I know of) are around for the Ka. So we calculate Ka by Kw/Kb since we CAN look up Kb for C2H5NH2