Acid-Base help....?

Calculate the concentrations of all species present in a 0.25M solution of ethylammonium chloride (C2H5NH3Cl)

what do i do???
thanks!

Answer:
This is a weak acid dissociation problem. The pKa of ethylammonium chloride is 10.6 (see reference). The reaction is (Et=ethyl):

EtNH3+ ↔ EtNH2 + H+

Notice the chloride is a spectator ion here. Its concentration will remain 0.25M. Since Ka = 10^(-pKa), Ka = 2.5E-11. (Here E-11 is shorthand for the scientific notation.) The equilibrium equation is:

Ka = 2.5E-11 = [EtNH2][H+]/[EtNH3+]

When first dissolved [EtNH3+] = 0.25, and [EtNH2] & [H+] are both = 0. After equilibrium, [EtNH2] = [H+] = x and [EtNH3+] = 0.25-x, where x is the change in ethylammonium concentration in going to equilibrium. The equation becomes 2.5E-6 = x²/(0.25-x) ≈ x²/0.25, since x will be small, so we can ignore it in the denominator. The equation rearranges to x = √((2.5E-11)(0.25)) = 2.5E-6.

So the answers are
[Cl-] = 0.25M
[EtNH2] = [H+] = x = 2.5E-6M (pH = 5.6 by the way)
[EtNH3+] = 0.25-x ≈ 0.25 (slightly lower concentration)
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