Calculate the pH of a 5.70×10-1 M aqueous solution of ethylamine hydrochloride (C2H5NH3Cl).

Question

(For ethylamine, C2H5NH2, Kb = 5.60×10-4.)

Answer 1

C2H5NH3Cl is a salt which dissociates in aqueous solution as
C2H5NH3^+ + Cl^-, then the C2H4NH3^+ hydrolyzes with water as follows:
....................C2H5NH3^+ + H2O ==> C2H5NH2 + H3O^+
I....................0.571 M.............................0......................0
C.........................-x.................................x.......................x
E..................0.571-x................................x.......................x
Ka for C2H5NH3^+ = (Kw/Kb) = (C2H5NH2)(H3O)/(C2H5NH3^+)
Ka = (1E-14/5.60E-4) = (x)(x)/(0.571 - x)
Solve for x = (H3O^+) and convert to pH.
Post your work if you get stuck.