....C2H5NH3^+ + HOH ==> C2H4NH3 + H3O^+
init...0.32...............0..........0
change..-x................x.........x
equil..0.32-x..............x.........x
Ka = (Kw/Kb) = (x)(x)/(0.32-x)
Solve for x and that will allow you to calculate all species except Cl^- and that is 0.32M.
Calculate the concentrations of all species present in a 0.32 M solution of ethylammonium chloride (C2H5NH3Cl).
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