The dissociation of C2H5NH3Cl is as follows:
C2H5NH3Cl(aq) → Cl–(aq) + C2H5NH3+(aq)
[C2H5NH3Cl] = [C2H5NH3+] = 0.25 M
The hydrolysis of C2H5NH3+:
C2H5NH3+(aq) + H2O(l) ⇌ H3O+(aq) + C2H5NH2(aq)
Kb of C2H5NH2 = 6.4 × 10–4
Ka = 1.56 x 10–11
Calculate the concentrations of all species present in a 0.25-M solution of ethylammonium chloride (C2H5NH3Cl).
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