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

FREE Expert Solution

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

Kw=KaKbKwKb=KaKbKbKa=KwKbKa=1.0×10-146.4×10-4

Ka = 1.56 x 10–11 

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Problem Details

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

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