The ionization constant of propanoic acid is 1.32×10−5. Calculate the degree of ionization of acid in its 0.05 M solution and also its pH.What will be its degree of ionization if the solution is 0.01 M in HCl also ?
Ch3CH−2COOH Propanoic acid
CH3CH2COOH+H2O⇋CH3CH2COO−+H3O+ Ka=1.32×10−5
At equlibrium
0.05−Cα Cα Cα
Ostwald's Dilution Law⟶α=√KdC=√1.35×10−50.05
α=0.016248
Degree of ionization is 16.25×10−3
[H3O+]=Cα=0.05×0.016248=8.124×10−4
pH=−log[H3O+]=C10.9098≅3.09
pH=3.09
Ka=[CH3CH2COO−][H+][CH2CH2COOH]
1.32×10−5=Cα(0.01)0.05−α
[H3O+]=0.0M From HCl
Cα=1.32×10−5(0.05−α)0.01
Cα=6.60×10−5
Degree of ionisation=6.66×10−50.05=1.32×10−3