Question

Calculate the degree of ionization of 0.05M acetic acid if its p$K_{a}$ value is 4.74. How is the degree of dissociation affected when its solution also contains 0.01M

Answer 1

The ionization of 0.05M acetic acid if its p$K_{a}$ value is 4.74.

How is the degree of dissociation affect when its solution also contains  0.01M solution is  pKa=−logKa=4.74

∴ Ka=Antilog(−4.74) =1.82×10−5

The degree of ionization of acetic acid may be calculated as: α=KaC−−−√=1.82×10−50.05−−−−−−−−−−√   = 0.01908

(a) In presence of 0.01 M HCl:

t=0teqCH3COOH0.050.05(1−α)≈0.05⇌CH3COO−00.05α+H+0.01(0.01+0.05α)≈0.1 Ka=[CH3COO−][H+][CH3COOH] 1.82×10−5=0.05α×0.01[CH3COOH] α=1.82×10−3