Calculate the degree of ionization of 0.05 M acetic acid if its pK_(a) value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01 M (b) 0.1 M HCl ?

Calculate the degree of ionization of 0.05 M acetic acid if its pK_(a)

1.	Calculate the degree of ionization of 0.05 M acetic acid if its pK_(a) value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01 M (b) 0.1 M HCl ?
Answer» Solution :`pK_(a) = 4.74, i.e., - log K_(a) = 4.74 or log K_(a) = - 4.74 = BAR(5).26 :. K_(a) = 1.82xx10^(-5)` `alpha=SQRT(K_(a)//C)=sqrt((1.82xx10^(-5))//(5xx10^(-2)))=1.908 xx 10^(-2)` In presence of HCl, DUE to high concentration of `H^(+)` ion, dissociation equilibrium will shift backward, i.e., dissociation of aceticacidwill decrease . (a) In presence of 0.01 MHCL, if x is the amount dissociated, then `{:(,CH_(3)CO OH ,hArr,CH_(3)CO O^(-) ,+,H^(+),,,),("Initial conc.",0.05 M,,,,,,,),("After disso.",0.05 - x ,,x,,0.01 + x,,,),(,~=0.05,,,,,~=0.01M,,),(,,,,,(0.01 M H^(+) "ions are obtained from 0.01 M HCl"),,,):}` `:. K_(a) = (x(0.01))/(0.05) or (x)/(0.05)=(K_(a))/(0.01)=(1.82xx10^(-5))/(10^(-2))=1.82xx10^(-3) . "But" (x)/(0.05)=alpha` Hence, `alpha=1.82xx10^(-3)`(`:' alpha = ("Amount dissociated")/("Amount taken")`) (b) In the presence of 0.1 M HCl, if y is the amount of acetic ACID dissociated, then at equilibrium, `[CH_(3)CO OH]=0.05-y~=0.05 M` `[CH_(3)CO O^(-)]=y, [H^(+)]=0.1M+y~=0.1M` `K_(a)=(y(0.1))/(0.05) or (y)/(0.05)=(K_(a))/(0.1)=(1.82xx106(-5))/(10^(-1))=1.82xx10^(-4), i.e., alpha = 1.82xx10^(-4)`