Calculate the degree of ionization of 0.05M acetic acid if its pK_a value is 4.74.(i) How is the degree of dissociation affected when its solution also contains 0.01M. (ii) How is the degree of dissociation affected when its solution also contains 0.1M HCl.

Calculate the degree of ionization of 0.05M acetic acid if its pK_a va

1.	Calculate the degree of ionization of 0.05M acetic acid if its pK_a value is 4.74.(i) How is the degree of dissociation affected when its solution also contains 0.01M. (ii) How is the degree of dissociation affected when its solution also contains 0.1M HCl.
Answer» Solution :`pK_a=ie..-logK_A=4.74` or `logK_a=-4.74=5.26thereforeK_a=1.82 times 10^-5` `a=sqrt(K_a//C)=sqrt((1.82times10^-5)//(5times10^-2))=1.908 times10^-2` In PRESENCE of HCl, due to high concentration of `H^+` ion,dissociation equilibrium will SHIFT BACKWARD i.e., dissociation of ACETIC acid will decrease (i) In presence of 0.01 M HCl if X is the amount dissociatedthen `CH_3COOH leftrightarrow CH_3COO^(-)+H^+` Initial 0.05 M After disso. 0.05-x x 0.01+x `approx 0.05 approx 0.01 M` `(0.01M H^+` ions obtained from 0.01 M HCl) `K_a=(x(0.01))/0.05 or x/0.05=K_a/0.01=(0.82 times10^-5)/10^-2=1.82 times10^-3` `a=1.82 times10^-3` `(because a=("AMount dissociated")/("amount taken"))` (ii) In the presence of 0.1 M HCl if y is the amount of acetic acid dissociated, then at equilibrium `[CH_3COOH]=0.05-y approx0.05M` `[CH_3COO^-]=y[H^+]=0.01M+y approx 0.1M` `K_a=(y(0.1))/0.05 or y/0.05=K_a/0.1=(1.82 times10^-5)/10^-1=1.82 times10^-4` `a=1.82 times 10^-4`