Question

Draw a graph of the relationship of the % ionization of acetic acid (pK – 4.76) and pH. Use the following pH values to calculate % ionization ( 3.06, 3.76, 3.86, 3.96, 4.06, 4.16, 4.26, 4.76, 4.86, 5.16, 5.56, 5.76, 6.66).

*Graph is not necessary, I can do that. Just need help finding one or two of the different answers with the different pH values

Answer 1

Given the pKa of the acid and its identity as a monoprotic acid, the percentage of ionization is calculated using the Ka of dissociation constant of the acid. Ka is given as 10^-pKa which here is 10^-4.76 = 1.7378x10^-5. Since acetic acid is a monoprotic acid, upon dissociation it produces as 1:1 ratio of protons to acetate anion. The expression for Ka is given as [H⁺][AcO^-]/[AcOH].

At any given pH, the concentration of protons in the medium is given by [H⁺] = 10^-pH. From the concentration of protons in the medium, we can calculate the concentration of acid in the medium [AcOH] as Ka = [H⁺]²/[AcOH] which rearranges as [AcOH] = [H⁺]²/Ka.

Then the percent of dissociation is calculated as [H⁺]/[AcOH] times 100.

For instance, for pH = 3.06, [H+] = 8.7096x10^-4M which gives [AcOH] = 7.5857x10^-7/1.7378x10^-5 = 4.3651x10^-2M. This gives the percentage of dissociation as (8.7096x10^-4M/4.3651x10^-2M)x100 = 1.9953%