Question

23. The pH of 250 ml of 0.2 M aspartic acid is 2.2. Sketch the titration curve obtained by titrating the aspartic acid solution with 0.5 M KOH show clearly how you obtain the values used to plot the curve. (pKa1 = 2.0 , pKa2 = 3.8 , pKa3 =9.8)

Answer 1

Aspartic acid is an acidic amino acid. In acidic conditions, its amino-terminal will be protonated (i.e. will exist as -NH₃⁺). As the solution containing it becomes more basic, its carboxy-terminal acid group will lose a proton and then its side-chain carboxyl group will lose a proton. That explains the existence of pKa₁, pKa₂, and pKa₃.

If pH is 2.2 and 0.5 M KOH is added drop-wise, its pH will not change much because it has two carboxyl groups and it will already start to act as buffer. This pH will remain almost within the same range till 0.5 equivalents of KOH is added. At this point, the solution will contain equal amounts of protonated aspartic acid and zwitter-ions. This will give pKa1 to plot against 0.5 equivalents of KOH. Similarly, when 1.0 more equivalents of KOH has been added, there will be equal amounts of dipolar ions with one COO-and one NH3+ and two COO^- and one NH³⁺. This will be pKa2 (=3.8) and will be plotted against 1.5 KOH equivalents.

After this further addition of OH^- will not change the pH much until all NH³⁺ start giving up their H⁺. This will happen after further 1.5 equivalents of KOH have been added (i.e at 2.5 equivalents). This will be pKa3.

Plotting the above points, the curve like the one drawn below will be obtained.

14.0 12.0 12.0 PK =9.8 10.07 10.0 8.0 pH . 8.04 6.0 pk = 3.9 4. 4.0 pKal =2.0 2.0 pl = 3.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0