A student was given a 0.10 M solution of an unknown diprotic acid H2A and asked
ID: 613474 • Letter: A
Question
A student was given a 0.10 M solution of an unknown diprotic acid H2A and asked to determine the Ka1 and Ka2 values for the diprotic acid. The student titrated 50.0 mL of the 0.10 M H2A with 0.10 M NaOH. After 25.0 mL of NaOH was added, the pH of the resulting solution was 6.70. After 50.0 mL of the NaOH was added, the pH of the resulting solution was 8.00. What are the values of Ka1 and Ka2 for the diprotic acid? Explain in words the approach you will use.Explanation / Answer
You have two equilibria to meet with the acid/mono salt/ and di-salt. [H+][HA-]/[H2A]= Ka1 and [H+] [A=]/HA-]=Ka2 as well as [H+][OH-]=Kw In the first part of the titration, we have 5 millimole of H2A (50mlx0.1M) being reacted with 2.5 millimole of NaOH. At this point, the Ka1 equilibrium is:: 10^-6.7 [2.5millimole/75ml]/[2.5 millimole/75ml]=Ka1. Then Ka1 = 10^6.7 or about 2x10^-7 In the second part, 5 millimoles of NaOH have been added, so we have essentially the salt NaHA as the species containing A. This slightly dissociates: HA- + H2O -> H2A + A= . To find Ka2, we have to get H2A into the second equation, which is [A=][H+]/[HA-]= Ka2. From the Ka1 eqtn, [HA-]= Ka1[H2A]/[H+]. Inserting this in the Ka2 eqtn for [HA-], [A=][H+]^2= Ka1Ka2[H2A]. From the stoichiometry, [H2A] = [A=] so the expression reduces to [H+]^2 = Ka1Ka2. Since [H+]=10-8, 2x10-7 Ka2 = 10x10^-17, and Ka2 = 5x10^-10.
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