In the titration curve, the initial pH of the triprotic acid H3A is given as approximately 1.8. This corresponds to the pH at which only H3A is present in solution.
The first equivalence point, where half of the H3A has been neutralized, occurs at approximately pH 4.3. This corresponds to the point where H2A^- is formed; therefore, Ka1 can be calculated as follows:
Ka1 = [H2A^-][H3O^+]/[H3A] = 10^(-pH1)
where pH1 is the pH at the first equivalence point.
Substituting the given values, we get:
Ka1 = [H2A^-][H3O^+]/[H3A] = 10^(-4.3)
At the second equivalence point, where H2A^- has been completely neutralized and HA^2- is formed, the pH is approximately 8.5. Thus, Ka2 can be calculated as follows:
Ka2 = [HA^2-][H3O^+]/[H2A^-] = 10^(-pH2)
where pH2 is the pH at the second equivalence point.
Substituting the given values, we get:
Ka2 = [HA^2-][H3O^+]/[H2A^-] = 10^(-8.5)
Lastly, at the third equivalence point, where HA^2- has been completely neutralized and A^3- is formed, the pH is approximately 12.3. Therefore, Ka3 can be calculated as follows:
Ka3 = [A^3-][H3O^+]/[HA^2-] = 10^(-pH3)
where pH3 is the pH at the third equivalence point.
Substituting the given values, we get:
Ka3 = [A^3-][H3O^+]/[HA^2-] = 10^(-12.3)
Thus, the values of pKa1, pKa2, and pKa3 for the triprotic acid H3A are:
pKa1 = -log(Ka1) = -log(10^(-4.3)) = 4.3
pKa2 = -log(Ka2) = -log(10^(-8.5)) = 8.5
pKa3 = -log(Ka3) = -log(10^(-12.3)) = 12.3