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Ch.16 - Acids and Bases
Chapter 16, Problem 116b

Consider a 0.10 M solution of a weak polyprotic acid (H2A) with the possible values of Ka1 and Ka2 given here: b. Ka1 = 1.0 * 10^-4; Ka2 = 1.0 * 10^-5. Calculate the contributions to [H3O+] from each ionization step. At what point can the contribution of the second step be neglected? c. Ka1 = 1.0 * 10^-4; Ka2 = 1.0 * 10^-6.

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<insert step 1> Identify the ionization steps for the polyprotic acid H2A. The first ionization step is H2A ⇌ H+ + HA-, and the second ionization step is HA- ⇌ H+ + A2-.
<insert step 2> For the first ionization step, use the expression for the acid dissociation constant: Ka1 = [H+][HA-]/[H2A]. Assume initial concentrations and solve for [H+].
<insert step 3> For the second ionization step, use the expression for the acid dissociation constant: Ka2 = [H+][A2-]/[HA-]. Use the [H+] from the first step to solve for the additional [H+] contributed by the second ionization.
<insert step 4> Compare the [H+] contributions from the first and second ionization steps. Determine when the contribution from the second step is negligible by comparing the magnitude of Ka1 and Ka2.
<insert step 5> For part c, repeat the process with the new value of Ka2 = 1.0 * 10^-6, and analyze the contributions to [H3O+] from each ionization step.>