A 0.10-mol sample of a diprotic acid, H2A, is dissolved in 250 mL of water. The Ka1 of this acid is 1.0 x 10-?

Question

A 0.10-mol sample of a diprotic acid, H2A, is dissolved in 250 mL of water. The Ka1 of this acid is 1.0 x 10-5 and Ka2 is 1.0 x 10-10. Calculate the concentration of A2- in this solution.

Answer 1

First of all find the concentration
C =0.1/0.25 = 0.4 M

The difference between Ka1 and Ka2 is big enough so that you can calculate the [H+] according to the first dissociation ignoring the second.
Thus
from Ka1= [H+][HA-]/[H2A] = [H+]^2/(C-[H+]) (because we consider only the first dissociation) and
[H+]^2 +Ka1[H+] - Ka1C =0 =>
x^2+ (10^-5)x -0.4*10^-5=0
Thus x=[H+]=0.001995

The concentration of A^-2 is

[A^-2]= (Ka1*Ka2*C) / ([H+]^2 + Ka1[H+] + Ka1*Ka2) =
= (0.4*10^-15) / (0.001995^2 +0.001995*(10^-5) + 10^-15)=
=10^-10 M

You could have calculated the [H+] taking both dissociation reactions into account (a 3rd degree equation), but the difference would have been extremely small.

At
http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-acid-base
you can see how to derive the equation for [A^-2], (there they write the product Ka1*Ka2 as Ka12 since it is true that for the overall reaction Katotal=Ka1*Ka2)
there you can also see the equation for finding the [H+] when taking into account both dissociation reactions. You can solve it with the free on-line tool at http://www.ifigure.com/math/algebra/algebra.htm

Answer 2

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