For the diprotic weak acid H2A Ka1=2.1x10^-5 and Ka2=5.5x10^-7. What is the pH of a 0.0800M solution of H2A? What are the equilibrium concentrations of H2A and A^2- in this solution?

Question

For the diprotic weak acid H2A Ka1=2.1x10^-5 and Ka2=5.5x10^-7.  What is the pH of a 0.0800M solution of H2A?  What are the equilibrium concentrations of H2A and A^2- in this solution?

DrBob222 · Answer

...........H2A ==> H^+ + HA^-
I.........0.0800....0.....0
C.........-x........x......x
E........0.0800-x...x.....x

k1 = (H^+)(HA^-)/(H2A)
Substitute and solve for x = (H^+) and convert to pH. Note that the k2 is about 100 times smaller than k1 so ignoring k2 doesn't cause much error.
I obtained 0.0787M for (H2A--you can round to correct sig figures) and for A^2- it is done this way.
k2 = 5.5E-7 = (H+)(A^=)/(HA^-). Since (H^+) = (HA^-), then (A=) = k2.