Question about calculating pH for polyprotic acids

[Zellmer, Robert]

I received a question about how to calculate the pH for polyprotic acids.
My answer is below.

I did an example for H3PO4 in class.  There are two EOCEs, 16.69 and
16.70, which cover this, with 16.70 being one of the "assigned" EOCEs.
There is at least one question in the Mastering due this weekend.  You
should know how to do a polyprotic acid problems (end of section 16.6)
to get the conc of all species in solution.  There have been questions
like this on past midterms.

For most weak polyprotic acids, the pH is determined by the first step
(1st H+ coming off).  This is especially true when the Ka's for the
ionization steps are widely different and the Ka's for subsequent
ionizations (protons coming off) after the first ionization step (1st H+
coming off ) are a lot smaller then Ka1.

This means you can usually just set up the ICE table for the first proton
coming off and get the pH from that step.  The H+ coming from any
more protons coming off in subsequent ionization steps don't normally
change the conc. of H+ in soln. and the pH appreciably (i.e. the [H+]
from protons coming from steps after the ionization step, the first
proton coming off, is inconsequential and won't affect the pH).

If a problem asks for any other species in solution you have to set up
subsequent ICE tables.  For example, look at a weak diprotic acid,
H2X, with Ka1 = 1.0 x 10^-3 and Ka2 = 1.0 x 10^-6 and [H2X] = 1.0 M.

H2X   <=>   H+   +   HX^-
1.0               0            0
  -x              +x          +x
-----------------------------------
1.0 - x          x            x

              x^2
Ka1 = ---------- = 1.0 x 10^-3
            1 - x

ignoring the "x" in the denominator,

x = [H+] = 3.16 x 10^-2 M       (3.16% error in ignoring "x")

pH = 1.50

This is also the conc. of the HX^- before any of it has a chance to
ionize.

Now set up the table for 2nd proton ionizing using the conc. of
[H+] and [HX^-] from step 1.

  HX^-   <=>     H+      +     X^2-
0.0316          0.0316            0
    -y                  +y             +y
----------------------------------------
0.0316 - y   0.0316+y        y

            (0.0316+y) y
Ka2 = ------------------- = 1.0 x 10^-6
            0.0316 - y

ignoring the "y" in the numerator and denominator,

y = [H+] = 1.0 x 10^-6 M         (.00316% error in ignoring "y")

This is the [H+] from step 2 and the [X^2-] conc.,

[H+]_step2 = 1.0 x 10^-6 M
[X^2-] = 1.0 x 10^-6 M.

The total conc. of H+ from steps 1 and 2 is,

[H+] = 3.16227 x 10^-2 M  + 1.0 x 10^-6 M = 3.16327 x 10^-2 M

The change compared to the [H+] from step one is in the 3rd
decimal place, and the number has only 2 s.f. so it rounds to
3.2 x 10^-2 M, the same as from step 1.

The pH from step 1 is 1.50.  The pH also counting the [H+] from
step 2 is 1.49998 which rounds to 1.50.  No change to the correct
number of s.f.

So, what are all the final conc.

[H2X] = 1 - 3.16 x 10^-2 M = 0.9683 M, which is 1.0 when rounded
            to the 1st decimal place.

[X^2-] = 1.0 x 10^-6 M

[H+]_total = 3.16327 x 10^-2 M      (3.2 x 10^-2 M, which is what you
                                                         get from using [H+] from step 1)

pH = 1.4999 = 1.50 (no different than that based on only the [H+]
                                 from step 1)

Dr. Zellmer
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