Florida International University ?pH Buffering in Natural Waters Worksheet
Question Description
pH Buffering in Natural Waters:
We have seen that the carbonic acid system plays a critical role in providing natural waters with both acid- and base-neutralizing capacity, i.e., in resisting changes in pH when strong acids or bases are added and maintaining a stable pH environment. The reasons that carbonic acid is so critical are because:
it is a weak acid (with pK1 = 6.35 and pK2 = 10.33), so it exhibits chemical speciation, and
it is an abundant species in natural waters (acetic acid and phosphoric acid would also contribute to pH buffering but are present at very low concentrations in natural waters)
What I want you to do for this quiz is to plot the fractions of the three carbonic acid species as a function of pH using a spreadsheet.
Unknown concentrations: [H2CO3], [HCO3–] and [CO32–]
So we need three equations:
(1)First acid dissociation constant, K1 = 10-6.35 = [H+][HCO3–]/ [H2CO3]
(2)Second acid dissociation constant, K2 = 10-10.33 = [H+][CO32–]/ [HCO3–]
(3)Mass balance equation, CT = [H2CO3] + [HCO3–] + [CO32–]
I suggest keeping pH as the master variable, i.e., define pH values in the range 0 to 14 in the first column of the spreadsheet; keep the second column for values of [H+] = 10-pH.
Then write the molar concentrations of each species as a function of hydrogen ions [H+]. Use equations (1) and (2) to write the [H2CO3] and [CO32–] in terms of [HCO3–]:
[H2CO3] = [H+][HCO3–] /K1 (4)
[CO32–] = K2[HCO3–] /[H+] (5)
Now substitute for [H2CO3] and [CO32–] using equations (4) and (5) in the mass balance equation (3):
CT = [H2CO3] + [HCO3–] + [CO32–] = [H+][HCO3–] /K1 + [HCO3–] + K2[HCO3–] /[H+]
CT = [HCO3–]
So that F1 = [HCO3–]/ CT = 1 /
Substitute for K1 and K2 to get values for F1 in the third column of your spreadsheet. Now use equations (4) and (5) to calculate the F0 and F2 fractions in columns four and five.
Heres a fragment of what your spreadsheet should look like:
pH |
[H+] |
F1 = [HCO3–]/CT |
F2 = [H2CO3]/CT |
F3 = [CO32–]/CT |
0 |
1 |
4.467E-07 |
1 |
2.089E-17 |
0.2 |
0.6309573 |
7.079E-07 |
1 |
5.248E-17 |
0.4 |
0.3981072 |
1.122E-06 |
1 |
1.318E-16 |
0.6 |
0.2511886 |
1.778E-06 |
1 |
3.311E-16 |
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