| Chem 109 |
|
Spring 2009 |
| Lecture Notes:: 6 May |
|
|
| PREVIOUS |
|
Acids and Bases, cont.
Calculate the concentrations of the various species of phosphate in a 0.10 M solution with a pH of 6.5.
H3PO4 H+ + H2PO4-
Rxn Ka pKa H3PO4 7.5 x 10-3 2.2 H2PO4- 6.2 x 10-8 7.2 HPO42- 4.8 x 10-13 12.7
From the pKa values we can see that the first titration is complete, and the third has not started - only the middle titration contributes to the hydrogen ion concentration, [H+] = 3.16 x 10-7 M
We can find the concentrations of the two phosphate species involved in this equilibria using the Henderson-Hasselbalch equation:
pH = pKa + log [HPO42-] / [H2PO4-]
log [HPO42-] / [H2PO4-] = pH - pKa= 6.5 - 7.2 = -0.7
[HPO42-] / [H2PO4-] = 0.20
but [HPO42-] + [H2PO4-] = 0.10 M,
[H2PO4-] = 0.10 - [HPO42-]
substituting for [H2PO4-] and rearranging
[HPO42-] = (0.2)(0.10 - [HPO42-])
1.2 [HPO42-] = 0.020
[HPO42-] = 0.017 M; [H2PO4-] = 0.083 M Even though the other two equilibria do not contribute significantly to the pH, they are still important for finding the concentrations of the other two phosphate species. And such species are frequently important in other aspects of the system of interest, such as potential for precipitating metals as phosphates etc.
Let's find phosphate ion first. From the table above:
HPO42-
and
Ka = 4.8 x 10-13 = [H+] [PO43-] / [HPO42-]
But we know [H+] = 3.16 x 10-7 M from the pH value given, and we know [HPO42-] = 0.017 M from above. Substituting
4.8 x 10-13 = (3.16 x 10-7) [PO43-] / (0.017), &
[PO43-] = 2.6 x 10-8 M Similarly we can find the phosphoric acid concentration. From the table above:
H3PO4
and
Ka = 7.5 x 10-3 = [H+] [H2PO4-] / [H3PO4]
Again know [H+] = 3.16 x 10-7 M, and [H2PO4-] = 0.083 M from above. Substituting:
7.5 x 10-3 = (3.16 x 10-7)(0.083) / [H3PO4],
We will not cover 15.6 - 15.8 in Zumdahl (Chem 110 material).
| Syllabus / Schedule | 
|
© R A Paselk
Last modified 4 May 2009
