In: Chemistry
For a 0.100 M solution of K2HPO4, find each of the following values: pH, [H+], [HPO42-], [PO43-], and [H2PO4-]. Use the minimum number of approximations.
HPO42- -------------> H+ + PO43-
0.100 0 0
0.100-x x x
Ka3 = x^2 / 0.100 - x
4.8×10–13 = x^2 / 0.100 - x
x = 2.19 x 10^-7
[H+] = 2.19 x 10^-7 M
pH = -log (2.19 x 10^-7)
pH = 6.66
[H+] = 2.19 x 10^-7 M
[HPO42-] = 0.100 M
[PO43-] = 2.19 x 10^-7 M
HPO42- + H2O --------------> H2PO4- + OH-
0.100 0 0
0.100 - x x x
Kb = x^2 / 0.100 - x
1.61 x 10^-7 = x^2 / 0.100 - x
x = 1.27 x 10^-4
[H2PO4-] = 1.27 x 10^-4 M