Question

The formal composition of an aqueous solution is 0.12 M K2HPO4 + 0.08 KH2PO4. The pKa's for triprotic phosphoric acid are 2.12, 7.21, and 12.32. Show the three equilibriums in solution and calculate the pH of the solution. Find concentrations of K+, H3PO4, H2PO4^-, HPO4^2-, PO4^3-, H+, and OH^- in the mixture.

Answer 1

use the Henderson-Hasselbalch equation and pKa2 to calculate the pH (and then [H+] of this solution:

pH = 7.21 + log [HPO42-]/[H2PO4-]
pH = 7.21 + log (0.12/0.08)

    = 7.38
[H+] = 10^-7.38 = 4.168 * 10^-8 M

Now, use the first pKa, the concentration of H2PO4- and the pH of the solution to calculate the concentration of H3PO4 in the solution:
7.38 = 2.12 + log [0.08]/[H3PO4]
[H3PO4] = 4.396 * 10^-7 M

From the pH, calculate pOH, and then [OH-]:
pH + pOH = 14.00
7.38 + pOH = 14.00
pOH = 6.62
[OH-] = 10^-pOH = 2.398*10^-7 M

From K2HPO4, [K+] = 2(0.12) = 0.24 M, and from KH2PO4, [K+] = 0.08 M.

So, total [K+] = 0.32 M

[H2PO4-] = 0.08 M and [HPO42-] = 0.12 M