Question

The formal composition of an aqueous solution is 0.12M K_2HPO_4 + 0.08 M KH_2PO_4. Using the data in Table 2-3, calculate the concentrations of all ionic and molecular species in the solution and the pH of the solution.

Answer 1

H3PO4 has three ionisations and each ionisation has different pKa values.
Given;
[H2PO4-] = 0.08 M (the initial value)
[HPO4-2] = 0.12 M (the initial value)

We can use the Henderson-Hasselbalch equation
and pKa2 to calculate the pH and then [H+] of this solution: (pKa2 =6.8; from liturature)

pH = 6.8 + log [HPO42-]/[H2PO4-]

pH = 6.8 + log (0.12/0.08) = 6.976

[H+] = 10^-6.976 = 1.05 x10^-7 M

Now, we can use the first pKa1, [H2PO4-] and the pH of the solution to calculate the concentration of H3PO4 in the solution:

6.976 = 2 + log [0.08]/[H3PO4]
[H3PO4] = 8.454 x 10^-7 M

From the pH, we can [OH-]:

pH + pOH = 14.00
6.976 + pOH = 14.00
pOH = 7.024
[OH-] = 10^-pOH = 9.46X10^-8 M

with K2HPO4 and KH2PO4 concentrations, we can calculate the [K+]

From K2HPO4, [K+] = 2(0.08) = 0.16 M;
from KH2PO4, [K+] = 0.12 M;
Therefore, total [K+] = 0.28 M