The 2 mL of 0.2 M solution of NaOH was added in 10 mL of 0.04 M solution of H3PO4.?

Question

Calculate the concentraiton HPO4^2- ions.

Answer 1

The three pKa values of H3PO4 are: 2.12, 7.21, 12.67.
Once 2 mL of 0.2 M solution of NaOH added in 10 mL of 0.04 M solution of H3PO4, the INITIAL concentration of NaOH is: (2/12)x0.2M = (1/30)M, and of H3PO4 is: (10/12)x0.04M = (1/30)M. We may consider the initial condition is after a complete (first step) reaction between NaOH and H3PO4 to form NaH2PO4:
NaOH + H3PO4 ==> NaH2PO4 ==> Na+ + H2PO4(-)
Therefore, we have 12ml of (1/30)M NaH2PO4. The question is what is [HPO4(2-)]?
Let a = 1/30. We may assume the concentraiton of HPO4(2-) ions at equilibrium to be X: [HPO4(2-)] = X, and the concentration of H3PO4 to be Y: [H3PO4] = Y. We also assume water self-dissociation contribution to [OH-] and [H3O+] to be Z. Thus in the following reaction:
OH- + H3PO4 <==> H2O + H2PO4- <==> H3O+ + HPO4(2-)
Y+Z......Y.................................a - X - Y............X+Z.......X
We have:
Ka1 = 10^(-2.12) = [H3O+]*[H2PO4-]/[H3PO4] = (X+Z)(a-X-Y)/Y
Ka2 = 10^(-7.21) = [H3O+]*[HPO4--]/[H2PO4-] = X(X+Z)/(a-X-Y)
Kw = 10^(-14) = [OH-]*[H3O+] = (X+Z)(Y+Z)
These are 3-variable 3 equations. Solve them we would get required value X.
But it is quite difficult to solve them analytically. Thus we will introduce some approximation. We believe that the pH would be acidic, thus X >>Z. Let us at this moment ignore the contribution of water self-dissociation to the concentration [OH-] and [H3O+], and we will check if this assumption is valid or not afterwards.
So the equations becomes:
10^(-2.12) = X(a-X-Y)/Y
10^(-7.21) = X*X/(a-X-Y)
10^(-14.0) = XY
Multiply all three equations together we get:
10^(-2.12-7.21-14.0) = X^4
X = 10^(-5.83) (M)
This proves our assumption X >> Z since Z will be no more than 10^(-14+5.83) M = 10^(-8.17)M.
The concentraiton of HPO4(2-) ions is 10^(-6) M.