Question

3 mL of 0.15 M solution of NaOH was added in 10 mL of 0.05 solution of H3PO4. Calculate the concentration of HPO4^2- ions. (pka of phosphoric acid - 2.15, 7.2, 12.3)

Answer 1

OH- + H3PO4 <==> H2O + H2PO4- <==> H3O+ + HPO4(2-)
Y+Z......Y............................... - X - Y............X+Z.......X
We have:
Ka1 = 10^(-2.15) = [H3O+]*[H2PO4-]/[H3PO4] = (X+Z)(a-X-Y)/Y
Ka2 = 10^(-7.2) = [H3O+]*[HPO4--]/[H2PO4-] = X(X+Z)/(a-X-Y)
Kw = 10^(-12.3) = [OH-]*[H3O+] = (X+Z)(Y+Z)

Since pH is in acid range, assume X >>> Z

Then,

10^(-2.15) = X(a-X-Y)/Y

10^(-7.2) = X*X/(a-X-Y)

10^(-12.3) = XY

Multiply all three equations together we get:
10^(-2.15-7.2-12.3) = X^4
X = 10^(-5.41) (M)

So, conc of  HPO4^2- == 3.87 x10^-6 M