pH question. Please help.?

Question

For H3PO4, a tripotic acid, Ka1 = 7.1 x 10–3, Ka2 = 6.3 x 10–8, and Ka3 = 4.5 x 10–13. What is the pH of a solution which contains 0.300 moles of Na2HPO4 and 0.400 moles of KH2PO4 per liter of solution?

Answer 1

Na2HPO4 is a strong electrolyte and is 100% dissociated in solution to make 0.300 mole HPO42-

KH2PO4 is a strong electrolyte and is 100% dissociated in solution to make 0.400 mole H2PO4-

Their concentrations are :
HPO42- = 0.300/1=0.300 M
H2PO4- = 0.400/1 =0.400 M

Let x=moles/L H2PO4- that dissociate

H2PO4- <> H+ + HPO4-
initial concentration
0.400...........0.........0.300
at equilibrium
0.400-x.........x.........0.300+x

K2=6.2 10^-8 = (x)(0.300+x) / 0.400-x

x =0.000000083 M

pH = - log0.000000083 = 7.08

Answer 2

You are forming a buffer solution.

So you can find the pH by applying the approximation of the Henderson-Hasselbalch equation

pH= pKa+log ([conj.base] / [acid])

Here we are talking about the second dissociation so
pH= pKa2+ log([HPO4(-2)]/[H2PO4(-)]) =
=-log(6.3*10^-8) +log(0.3/0.4) =7.08

This solution makes several assumptions. If you look at the pH value, it is so close to 7 that we ought to take the self-ionization of water into account if we want more accurate results. Then we would have to set up ICE tables for both the phosphate and water and solve the system of equations.

If you want that (though I doubt it is necessary) please e-mail me so that I can edit the answer accordingly