Can someone help me with this pH problem.?

Question

Consider 50 mL of a solution that is 0.25 M in both H2PO41- and HPO42-.

What is the pH of the solution?

Answer 1

You can find the Ka constant for H3PO4 any place:
H2PO4(–)(aq)+ H2O(l) <==> H3O+(aq) + HPO4(2–)(aq) Ka2= 6.2×10−8
That means: Ka2= 6.2×10−8 = [H3O+]*[HPO4(2–)]/[H2PO4(–)] = [H3O+]
Thus pH = - log[H3O+] = - log(6.2×10−8) = 7.2