You have .1 moles of H3PO4 in a soln with a total volume of 250 mL. You then add .15 moles of NaOH and dilute the soln to a total volume of 1L calculate the pH of the resulting soln and the concentrations of all aqueous species. Thanks!!!!
Perhaps I can get you started.
.......H3PO4 + OH^- ==> H2PO4^- + H2O
I........0.1...0.15.......0.........0
C.......-0.1..-0.10......+0.1....+0.1
E..........0....0.05.....+0.1....+0.1
Now you have a solution of 0.05 mol NaOH and 0.1 mol H2PO4^- which reacts similarly.
.......H2PO4^- + OH^- ==> HPO4^= + H2O
I.......0.1......0.05......0.......0
C......-0.05....-0.05....+0.05...+0.05
E.......0.05....0.0......0.05....0.05
The pH can be determined from the Henderson-Hasselbalch equation. Note that all of the numbers are mols; you must convert to M by mols/1L = which doesn't change anything. This gives you directly (H2PO4^-), (HPO4^=), and (H^+). You can find OH^- from (H^+)(OH^-) = Kw and you can find (PO4^3-) from k3 for H3PO4.