Calculating pH of solutions?

Calculating pH and Buffer Solutions?

• Ok so i have the following question for chemistry Ka for HPO4^2- = 4.4x10^-13. A buffer system of 1.00 mole of Na2HPO4 and 1.00 mole Na3PO4 to which 0.40 mol of OH- has been added (all in 1.00L of solution) would have a pH of what? I'm not sure how start to work this out, any help much appreciated!

• Answer:

  1 Na2HPO4 & 1 NaOH --> 1 Na3PO4 & 1 H2O 1.00 mol Na2HPO4 reacts with 0.40 mol of OH- producing an additional 0.4 0moles of Na3PO4 however it leaves behind 0.60 moles of Na2HPO4 0.60 mol of Na2HPO4 / litre releases 0.60 Molar (HPO4-) 1.00 mole Na3PO4 & an additional 0.4 mol Na3PO4 / litre releases a total of 1.40 Molar (PO4)-3 (HPO4)-2 <=> H+ & (PO4)-3 Ka = [H+] [(PO4)] / [HPO4] 4.4x10^-13 = [H+] [1.40M] / [0.60M] [H+] = (4.4x10^-13) / 2.333 [H+] = 1.89 e-13 Molar pH = 12.72