A buffer solution of volume 100.0 mL is 0.140 M Na2HPO4(aq) and 0.120 M KH2PO4(aq). What are the pH and the pH change resulting from the addition of 55.7 mL of 0.0100 M NaOH(aq) to the buffer solution?

Answer 1

pH = pKa2 + log (base)/(acid)
base is 0.140M Na2HPO4
acid is 0.120 M KH2PO4
Substitute and solve for pH.

I prefer to work in millimols (mmols) for the second part of the problem.
100.0 mL x 0.140 = 14.0 mmols base Na2HPO4
100.0 mL x 0.120 = 12.0 mmols acid KH2PO4
added 55.7 mL x 0.0100 MNaOH = 0.557 mmols.

.........H2PO4^- + OH^- ==> HPO4^- + H2O
I........12.0.......0.......14.0
add.............0.557................
C......-0.557..-0.557.......+0.557
E.......11.44.......0......14.557.....

Then substitute the E line into the HH equation and solve for pH.
Then delta pH is difference between beginning and end.

Answer 2

How do we figure out which one is a base and acid between Na2HPO4 and KH2PO4?