Question

Prepare 250ml of 0.2M phosphate buffer of pH 6.4.starting from Na2HPO4,NaH2PO4,0.129N NaOH solution.

Answer 1

I used a pKa2 for H3PO4 of 7.21. You should use the value in your text/notes.
pH = pKa2 + log (base)/(acid)
6.4 = 7.21 + log b/a
b/a = about 0.15 but you should do all of this more accurately.
Then b = 0.15*a. That's equation 1.
equation 2 is a + b = 0.2 M
Solve these two equation simultaneously for a and b. I obtained values of approx a = 0.17 M and b = about 0.03 M. Again you should do it more accurately.

You want 250 mL; therfore,
millimols acid = 250 x 0.17 = about 43
mmols base = 250 x 0.03 = about 7
mmols buffer = 0.2 x 250 = 50.

So you want to weigh out 50 millimols H2PO4^- (convert to mmols NaH2PO4 and weigh out that much NaH2PO4).
Calculate mL 0.129 M NaOH to add to give you the approx 7 mmols HPO4^- and voila! Place all of this in a 250 mL flask, add water to the mark and mix. I like to check these things by starting with 0.129 M NaOH and the grams NaH2PO4 and see if that actually gives you a pH of 6.4.