What is the pH of the resulting buffer?Help please? i need to know how to do this

500.0 mL of 0.150 M NaOH is added to 625 mL of 0.200 M weak acid (Ka = 6.22 × 10-5).

HA(aq)+OH-(aq)--->H2O(l)+A-(aq)

Moles NaOH added = .150 Molar times .500 Liters = .075 moles

Moles weak acid HA .200 Molar times .625 liters = .125 moles

HA + NaOH = H20 + Na+ + A- By observation you have more moles of HA than you have of NaOH, so all of the NaOH will be neutralized and you will be left with

.125 moles HA minus .075 moles NaOH = .05 moles of HA

and the moles of A- created in this neutralization process will be the same as the moles of NaOH because each mole of OH- reacted with a mole of HA to make a mole of A- and a mole of water.

so moles of A- present would be .075..

so we have a solution with a total volume of 500 ml plus 625 ml or ll25 ml or l.125 liters.

New molarity of HA = .05 moles over l.125 Liters = .044

Now molarity A- = .075 moles over l.125 liters = .066.

Buffer equation is (H+) = Ka times (HA) over (A-)

so (H+) = 6.22 x l0^-5 ) times (.044) over (.066)

(H+) = 4.15 x l0^-5

pH = -log(H+) = -log 4.15 x l0^-5 = 4.38