Calculate the ph? Need help!

"500 mL of 0.100 M NaOH is added to to 625 mL of 0.200 M weak acid (Ka=5.11 * 10^-5). What is the pH of the resulting buffer?"

I know I have to use the Henderson-Hasselbach equation, which is ph=pKa + log(base/acid)
I have to find the moles of the weak acid and NaOH.

Moles NaOH -----> 0.5 L x 0.1 M= 0.05 mol NaOH. I think this is the limiting reagent?
Moles W.A. -------> 0.625 L x 0.2 M= .125 mol of the weak acid

pKa= -log(5.11 x 10^-5)= 4.29

I'm not sure what to do next. I guess I need to find the amount of base and acid left over and find the ratio? And then use the H.H equation from there? Please help!

When you add the NaoH to the weak acid you produce the sodium salt of the acid. We assume that the weak acid is monoprotic:

Mol NaOH in 500mL of 0.10M solution = 0.5*0.1 = 0.05 mol NaOH
Mol acid in 625mL of 0.20M solution = 0.625*0.2 = 0.125 mol acid

On mixing you produce 0.05 mol Na salt and 0.125-0.05 = 0.075 mol acid remaining unreacted
These are dissolved in 500+625 = 1,125mL = 1.125L solution
Molarity salt = 0.05/1.125 = 0.0444M
Molarity acid = 0.075/1.125 = 0.0666M
pKa of acid = -log (5.11*10^-5) = 4.29

Now apply the H-H equation:
pH = pKa + log ([salt]/[acid])
pH = 4.29 + log ( 0.0444*0.0666)
pH = 4.29 + log 0.666
pH = 4.29+ (-0.18)
pH = 4.11