Hello every body I have a home work but I didn't understand it because I was absent for while from school and now please please I need your help to understand this one:

0557 mol of a weak acid  HA was placed, and 13.1 g NaOH in enough water to produce 1.00 L of solution. The final pH of this solution is 4.68. What is the pKa of HA?

All what I did is I've converted 13.1g to mol and tried Ice table but I cannot find the pKa 


Thank you for your time.

You can use Henderson–Hasselbalch equation
pH =pKa  + log([A-]/[HA])
now NaOH is a strong base
take out the concentration of NaOH
[NaOH]  =  moles / volume in L
moles = mass/molar mass
Volume = 1L
[NaOH]  = [A-] (after the neutralization reaction)
[HA] left after the neutralization reaction
now you can plug in the values in the above equation