g12 chem part way through a titration question?

Initial moles of NaOH = (0.10 mol/L) × (20/1000 L) = 0.002 mol
Initial moles of HCl = (0.10 mol/L) × (30/1000 L) = 0.003 mol

Equation for the reaction in titration:
NaOH + HCl → NaCl + H₂O
Mole ratio NaOH : HCl = 1 : 1
As (Initial moles of HCl) > (Initial moles of NaOH), HCl is in excess.

Moles of excess HCl = (0.003 - 0.002) mol = 0.001 mol
Volume of the final solution = (20 + 30) ml = 50 ml = 0.05 L
Molarity of excess HCl, [HCl]ₒ = (0.001 mol) / (0.05 L) = 0.02 mol/L
[H⁺] = [HCl]ₒ = 0.02 mol/L
pH = -log[H⁺] = -log(0.02) = 1.7

have you tried
.. (1) writing down the equation for pH?
.. (2) calculating moles OH- added vs moles H+ in solution?
.. . .. to see which is in XS?

should take you about < 5 min to solve.  I know that because I can solve it in.. about 30 seconds.