aicd base help?

Question

A 20.0-mL sample of 0.25 M HNO3 is titrated with 0.15 M Na OH. What is the pH of the solution after 30.0 mL of NaOH have been added to the acid?
2.00
1.60
1.05
1.00
none of the above

plz show me work so i can understand, thanks in advance

Answer 1

20 mL of 0.25 M HNO3 contains 0.005 moles H+

30 mL of 0.15 M NaOH contains 0.0045 moles OH-

0.005-0.0045 = 0.0005 mole H+ in excess

Total volume = 50 mL= 0.050 L

M= 0.0005/0.050= 0.01 M

pH= -log 0.01 = 2