Could someone please help me with a pH calculation?

First off, all of the NaOH added will neutralize that much HCl. You're adding 0.005 L x 0.1 mole/L = 0.0005 moles NaOH to 0.02 L x 0.1 mole/L = 0.002 moles HCl. After the reaction you'll have 0.0015 moles HCl left. The new volume will be 25 ml, or 0.025 L, so the new concentration will be 0.0015 moles/0.025 L = 0.060 M. Since M H3O+ = M HCl, you have a 0.060 M solution of H3O+, and pH = -log(0.060) = 1.22.