PH buffer question plz help and explain thnx :D?

Question

you have 1 L of buffer solution that contains 9.0 mM weak acid and 1.0 mM of its conjugate base. the pKa of this weak acid is 3.95. you add .06 g NaOh to this buffer solution. what is the resultant pH?

Answer 1

0.06 g NaOH = 1.5 mmol NaOH

9.0 mM weak acid + 1.5 mM NaOH --> 7.5 mM weak acid

1.0 mM conjugate base + 1.5 mM (from NaOH reaction with weak acid) = 2.5 mM

pH = pKa + log(A-/HA) = 3.95 + log(2.5/7.5)

pH = 3.95 + (-0.48) = 3.47

pH before the addition of NaOH was-

pH = 3.95 + log(1.0/9.0) = 3.95 + (-0.95) = 3.00