Assuming that the reaction is unimolecular on both reactant and product sides such that:
HA + OH- --> A- + H2O
Therefore the amount of A- left is:
mmol HA = 9.00 - 3.00 = 6.00
mmol A- = 3.00
pKa = - log Ka = 5.25
Calculating for pH:
pH = 5.25 + log 3.00/6.00