"Calculate the number of moles of NaHCO3 that were required to neutralize the CH3COOH in the vinegar."

I had to add NaHCO3 to a bowl of vinegar. It took an average of 142.3 drops of NaHCO3 using an eyedropper to neutralize the CH3COOH in the vinegar. The volume of the NaHCO3 was 7.3 mL, while the volume of the CH3COOH was 9.86 mL.
I'm not sure how to find the answer to this question though. Any help would be appreciated!