Chemistry molarity

Question

Acid content in vinegar. 9.65 ml of 1.05 M NaOH is required to neutralize 10.00 ml of vinegar. Calculate the moles of NaOH.

Answer 1

The acid in vinegar is acetic acid (HAc). It has the structure of

CH3COOH. The reaction between NaOH and HAc is...

NaOH + HAc ==> H₂O + NaAc

moles NaOH used = 0.00965 L x 1.05 mol/L = 0.0101 moles (to 3 sig figs).

Maybe your question meant to ask for moles or molarity of the acid.