volume of 0.0924 M NaOH

Question

What volume of 0.0924 M NaOH would be required to exactly neutralize 6.25 mL of a vinegar solution which is 3.54% acetic acid by mass? The density of vinegar is 1.005 g/mL. Show all calculations.

Answer 1

The chemical equation describing this reaction is as follows:

NaOH + CH3COOH ==> CH3COONa + H2O

(1). find moles of CH3COOH present:

6.25 ml x 1.005 g/ml x 3.54 g/100 g = 0.222 g CH3COOH present

(2). convert g to moles using the molar mass of CH3COOH

0.222 g CH3COOH x 1 mol / 60.1 g = 0.003699 mols

(3). find moles of NaOH needed:

0.003699 mols CH3COOH x 1 mol NaOH / mol CH3COOH = 0.003699 mols NaOH needed

(4). find volume of NaOH needed:

0.003699 mols NaOH x 1 L / 0.0924 mols = 0.0400 L NaOH = 40.0 mls of NaOH needed