A 1.00mL vinegar sample was completely neutralized by 22.5mL 0.2 M NaOH solution. Calculate molarity and...

A 1.00mL vinegar sample was completely neutralized by 22.5mL 0.2 M NaOH solution. Calculate molarity and % of Acetic acid in vinegar.

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Expert Solution

One mole NaOH reacts with one mole acetic acid.

So, the number of moles of acetic acid is equal to the number of moles of NaOH.

Moles of NaOH = Volume of NaOH x Molarity of NaOH = 22.5 mL x 0.2 M = 4.5 mmol

So, the number of moles of acetic acid is 4.5 mmol.

Molarity of acetic acid = moles of acetic acid/volume of vinegar = 4.5 mmol / 1.00 mL = 4.5 M

Mass of acetic acid = moles of acetic acid x molar mass of acetic acid

= 4.5 mmol x (60.05 g/mol) = 270 mg = 0.270 g

% of acetic acid in vinegar =( mass of acetic acid / volume of vinegar)x100

= 0.270 g/1.00 mL) x100

= 27%

queen_honey_blossom answered 1 year ago

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