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The distinctive odor of vinegar is due to acetic acid, HC₂H₃O₂, which reacts with sodium hydroxide. If a 3.45 mL sample of vinegar needs 42.5 mL of 0.115 M NaOH to be neutralized, how many grams of acetic acid are in a 1.00 L bottle of this vinegar?​

Respuesta :

Answer:

Explanation:

CH₃COOH + NaOH = CH₃COONa + H₂O .

42.5 mL of .115 M of NaOH will contain .0425 x .115 moles of NaOH

= 48.875 x 10⁻⁴ moles NaOH

It will react with same number of moles of acetic acid

So number of moles of acetic acid in 3.45 mL = 48.875 x 10⁻⁴

number of moles of acetic acid in 1000 mL = 48.875 x 10⁻⁴ x 10³ / 3.45 moles

= 1.4167 moles

= 1.4167 x 60 gram

= 85 grams .

So 85 grams of acetic acid will be contained in one litre of acetic acid.

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