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The aim of this exercise is to determine the percentage (degree) of acetic acid in vinegar.

For this purpose 10 mL of vinegar requires 8mL of sodium hydroxide of molar concentration 0.1mol/L to react with the acetic acid present.

The equation of the reaction is :

$\ce{CH3COOH + HO^- -> CH3COO^- + H2O}$

a- Find the mass of acetic acid present in 10mL vinegar.

b- Find the percentage of acetic acid in vinegar.

Given: M (CH3COOH) = 60 g/mol Density of vinegar = 1.02g/mL

I solved question (a) in which the answer is mass = 0.048g, however I'm confused with question (b)

I think I should use this formula but I'm not sure about it.

$C = \frac{\%d}{100M} $ where d is the density and M is the Molar mass.

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Instead of using a formula, think ! How much does $\pu{10 mL}$ vinegar weigh ? Answer: $\pu{10.2 g}$.

You have found that this sample contains $\pu{0.048 g}$ acetic acid. As a consequence, the percentage of acid in vinegar is : $\frac{\pu{0.048 g}}{\pu{10.2 g}} = \pu{4.7E-3} = 0.47$%

P.S. Are you sure of these numerical values ? Usual vinegars are at least $12$ times more concentrated !

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