COLLEGE CHEMISTRY QUESTION: How to execute experiment to determine PERCENT BY MASS OF ACETIC ACID IN VINEGAR?
heres what i did:
i got a solution of household vinegar, which was just a mixture of acetic tart and water. i measured out 10.0 mL of that vinegar and diluted it with distilled water until the total solution be 250. mL. then, i took 25.0 mL of this diluted solution and titrated it with NaOH.
the reaction is this:
HC2H3O2 (aq) + OH- (aq) ---> C2H3O2- (aq) + H2O (l)
in the order of 16.7 mL of .0500 M NaOH was needed to titrate 35.0 mL of the diluted vinegar. the density of the original househod vinegar is 1.05 g/mL.
my question is: what is the percent by mass of acetic tart in the original household vinegar?
thanks
Answers:
10/240=.04106(rep)
.04106x100=4.106 Source(s): rn
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i got a solution of household vinegar, which was just a mixture of acetic tart and water. i measured out 10.0 mL of that vinegar and diluted it with distilled water until the total solution be 250. mL. then, i took 25.0 mL of this diluted solution and titrated it with NaOH.
the reaction is this:
HC2H3O2 (aq) + OH- (aq) ---> C2H3O2- (aq) + H2O (l)
in the order of 16.7 mL of .0500 M NaOH was needed to titrate 35.0 mL of the diluted vinegar. the density of the original househod vinegar is 1.05 g/mL.
my question is: what is the percent by mass of acetic tart in the original household vinegar?
thanks
Answers:
10/240=.04106(rep)
.04106x100=4.106 Source(s): rn
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