Titration of vinegar question

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Homework Statement


A solution of household vinegar (a mixture of acetic acid and water) is to be analyzed. A pipet is used to measure out 10.0 mL of the vinegar, which is placed in a 250 mL volumetricflask. Distilled water is added until the total volume of solution is 250 mL. A 25.0 nL portion of the diluted solution is measured out with a pipet and titrated with a standard solution of sodium hydroxide.

The neutralization reaction is as follows:

HC2H3O2 (aq) + OH- (aq) --> C2H3O2- (aq) + H2O (l)

1) It is found that 16.7 mL of 0.0500 M NaOH is needed to titrate 25.0 mL of the diluted vinegar. Calculate the molarity of the diluted vinegar.
.0167 L NaOH x .0500 mol NaOH / 1 L NaOH = 8.35 x 10-4 mol NaOH

8.35 x 10-4 mol NaOH x 1 mol HC2H3O2 / 1 mol NaOH = 8.35 x 10-4 mol HC2H3O2


8.35 x 10-4 mol HC2H3O2 / .0250 L = 0.0334 M HC2H3O2

2) Calculate the molarity of the household vinegar.

250 / 25 = 10

8.35 x 10-4 mol HC2H3O2 x 10 = 8.35 x 10-3 mol HC2H3O2

8.35 x 10-3 mol HC2H3O2 / .0100 L =
0.835 M HC2H3O2

3) The household vinegar has a density of 1.05 g/mL. Calculate the percent by mass of acetic acid in the household vinegar.

1.05 g/mL x 10 mL = 10.5 g HC2H3O2 + H2O

8.35 x 10-3 mol HC2H3O2 x 60.04 g/mol =
0.501 g HC2H3O2

(0.501 g HC2H3O2/ 10.5 g) x 100 = 4.77% HC2H3O2
 
Last edited:

Answers and Replies

  • #2
Borek
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Looks OK to me.
 

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