Acetic acid (HC2H3O2) is an important component of vinegar. A 10.00mL sample of vinegar is titrated with .5052 M NaOH, and 16.88 mL are required to neutralize the acetic acid that is present.

Question

a.write a balanced equation for this neutralization reaction
b.what is the molarity of the acetic acid in this vinegar?
c.If the density of the vinegar is 1.006 g/mL, what is the mass percent of acetic acid in the vinegar?

Answer 1

1/
HC2H3O2 + NaOH ==> HOH + NaC2H3O2

2.
Calculate mols NaOH from L x M = ??
Convert mols NaOH to mols HC2H3O2 using the coefficients in the balanced equation.
M acetic acid= mols/L.

3.
Determine mols acetic acid in the 10 mL sample and from that grams acetic acid, then calculate percent from
[grams acetic acid/mass sample]*100

Post your work if you get stuck.

Answer 2

ok for part 2
moles of NaOH=L*M
.01*.5052=.005052 moles of NaOH
then moles of NaOH =moles of HC2H3O2 since 1-1 ratio from equation that makes .005052 moles of HC2H3O2 correct?
then M acetic acid=.005052 moles/.01688L which =2.9*10^-1 is that correct?

and for part 3 is it mols of acetic acid=.01688L*.5052M and then I'm kinda lost

Answer 3

Pay attention.
You must have used 0.010 for L NaOH but the problem says volume NaOH is 16.88 mL.

Answer 4

Assume that your vinegar contained a small amount of citric acid (a triprotic acid).
Using the same experimental data, would you expect the molarity of this sample to be
the same as or different than a sample which contained only pure acetic acid?

Answer 5

no

Answer 6

What is the weight per volume percent (i.e. % density) of 0.5522 g acetic acid in the 10.00 mL sample?

Answer 7

Lashio Education College
Chemistry Department

Answer 8

1:b
2:d
3:a
4:b
5: a
6: b
7:d
8:b
9:b
10:a
11: d
12:b
13: d
14:c
15 b
The rest you will have to do it yourself!

Answer 9

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