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.
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?
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
Pay attention.
You must have used 0.010 for L NaOH but the problem says volume NaOH is 16.88 mL.
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?
What is the weight per volume percent (i.e. % density) of 0.5522 g acetic acid in the 10.00 mL sample?
Lashio Education College
Chemistry Department
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!
ВКЭн
a. To write a balanced equation for the neutralization reaction between acetic acid (HC2H3O2) and sodium hydroxide (NaOH), we need the stoichiometric coefficients of both reactants and products. The general equation for the reaction can be written as follows:
HC2H3O2 + NaOH -> NaC2H3O2 + H2O
b. To find the molarity of the acetic acid in the vinegar, first, we need to calculate the number of moles of NaOH used in the titration.
Number of moles of NaOH = Molarity of NaOH × Volume of NaOH used (in liters)
= 0.5052 mol/L × 0.01688 L
Next, we can use the stoichiometry of the balanced equation to find the moles of acetic acid used in the reaction. The ratio between NaOH and acetic acid is 1:1.
Number of moles of acetic acid = Number of moles of NaOH used
Finally, we can calculate the molarity of the acetic acid in the vinegar.
Molarity of acetic acid = Number of moles of acetic acid / Volume of vinegar (in liters)
= Number of moles of acetic acid / 0.01000 L
c. To calculate the mass percent of acetic acid in the vinegar, we need to determine the mass of the acetic acid present in the 10.00 mL sample. The mass can be calculated using the density of the vinegar.
Mass of acetic acid = Volume of vinegar × Density of vinegar × Mass percent of acetic acid
Mass percent of acetic acid = (Mass of acetic acid / Mass of vinegar) × 100%
Considering that the density of the vinegar is 1.006 g/mL, we can rearrange the equation to find the mass percent:
Mass percent of acetic acid = (Mass of acetic acid / (Volume of vinegar × Density of vinegar)) × 100%
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.