Volumetric Analysis- Acid-Base Titration Concentration of Acetic Acid in Commercial sample of Vinegar 1. Brand name of vinegar sample: America's Choice 2. Volume of Vinegar sample used: 3.0ml mL 0.0030L %3D 3. Percent (%) of acetic acid on label: 4. Concentration (in Molarity, M) of NaOH solution:_0,28 5 M Trial 1 Trial 2 Trial 3 Initial level of NaOH solution in the burette (V1) 0.8 0-0 1.0 Final Level of NaOH solution in the burette (V2) 48-1 47.4 48.3 Volume (in mL) of NaOH solution used (V2-V1) 47.3ml 47.4ml 47.3ml Average Volume (in mL) of NaOH solution used 47.3 ml Average volume (in L) of NaOH solution used 0.0473 5. Moles of NaOH used in titration (show your calculations) 0.0135 moles of NaOH 0.285 H x O-0473=0.0135 6. Moles of Acetic acid (HC2H3O2) neutralized by NaOH O -0135 moles of HC2H3O2 7. Molarity of acetic acid, HC2H3O2 in vinegar sample (show your calculations) TZ vinerar ontain= Q0135 니.S M 4.5 moles of ctic acid 8 O-003 0.81 g HC2H3O2 8.Grams of Acetic Acid, HC2H3O2 (show your calculations) 0.013 SX 60.052= 0.810 O.819 acelie acid presentin 3ml of Vinegar sample 270 % 9. Percent (m/v) of HC2H3O2 in vinegar sample (show your calculations) 0.81 = 0.27 (glml) 3.0 - 0+27 x10?(g/) = 270 Question 1: Is your experimentally calculated value of percent of HC2H3O2 is high or low compare to the percent of acetic acid value listed on the label of your vinegar sample bottle? Explain, why do you have a different calculated percent concentration?
Volumetric Analysis- Acid-Base Titration Concentration of Acetic Acid in Commercial sample of Vinegar 1. Brand name of vinegar sample: America's Choice 2. Volume of Vinegar sample used: 3.0ml mL 0.0030L %3D 3. Percent (%) of acetic acid on label: 4. Concentration (in Molarity, M) of NaOH solution:_0,28 5 M Trial 1 Trial 2 Trial 3 Initial level of NaOH solution in the burette (V1) 0.8 0-0 1.0 Final Level of NaOH solution in the burette (V2) 48-1 47.4 48.3 Volume (in mL) of NaOH solution used (V2-V1) 47.3ml 47.4ml 47.3ml Average Volume (in mL) of NaOH solution used 47.3 ml Average volume (in L) of NaOH solution used 0.0473 5. Moles of NaOH used in titration (show your calculations) 0.0135 moles of NaOH 0.285 H x O-0473=0.0135 6. Moles of Acetic acid (HC2H3O2) neutralized by NaOH O -0135 moles of HC2H3O2 7. Molarity of acetic acid, HC2H3O2 in vinegar sample (show your calculations) TZ vinerar ontain= Q0135 니.S M 4.5 moles of ctic acid 8 O-003 0.81 g HC2H3O2 8.Grams of Acetic Acid, HC2H3O2 (show your calculations) 0.013 SX 60.052= 0.810 O.819 acelie acid presentin 3ml of Vinegar sample 270 % 9. Percent (m/v) of HC2H3O2 in vinegar sample (show your calculations) 0.81 = 0.27 (glml) 3.0 - 0+27 x10?(g/) = 270 Question 1: Is your experimentally calculated value of percent of HC2H3O2 is high or low compare to the percent of acetic acid value listed on the label of your vinegar sample bottle? Explain, why do you have a different calculated percent concentration?
Chapter8: Polyfunctional Acids And Bases
Section: Chapter Questions
Problem 10P
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Please anwer the last questions. Question 1 and 2
Transcribed Image Text:Volumetric Analysis- Acid-Base Titration
Concentration of Acetic Acid in Commercial sample of Vinegar
1. Brand name of vinegar sample:
America's Choice
2. Volume of Vinegar sample used:
3.0ml
mL
0.0030L
%3D
3. Percent (%) of acetic acid on label:
4. Concentration (in Molarity, M) of NaOH solution: 0.285 M
Trial 1
Trial 2
Trial 3
Initial level of NaOH solution in
the burette (V1)
0.8
0.0
1.0
Final Level of NaOH solution in
the burette (V2)
48-1
47.4
48.3
Volume (in mL) of NaOH solution
used (V2-V1)
Average Volume (in mL) of
47.3ml
47.4ml
47.3ml
NaOH solution used
47.3 ml
Average volume (in L) of NaOH
solution used
0.0473
5. Moles of NAOH used in titration
(show your calculations)
10.0135
moles of NaOH
0.285 H x O.0473=0.0135
6. Moles of Acetic acid (HC2H3O2) neutralized by NaOH
O-0133
moles of HC2H3O2
7. Molarity of acetic acid, HC2H3O2 in vinegar sample
(show your calculations)
12 vinenar contain=0.0135
4.5
- 4.5 moles of cctic acid
O-003
8.
0.81
g HC2H3O2
8.Grams of Acetic Acid, HC2H3O2
(show your calculations)
0.013 SX60.052= 0.810
O. 81q ocedie aid presentin 3ml of Vinegar sample
270
%
9. Percent (m/v) of HC2H3O2 in vinegar sample
(show your calculations)
O.81
= 0.27 (g/ml)
3.0
-0+27 x10
=270
Question 1: Is your experimentally calculated value of percent of HC2H3O2 is high or low compare
to the percent of acetic acid value listed on the label of your vinegar sample bottle? Explain, why
do have a different calculated percent concentration?
you
Question 2: You have added distilled water to the sample of vinegar solution before you started
the titration. Does this addition of water to the vinegar sample affect the calculated molarity of
acetic acid? Support your answer with proper reasoning.
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