inegar? required to neutralize the acetic acid in each sample of vinegar. 16. Complete the calculations to find the molarity and percentage of acetic acid in vinegar. 15. Subtract the initial volume from the final volume to calculate the volume of NaOH Average volume of NaOH required to reach endpoint in mL Convert the average volume of NaOH required to reach endpoint from mL to L I DATA 269 12.55 TITRATION DATA (REPORT YOUR VOLUMES TO 1/100TH M TRIAL 1 TRIAL 2 Initial base buret reading Final base buret reading Volume of sodium hydroxide (N2OH) required to reach endpoint in mL Volume of vinegar used O15 12.55 TRIAL 3 12:55 24.11 EXTRA TRIAL 24.11 35.89 11.78 12.05 11.56 15.00 mL 15.00 mL 15.00 mL 15.00 mL (Show calculations here.) 12.05711-56+1178 3) 27.54ML -27.54 a7.54ML X L (Show calculations here.) :0-02754L 0:087541 d ba b00 Molarity of NAOH (recorded from the standard solution bottle) 0.1 I CALCULATIONS FOR MOLARITY OF ACETIC ACID IN VINEGAR 1. Using the known molarity of NaOH(aq) and the average volume of NaOH(aq) in liters Ol= required to reach the endpoint, calculate the moles of NaOH used in the titration. Show your calculations using dimensional analysis or the following equation: moles of NaOH a M = Liters of solution 0.02754L X 0.1 6-002754 moles of NaOH used to reach endpoint of titration 2. Since the acid/base mole ratio is 1:1 for this neutralization reaction, how many moles of acetic acid (HC,H,O,) are present in a 15.00 mL sample of diluted vinegar? moles of HC,H,O, in diluted vinegar sample 0:002754 K-11

World of Chemistry, 3rd edition
3rd Edition
ISBN:9781133109655
Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
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Chapter15: Solutions
Section: Chapter Questions
Problem 63A
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Question
Vinegar?
required to neutralize the acetic acid in each sample of vinegar,
16. Complete the calculations to find the molarity and percentage of acetic acid in vinegar.
Convert the average volume of NaOH required to reach endpoint from mL to L
Average volume of NaOH required to reach endpoint in mL
15. Subtract the initial volume from the final volume to calculate the volume of NaOH
I DATA
269
12.55
TITRATION DATA (REPORT YOUR VOLUMES TO 1/100TH MI
TRIAL 1
TRIAL 2
Initial base buret reading
Final base buret reading
Volume of sodium hydroxide
(NaOH) required to reach
endpoint in mL
Volume of vinegar used
O15
12.55
TRIAL 3
12:55
24.11
EXTRA TRIAL
24.11
35.39
11.78
12.05
11.56
15.00 mL
15.00 mL
15.00 mL
15.00 mL
(Show calculations here.)
12.05711-56+1178
-27.54
27.54ML
(Show calculations here.)
a7.54ML X IL
0-02754L
:0.02754L
nd bon
1600
Molarity of NAOH (recorded from the standard solution bottle)
0.1
I CALCULATIONS FOR MOLARITY OF ACETIC ACID IN VINEGAR
1. Using the known molarity of NaOH(aq) and the average volume of NaOH(aq) in liters O1=
required to reach the endpoint, calculate the moles of NaOH used in the titration. Show
your calculations using dimensional analysis or the following equation:
moles of NAOH CỢ.Ô2754L X 0. /
M =
Liters of solution
6-002754
2. Since the acid/base mole ratio is 1:1 for this neutralization reaction, how many moles of
acetic acid (HC,H;O,) are present in a 15.00 mL sample of diluted vinegar?
0.002754
moles of NaOH used to reach endpoint of titration
moles of HC,H,02 in diluted vinegar sample
K-11
Nowimuc
Transcribed Image Text:Vinegar? required to neutralize the acetic acid in each sample of vinegar, 16. Complete the calculations to find the molarity and percentage of acetic acid in vinegar. Convert the average volume of NaOH required to reach endpoint from mL to L Average volume of NaOH required to reach endpoint in mL 15. Subtract the initial volume from the final volume to calculate the volume of NaOH I DATA 269 12.55 TITRATION DATA (REPORT YOUR VOLUMES TO 1/100TH MI TRIAL 1 TRIAL 2 Initial base buret reading Final base buret reading Volume of sodium hydroxide (NaOH) required to reach endpoint in mL Volume of vinegar used O15 12.55 TRIAL 3 12:55 24.11 EXTRA TRIAL 24.11 35.39 11.78 12.05 11.56 15.00 mL 15.00 mL 15.00 mL 15.00 mL (Show calculations here.) 12.05711-56+1178 -27.54 27.54ML (Show calculations here.) a7.54ML X IL 0-02754L :0.02754L nd bon 1600 Molarity of NAOH (recorded from the standard solution bottle) 0.1 I CALCULATIONS FOR MOLARITY OF ACETIC ACID IN VINEGAR 1. Using the known molarity of NaOH(aq) and the average volume of NaOH(aq) in liters O1= required to reach the endpoint, calculate the moles of NaOH used in the titration. Show your calculations using dimensional analysis or the following equation: moles of NAOH CỢ.Ô2754L X 0. / M = Liters of solution 6-002754 2. Since the acid/base mole ratio is 1:1 for this neutralization reaction, how many moles of acetic acid (HC,H;O,) are present in a 15.00 mL sample of diluted vinegar? 0.002754 moles of NaOH used to reach endpoint of titration moles of HC,H,02 in diluted vinegar sample K-11 Nowimuc
EXPERIMENTK Acid-Base Titration: How Much Acid Is In Vinegar?
Discussion
1. The label on the bottle of commercial vinegar states that it contains 5.0% acetic acid.
According to your calculations, has the manufacture truthfully reported the percent of
acetic acid in vinegar? Support your answer with data.
2. A grocery store sells vinegar in 1-L bottles. Based on your answer to question 4 on the
previous page.
0.1 mole
6.09
a. how many moles of acetic acid are in a 1-L bottle of vinegar?
b. how many grams of acetic acid are in a 1-L bottle of vinegar?
Would the calculated molarity of acetic acid in vinegar be
diluted vinegar had been used in place of the 15.00 mL sample. Explain your answer.
3.
if 10.00 mL of
a.
the same
higher
c. lower
4. If you over-titrate, a significant error will occur in your calculations. Determine how
each of the following values would be affected if you over-titrate? Circle the cor-
rect response.
The volume of NaOH recorded to reach the equivalence point would be (too
high or too low).
a.
b. The calculated moles of NaOH used to reach the equivalence point would be (too
(high or too low).
с.
The calculated moles of acetic acid neutralized by the NaOH would be (too
(high
or too low).
d. The calculated molarity of acetic acid of the vinegar sample would be (too
high or too low).
Transcribed Image Text:EXPERIMENTK Acid-Base Titration: How Much Acid Is In Vinegar? Discussion 1. The label on the bottle of commercial vinegar states that it contains 5.0% acetic acid. According to your calculations, has the manufacture truthfully reported the percent of acetic acid in vinegar? Support your answer with data. 2. A grocery store sells vinegar in 1-L bottles. Based on your answer to question 4 on the previous page. 0.1 mole 6.09 a. how many moles of acetic acid are in a 1-L bottle of vinegar? b. how many grams of acetic acid are in a 1-L bottle of vinegar? Would the calculated molarity of acetic acid in vinegar be diluted vinegar had been used in place of the 15.00 mL sample. Explain your answer. 3. if 10.00 mL of a. the same higher c. lower 4. If you over-titrate, a significant error will occur in your calculations. Determine how each of the following values would be affected if you over-titrate? Circle the cor- rect response. The volume of NaOH recorded to reach the equivalence point would be (too high or too low). a. b. The calculated moles of NaOH used to reach the equivalence point would be (too (high or too low). с. The calculated moles of acetic acid neutralized by the NaOH would be (too (high or too low). d. The calculated molarity of acetic acid of the vinegar sample would be (too high or too low).
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