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Analysis of vinegar pre lab answers, Lab Reports of Chemistry

In an acid-base neutralization reaction, an acid reacts with a base to produce water and a salt.

Typology: Lab Reports

2021/2022

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GCC CHM152LL: SAMPLE LAB WRITE-UP 1 of 4
Note: This is not an experiment for CHM152 students – this is an
example to illustrate the proper format for a lab report.
Pages 1-2 provide the experiment that was performed
Pages 3-4 show the sample lab report that a student would
write and submit to be graded.
*Formal Lab reports must be neatly handwritten or typed, or a
combination of the two. Reports cannot be pages torn from your
lab notebook.
ANALYSIS OF VINEGAR VIA TITRATION
INTRODUCTION
In an acid-base neutralization reaction, an acid reacts with a base to produce water and a salt:
HX (aq) + MOH (aq) H2O (l) + MX (aq) (1)
acid base salt
The protons (H+) from the acid react with the hydroxide ions (OH) from the base to form the
water. The salt forms from the cation from the base and the anion from the acid. Because
water is always formed, acids will always react with bases; whether the salt is soluble or
insoluble does not determine whether the reaction occurs.
In this experiment, you will use a dilute solution of sodium hydroxide, NaOH (aq) to determine
the molar concentration of acetic acid present in a sample of vinegar. You will measure out a
small volume of vinegar and use a buret to determine the volume of sodium hydroxide required
to completely neutralize the vinegar. The process of slowly adding one solution to another until
the reaction between the two is complete is called a titration. The reaction between acetic
acid, HC2H3O2 (aq), and sodium hydroxide, NaOH (aq), is shown below:
HC2H3O2 (aq) + NaOH (aq) H2O (l) + NaC2H3O2 (aq) (2)
When carrying out an acid-base neutralization reaction in the laboratory, you observe that
most acid solutions and base solutions are colorless, and the resulting water and soluble salt
solutions are also colorless. Thus, it is impossible to determine when a reaction has occurred,
let alone when it is complete. To monitor the progress of a neutralization reaction, you use an
acid-base indicator, a solution that changes color depending on the pH (or acid-content) of
the solution. One commonly used indicator is phenolphthalein, which is colorless in acidic and
neutral solutions and pink in basic (or alkaline) solution. During a titration, the indicator is
added to the solution being titrated (the analyte). The titrant (or standard) is slowly added to
the analyte until the endpoint, when the indicator changes color, signaling that the reaction
between the two is complete. Note that phenolphthalein turns pink only when excess sodium
hydroxide, NaOH (aq), has been added.
The equivalence point in a titration is when there are equal numbers of moles of acid and
base present in the solution. Thus, for the reaction between KHP and sodium hydroxide, at the
equivalence point
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GCC CHM152LL: SAMPLE LAB WRITE-UP 1 of 4

Note: This is not an experiment for CHM152 students – this is an

example to illustrate the proper format for a lab report.

• Pages 1-2 provide the experiment that was performed

• Pages 3-4 show the sample lab report that a student would

write and submit to be graded.

*Formal Lab reports must be neatly handwritten or typed, or a

combination of the two. Reports cannot be pages torn from your

lab notebook.

ANALYSIS OF VINEGAR VIA TITRATION

INTRODUCTION

In an acid-base neutralization reaction, an acid reacts with a base to produce water and a salt:

HX (aq) + MOH (aq) ⇒ H 2 O (l) + MX (aq) (1)

acid base salt

The protons (H+^ ) from the acid react with the hydroxide ions (OH–^ ) from the base to form the water. The salt forms from the cation from the base and the anion from the acid. Because water is always formed, acids will always react with bases; whether the salt is soluble or insoluble does not determine whether the reaction occurs.

In this experiment, you will use a dilute solution of sodium hydroxide, NaOH (aq) to determine the molar concentration of acetic acid present in a sample of vinegar. You will measure out a small volume of vinegar and use a buret to determine the volume of sodium hydroxide required to completely neutralize the vinegar. The process of slowly adding one solution to another until the reaction between the two is complete is called a titration. The reaction between acetic acid, HC 2 H 3 O 2 (aq), and sodium hydroxide, NaOH (aq), is shown below:

HC 2 H 3 O 2 (aq) + NaOH (aq) ⇒ H 2 O (l) + NaC 2 H 3 O 2 (aq) (2)

When carrying out an acid-base neutralization reaction in the laboratory, you observe that most acid solutions and base solutions are colorless, and the resulting water and soluble salt solutions are also colorless. Thus, it is impossible to determine when a reaction has occurred, let alone when it is complete. To monitor the progress of a neutralization reaction, you use an acid-base indicator , a solution that changes color depending on the pH (or acid-content) of the solution. One commonly used indicator is phenolphthalein , which is colorless in acidic and neutral solutions and pink in basic (or alkaline) solution. During a titration, the indicator is added to the solution being titrated (the analyte ). The titrant (or standard ) is slowly added to the analyte until the endpoint , when the indicator changes color, signaling that the reaction between the two is complete. Note that phenolphthalein turns pink only when excess sodium hydroxide, NaOH (aq), has been added.

The equivalence point in a titration is when there are equal numbers of moles of acid and base present in the solution. Thus, for the reaction between KHP and sodium hydroxide, at the equivalence point

ANALYSIS OF VINEGAR VIA TITRATION PAGE 2

moles of HC 2 H 3 O 2 = moles of NaOH. (4)

You should recognize that the end point is determined by observation of a color change in the lab while the equivalence point is a theoretical point one can calculate or estimate given the amount of acid or base present to titrate.

PROCEDURE

Analysis of Vinegar

  1. Obtain ~50 mL of vinegar in your 100-mL beaker. Record the brand of vinegar and the percent acetic acid indicated on the label.
  2. Condition a 10-mL volumetric pipet with the vinegar solution. If you are unfamiliar with the operation of a pipet, ask your instructor for help. Obtain and label three Erlenmeyer flasks #1, #2, and #3. Pipet 10 mL of vinegar to each of three Erlenmeyer flasks. Record the volume for each trial as 10.00 mL. Add ~25 mL of distilled water to each flask
  3. Add 2-3 drops of phenolphthalein indicator to each of your vinegar solutions.
  4. Obtain about 150 mL of the dilute NaOH solution, and record the molar concentration for the NaOH. Condition a 50-mL buret with the NaOH solution.
  5. Fill the buret close to 0.00 mL with the dilute NaOH solution, and record the initial volume for trial #1 for vinegar on your Data Sheet. Slowly lower the stir bar into flask #1, and carry out the titration as before. Record the final volume of NaOH on your Data Sheet.
  6. Repeat the titration for trials #2 and #3.
  7. Dispose of all the solutions in the proper waste containers. Wash and rinse your buret and flasks, and return them to their proper places.

BE SURE TO WASH AND DRY YOUR LAB BENCH AFTER COMPLETING THE EXPERIMENT TO REMOVE ALL TRACES OF ANY SPILLED CHEMICALS.

CALCULATIONS

  1. Calculate the molarity of the acetic acid in each vinegar solutions. Next, calculate the average molarity of the acetic acid for all three solutions.
  2. Determine the density of the vinegar solution.
  3. Calculate the mass percentage of acetic acid in the vinegar.

QUESTIONS

  1. Compare the mass percentage you determined with the mass percentage reported on the bottle of vinegar.

ANALYSIS OF VINEGAR VIA TITRATION PAGE 4

Sample Calculations:

Molarity of HC 2 H 3 O 2 in vinegar:

Trial 1:

2 3 2

2 320. 842 MHCHO

0. 01000 L

1 molNaOH

1 molHCHO L

0.216molNaOH

  1. 03900 LNaOH× × × =

Trials 2 and 3: see results table

Average [HC 2 H 3 O 2 ] = 0. 839 M 3

0. 842 M 0.838M 0.838M

Density of vinegar: density = 1. 005 g/mL 10.00mL

10.0503g volume

mass = =

Mass percent of HC 2 H 3 O 2 in vinegar:

1.005gsolution

mLsolution

3

O

H

molHC

O

H

  1. 052 gHC 1000 mLsolution

O

H

0.839molHC × × × =

Vinegar Titration Results

Trial 1 Trial 2 Trial 3

Molarity of HC 2 H 3 O 2 in vinegar 0.842 M 0.838 M 0.838 M

Average Molarity of HC 2 H 3 O 2 0.839 M

Density of vinegar solution 1.005 g/mL

Mass percent of HC 2 H 3 O 2 in vinegar 5.01%

Questions:

#1. The mass percentage of the acetic acid in the vinegar solution was found to be

5.01%, which is in close agreement with the manufacturer’s reported acetic acid content

of 5.0%.

Conclusions:

In this experiment, a sample of Smiths brand vinegar was analyzed. Three trials were

carried out, and the average molarity of the vinegar was determined to be 0.839 M. The

mass percentage of the acetic acid, HC 2 H 3 O 2 (aq) , in the vinegar was found to be 5.01%,

which is only slightly higher than the manufacturer’s reported acetic acid content of

5.0%. One possible source of error was that NaOH was added past the endpoint for trial

#1, for which 0.20 mL more NaOH was used compared to trials #2 and #3 even though

the same volume of vinegar was titrated for all three trials.