Chem Lab Report Lab 10 - Vinegar Analysis PDF

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Vinegar Analysis...

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Chem25 Lab – Section 8 Vinegar Analysis Purpose: The purpose of the vinegar analysis lab is to use the titration technique and determine the percent by mass of acetic acid ( C H 3 COOH ¿ in vinegar.

Source: Laboratory Lab Manual For Principles of General Chemistry, J.A. Beran, John Wiley & Sons, Inc., 2014 and p.154

Introduction: The purpose of the vinegar analysis lab is to determine the percent by mass of acetic acid ( C H 3 COOH ¿ in vinegar. Household vinegar is said to be four to five percent acetic acid. This experiment will attempt to verify that. Acid base titration uses the concentration of one known solution (the titrant) to find the concentration of the unknown solution (the analyte). The experiment uses sodium hydroxide ( NaOH ) as the titrate, which is a basic solution with the known concentration to titrate the analyte which is acetic acid with the unknown concentration. +¿ When the H ¿ from the acetic acid completely mixes with

−¿ ¿ O H from the acetic acid, the

amount of acid and base is equal meaning and a neutralization reaction occurred and the equivalence point is reached. Phenolphthalein is an indicator that is colorless when dissociated in water but turns pink in basic solution. By adding it into the analyte (vinegar), it helps during the titration to determine when the solution is neutralized and reached its endpoint. The endpoint will be determined once there is a color change from the solution being colorless to a light pink hue. In mixing the acid and base, salt and water is produced. The balanced equation for this experiment is

1

C H 3 COOH ( aq )+NaOH ( aq) → NaC H 3 C O2 ( aq ) + H 2 O (l) .

Chem25 Lab – Section 8

Chemical Properties and Safety Melting

Boiling

Density

Molar

point (C)

point

(g/cm3)

Mass

16.6

118.1

Safety Issues

(g/mol)

(C)

CH3COO

pKa

1.05

60.052

4.8

H

Corrosiv e and flammab le

NaOH

318

1388

2.13

39.997

14

Corrosiv e

NaCH3C

324

881.4

1.53

82.0343

4.76

NA

0

100

1

18.01528

14

NA

O2

H2O

Procedure

2

Chem25 Lab – Section 8 A. Preparation of Vinegar Sample

Observations

1. Clean out two 125 or 250 mL Erlenmeyer flasks. 2. Calculate the volume of vinegar that would be needed to neutralize 25 mL of NaOH solution. Assume that the density of vinegar is 1g/mL and acetic acid is 5% mass of vinegar. 3.

Prepare the vinegar sample by adding



Solution stays clear when indicator is added.

approximately 3g/mL into a clean Erlenmeyer flask. Record the mass of the vinegar sample then add two drops of indicator (phenolphthalein). Rinse the wall of the flask with 20 mL of deionized water that has previously been boiled. 4. Prepare the buret and titration step.



Buret NaOH solution is clear.



It’s easier to differentiate if a color

Rinse a clean 50 mL buret with NaOH. Once clean, fill up the buret with NaOH solution and read and record the starting volume. 5. Adding a clean white paper beneath

change occurs.

the flask. B. Analysis of Vinegar Sample 1. To begin titrating the vinegar sample,



If not stirred, the solution was pink.

slowly release NaOH from the buret

Once stirred, the solution changed

while swirling the flask.

back to clear were able to mix and change back to clear.

2. Occasionally rinse the flask of the wall to ensure the NaOH is fully mixing with the vinegar. 3. Continue adding NaOH until there is a 3



The endpoint is reached.

Chem25 Lab – Section 8 light pink hue even with swirling. 4. Repeat with the same vinegar. Fill up the buret again and repeat the titration. 5. Do calculations and determine the average percent by mass of acetic acid in the vinegar.

Data Mass of flask (g) Mass of flask +

Trial 1 86.085 88.168

Trial 2 86.101 88.167

Trial 3 86.139 88.275

Trial 4 81.964 84.023

vinegar (g) Mass of vinegar (g) Initial buret reading

2.083 0.00

2.066 0.00

2.136 0.00

2.059 0.00

of NaOH (mL) Final buret reading

11.51

11.45

11.75

11.45

of NaOH (mL) Volume of NaOH

11.51

11.45

11.75

11.45

used (mL) Molar concentration of NaOH (mol/L) Moles of NaOH

0.0994 0.00114

0.00114

0.00117

0.00114

added (mol) Moles of CH3COOH 0.00114

0.00114

0.00117

0.00114

in vinegar (mol) Mass of CH3COOH

0.0841

0.0685

0.0703

0.0685

in vinegar (g) Percent by mass of

3.30

3.32

3.29

3.33

CH3COOH in vinegar (%) Average percent by

3.32%

3.31%

mass of CH3COOH in vinegar (%) Standard Deviation 4

0.0880

Chem25 Lab – Section 8 Relative Standard

2.68

deviation Trial 1 Calculations: Mass of vinegar = (mass of flask + vinegar) – (mass of vinegar) = 88.168g - 86.085gg = 2.08g Volume of NaOH used = (volume of NaOH used) – (initial buret reading of NaOH) = 11.51 – 0 = 11.51mL Moles of NaOH added = (volume of NaOH used) x (molar concentration of NaOH) =

11.51 mL ×

0.151mol 1L =0.0014 mol × 1L 1000 mL

Moles of CH3COOH in vinegar = one to one ratio of NaOH to CH3COOH = 0.00114 mol Mass of CH3COOH in vinegar = (moles of CH3COOH in vinegar) x (molar mass of CH3COOH) =

(0.00114 mol)×(60.052 g/mol)=0.0841 g

Percent by mass of CH3COOH in vinegar = (mass of CH3COOH/mass of vinegar) x 100 = (0.0841g/2.083g) x 100 = 3.30% Average percent by mass of CH3COOH in vinegar = (sum of mass of CH3COOH in vinegar)/(# of trails) = (3.30%+3.32%+3.29%+3.33%)/4 = 3.31% Standard deviation =

√

N

1 ∑ (x −mean)2 N i=1 i

= 0.0880 Relative standard deviation = (standard deviation)/(average)x100 = 0.0880/3.28% x 100 = 2.68% Discussion 5

Chem25 Lab – Section 8 This experiment is trying to find the percent by mass of acetic acid in vinegar. Household vinegar is said to be four to five percent acetic acid and this experiment tests that. The experiment went smoothly but required much tediousness. It became increasingly difficult near the equivalence point to not go over endpoint of the titration. The solution had to be a light pink for the most accurate results. From the trails, the color changed occured when the amount of NaOH added was nearing 11.50mL. There needed to be around 11.50mL of NaOH added to titrate the 2g of vinegar. The results made sense and are in agreement with expected/theorical values. Once the amount of sodium hydroxide needed to reach the titration end point is known, it can be used to calculate and determine the percent by mass of acetic acid in vinegar. When the volume is known, it can be converted to moles by using the molar concentration of sodium hydroxide. From the balanced equation, C H 3 COOH ( aq )+NaOH ( aq) → NaC H 3 C O2 ( aq ) + H 2 O (l) , it’s seen that the mole ratio between sodium hydroxide and acetic acid is 1:1, meaning the moles of sodium hydroxide equals the moles of acetic acid. With the moles of acetic acid, multiplying it by its molar mass (60.052g) gives the mass of CH3COOH in vinegar. All four experiments showed an average of 3.31% acetic acid in vinegar. The standard deviation was 0.0880 and the relative standard deviation was 2.68%. Having a small standard deviation shows that the trail results were fairly close to the average. When formation the experiment, adding phenolphthalein as an indicator did not turn the solution pink as it was still an acidic solution. It was not until a sufficient volume of sodium hydroxide was added that caused the solution to turn basic and turn pink. Before reaching the end point, adding sodium hydroxide would cause the solution to turn pink until stirred back to clear. The solution stops clearing up and stays pink once the end point is reached. The goal when adding NaOH is to get the lightest shade of pink possible and get as close to the equivalence point as possible. An easy error to make during this experiment is overshooting the endpoint of the titration. When it is over titrated, the color of the solution would be dark pink showing that the solution turned too basic and too much NaOH was added. The smallest volume of NaOH has the power to go past the neutralization point and over titrate the solution. Over titrating causes the calculation of the percentage of acetic acid in vinegar to be higher than its actual.

Conclusion 6

Chem25 Lab – Section 8 The purpose of the vinegar analysis lab was to use the titration technique and determine the percent by mass of acetic acid ( C H 3 COOH ¿ in vinegar. After doing four trails, the average result from the experiment showed that there was a 3.31% by mass of acetic acid in vinegar. The relative standard deviation was 2.68% percent.

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