ANALYSIS OF VINEGAR VIA TITRATION

The experiment aims to determine the mass percent of acetic acid in vinegar. In two trials, vinegar is titrated against NaOH. The titre values obtained are then used to quantify the mass percent of acetic acid in the vinegar sample, which is then compared to federally agreed values.

Household vinegar contains 4-5 percent by mass of acetic acid (CH3COOH) solution, while the federal minimum is 4 percent. The liquid is then enhanced in both taste and appearance by the addition of caramel flavoring and coloring.

We use volumetric analysis and the titration process to calculate the percentage of acetic acid in the vinegar. A measured mass of vinegar is titrated to the endpoint against a measured volume of sodium hydroxide solution using phenolphthalein indicator. Since the volume and molar concentration of NaOH is known, then we can calculate the moles of NaOH used and use it for our analysis;

Mol of NaOH= L NaOH solution x (10.2)

The balanced equation between the acetic acid and NaOH is;

CH3COOH(aq) + NaOH(aq) NaCH3CO2(aq) + H2O(l)

Therefore, mol of CH3COOH = mol of NaOH

The mass of CH3COOH in the vinegar is then calculated using the measured moles of CH3COOH neutralised in the reaction and the molar mass of 60.05 g/mol.

Mass (g) of CH3COOH = mol CH3COOH x

Finally, we calculate the percentage by mass of CH3COOH in the vinegar

% by mass of CH3COOH =

Procedure

About 150 ml of standardized NaOH solution was prepared from stock solution

Preparation of vinegar sample

We obtained two 10 ml of each of the two vinegar samples in a separate 10-ml graduated cylinders. Also, we prepare two clean 250-ml Erlenmeyer flasks. The volume of vinegar required for complete neutralization of 25 ml of standardised was calculated, assuming that vinegar has density of 1 g/ml and 5% acetic acid by mass, and the standardised solution is 0.1 M NaOH. The calculated volume of vinegar was added to 250-ml Erlenmeyer flask that has already been weighed, and the mass recorded. Then, two drops of phenolphthalein indicator were added and the sides of the container rinsed with 20 ml of previously boiled and deionized water. A 50 ml burette was rinsed twice with standardised NaOH solution and finally filled with the standardized NaOH solution ensuring that there are no bubbles at the tip of the burette. The vinegar sample was then titrated using the NaOH solution from burette to the acid swirling the flask after each addition. NaOH is added until the endpoint is reached and the volume of the applied NaOH recorded. The procedure is then repeated for another sample of vinegar, marking the final amount NaOH titrant. Finally, the values obtained are used to calculate the average percent by mass of acetic acid in the vinegar.

Results

Preparation of vinegar sample

Calculate the approximate volume of the vinegar sample needed for the analyses

0.02500 L NaOH () ( ) () = 0.150g CH3COOH

0.500(x) = 0.150g

X= x= 3.00 g

Trial 1

Trial 2

Mass of flask (g)

91.748 g

81.455 g

Mass of flask + vinegar (g)

92.869 g

83.099 g

Mass of vinegar (g)

1.121 g

1.644 g

Analysis of vinegar sample

Trial 1

Trial 2

Initial burette reading (mL)

0.00

7.00

Final burette reading (mL)

7.00

16.5

Volume of NaOH used (mL)

7.00

9.50

Molar concentration of NaOH (mol/L)

0.113 mol/L

Moles of NaOH added

7.91 x 10-4

0.00107

Moles of CH3COOH in the vinegar

7.91 x 10-4

0.00107

Mass of CH3COOH in the vinegar (g)

0.0475

0.0645

Percent by mass of CH3COOH in vinegar (%)

4.246

3.926

Average percent by mass of CH3COOH in vinegar (g)

4.08 %

Calculations for trial 1

7 x x 0.113 mol/L = 7.91 x 10-4,

(7.91 x 10-4) ( 60.05 g)= 0.0475

100 x = 4.29 %

Trial 2

9.5 x x 0.113 mol/L = 0.00107

(0.00107) (60.05 g) = 0.0645

100 x () = 3.92 %

Discussion

Questions

Assuming the density of a 5% acetic acid by mass solution is 1.0 g/mL, determine the volume of the acetic acid solution necessary to neutralize 25.0 mL of 0.10 M NaOH (Xie et al., 421).

(5 x 1000)/ 100 g = 50 g of acetic acid

(1000 g)/ 1 g/mL) = 1000 mL= 1L

Mm of acetic acid = 60.05 g/mol

(50 g) x ( 1 mol/60.05) = 0.8326 moles

Hence,

(25 x 0.10/ 0.8326) = 3.002

Therefore, volume of acetic acid required is 3.0 mL

a. A chemist often uses a white card with a black mark to aid in reading the meniscus of a clear liquid. How does this technique make the reading more accurate? Explain.

It is easy to identify a white mark for reading against a black background. Hence, a white card with a black mark is used to make the reading more relaxed and more prominent.

b. A chemist should wait 10-15 seconds after dispensing a volume of titrant before reading is done. Explain why the wait is reasonable laboratory technique.

The titrant stick to the walls of the burette from which titration is carried out. Thus, the wait is ideal to make sure that the titrant droplets have uniformly been distributed and dropped into the solution in the burette (Lenghor, pg. 1139). If the titration is started before all the droplets have fallen into the burette, then erroneous results can be obtained as a result of droplets that stuck to the walls dropping into the titrant after titration has been done.

c. the colour change of endpoint should persist for 30 seconds. Explain why the time lapse is a proper titration technique.

A temporary colour change possibly occurs due to a local reaction between the reactants to form a complex. Most titration reactions are pH dependent, hence if correct pH is not attained the complex will decompose (McMills et al., 958). At the equivalence point, the solution has the proper pH to stabilise the complex and consequently, it will no longer decompose forming a permanent colour change. Thus, the 30 seconds are required for ascertaining the correct equivalence point.

Lemon juice has a pH of about 2.5. Assuming that the acidity of lemon juice is due to solely citric acid, and the density of lemon juice is 1.0 g/mL, then the citric acid concentration calculates to 0.5% by mass. Estimate the volume of 0.01 M NaOH required neutralising a 3.71 g sample of lemon juice. The molar mass of citric acid is 190.12 g/mol.

(3.71g) x (0.5 g/100 g) = 0.01855 g citric acid

(3.71 g)/ 1.0g/mL) = 3.71 mL

3.71 g of lemon juice contains 0.01855 g citric acid

Mm of citric acid = 190.12 g/mol

(0.01855g) x (1 mole/190.12 g) = 9.75 x 10 -5 mol

(9.757 x 10-5 x 1000)/0.0100 mL =9.757 mL ̴9.8 mL

Explain why it is quantitatively not acceptable to titrate each of the vinegar samples with the NaOH titrant to the same dark pink endpoint.

If we titrate to a dark pink endpoint, we will have more moles of NaOH than there is CH3COOH. Assuming the moles to be equal, we will report a higher CH3COOH concentration in the vinegar that is.

Commercial vinegar is analysed for the percent acetic acid present. The data for trial one is listed in the table below. Complete the table to determine the percentage. Record the calculated values with a correct number of significant fugues.

Preparation of vinegar sample.

Mass of vinegar (g) 3.06

Analysis of vinegar sample

Initial burette reading (mL)

3.70

Final burette reading (mL)

25.40

Volume of NaOH used (mL)

21.70

Molar concentration of NaOH solution (mol/L)

0.0940

Moles of NaOH added (mol)

2.04 x 10-3 = 21.70 x (1L/100mL) x (0.0940mol/L)= 2.0398 x 10-3

Moles of CH3COOH in vinegar (g)

2.04 x 10-3

Mass of CH3COOH in vinegar (g)

0.1224 (2.04 x 10-3) x (60.05g/mol)= 0.1224g

Percent mass of CH3COOH in vinegar (g)

4.00 % (0.1224/ 3.06) x 100 = 4.00 %

b. for trials 2 and 3, the percent of CH3COOH in the vinegar was 5.01% and 4.66% respectively.

I. what is the average percent of CH3COOH in the vinegar sample

= (4.00 + 5.01 + 4.66)/3

= 4.56 % CH3COOH

II. What are the standard deviation and the relative standard deviation for the percent of CH3COOH in the vinegar sample?

Standard deviation= 0.5117

Relative standard deviation = (0.5117/456)x 100

= 11.22%

Conclusion

In the experiment, two samples of vinegar were analysed using the same parameters such as the molarities of the titrant and two trials carried out. In the trial one, the mass parentages of the acetic acid in the vinegar was found to be 4.29 %. It is a little higher than the recommended percent of acetic acid but within the range of the accepted standards (4-5%). In the second trial, the obtained percent of the acetic acid in the vinegar by mass is 3.92%, lower than in trial one and lower than the accepted standards. The possible reason for such an error could be that NaOH was added past the endpoint for the trial 1, despite using the same of a volume of vinegar in the two trials.

References

Lenghor, Narong, et al. “Sequential injection redox or acid-base titration for determination of ascorbic acid or acetic acid.” Talanta 58.6 (2002): 1139-1144.

McMills, Lauren, Frazier Nyasulu, and Rebecca Barlag. “Comparing mass and volumetric titrations in the general chemistry laboratory.” Journal of Chemical Education 89.7 (2012): 958-959.

Xie, Wei-Qi, and Xin-Sheng Chai. “Determination of Total Acid Content in Vinegars by Reaction-Based Headspace Gas Chromatography.” Food Analytical Methods 10.2 (2017): 419-423.