Determination of Acetic Acid

2021-05-07 15:34:18
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First, the buret was cleaned using a detergent solution and a long thing brush, then washed out with tap water followed by distilled water. The stopcock of the buret was examined carefully to ensure it is functioning well. Afterwards, the buret was washed thoroughly with 10 mL of sodium hydroxide solution (NaOH). Then, the stopcock was closed and 150 mL of NaOH solution were poured into the buret using a funnel. A small amount of NaOH was allowed to flow out to ensure there were no bubbles and the NaOH level was set at 0.5 0.01 mL- mark on the buret scale. The reading was done accurately using the bottom of the meniscus- the curved surface at the top of the liquid level.

Second, five mL (5 mL) of vinegar were transferred into a 250 mL Erlenmeyer flask using a clean 5 mL pipet, then 50 mL of distilled water and 1 mL phenolphthalein (an indicator) were added. The mixture was shaken well and no colour was observed (colourless solution).

Third, the titration was done by slowly adding the NaOH solution from the buret to the vinegar in the Erlenmeyer flask. The flask was continuously swirled during the titration process. Some pinkness appeared briefly in the flask as the base is added, but it quickly disappeared as the flask is swirled. The equivalence point was reached when NaOH solution gave a constant pink colour that did not disappear with swirling. The used volume of NaOH was recorded for later calculations. The same procedure was run another time and the values of the two runs were recorded for later calculations.

Determination of the molar mass of an unknown acid

The last step of this experiment included titration of unknown monoprotic acid with the standard NaOH solution. A small plastic dish was put on the scale and the scale was set to 0.0000 g, then 0.4968 and 0.4847 g of the unknown monoprotic acid (sample F) were accurately weighed and transferred into a 250 mL Erlenmeyer flask. The plastic dish was washed with a small amount (approx. 1 mL) of distilled water and then the washout was transferred to the flask. Fifty millilitres (50 mL) of distilled water and 1 mL of phenolphthalein were added to the flask and mixed well. Thereafter, standard NaOH was slowly added from the buret with continuous swirling, until a stable pink colour occurred. The volume of NaOH used for generating the colour was recorded.

DATA

Determination of acetic acid

Run #1 Run #3

Volume of vinegar solution used 5.0 mL 5.0 mL

Initial reading of buret with NaOH0 68 mL 0.56 mL

Final reading of buret with NaOH 15.49 mL 15.16 mL

Volume of NaOH solution used 14.81 mL 14.63 mL

Molarity of acetic acid in vinegar 0.611 M 0.603 M

Average molarity of acetic acid in vinegar 0.606 M

Grams of acetic acid per liter of vinegar 36.3 g/L

Grams of vinegar per liter of vinegar 1005 g/L

Mass% of acetic acid in vinegar sample 3.62 %

Determination of the molar mass of an unknown acid

Molarity of NaOH solution used 0.2061 M

Code of unknown sample used F

Run #1 Run #2

Initial weight of weigh dish 0.0000 g 0.0000 g

Final weight of weigh dish 0.0000 g 0.0000 g

Mass of unknown acid used 0.4968 g 0.4877 g

Initial reading of buret with NaOH0.58 mL 0.48 mL

Final reading of buret with NaOH35.98 mL 35.68 mL

Volume of NaOH delivered 35.4 mL 35.2 mL

Moles of NaOH0.007296 moles 0.007255 moles

Molar mass of unknown acid 68.09 g/mol66.81 g/mol

Average molar mass 67.45 g/mol% deviation 0.95 %

CALCULATIONS

Determination of acetic acid

Volume of NaOH solution used for the titration for run #1 (V1)

V1= Vf Vi

Where Vf is the final reading of buret with NaOH (fter the end of the titration) and Vi is the initial reading of buret with NaOH.

V1= 15.49 0.64 = 14.81 mL

Similarly, volume of NaOH solution used for run #2 (V2)

V2 = 15.66 0.58 = 15.08 mL

Since V2 V1 = 15.08 14.81 = 0.27 mL > 0.2 mL (Not acceptable)

A third run was done and the volume of NaOH solution used (V3) was calculated

V3 = 15.16- 0.56 = 14.63 mL

V1 V3 = 14.81 14. 63 = 0.18 mL < 0.2 mL (Acceptable)

To calculate Molarity of acetic acid in vinegar (Ma)

Ma Va = Mb Vb Ma= Mb- VbMaWhere Va is the volume of acetic acid, Mb is the molarity of NaOH solution and Vb is the volume of NaOH used in the titration

Run #1: Ma = (0.2061) (14.81)5=0.611 MRun #3: Ma = (0.2061) (14.63)5=0.603 MThe average Molarity Mavg = 0.611+0.6032 = 0.607

grams of acetic acid per liter of vinegar = Average molarity X Molecular mass

= Mavg (g/mol) = 0.606 (60.05) = 36.5 g/L

Where the molecular mass of acetic acid (CH3COOH) = 2 carbon (12*2) + 2 oxygen (16*2) + 4 hydrogen (1*4) = 60.05 g/L

Mass % of acetic acid in vinegar sample = grams of acetic acid per liter of vinegar grams of vinegar per liter of vinegar = 36.51005 = 3.63%

Determination of the molar mass of an unknown acid

Volume of NaOH solution used for the titration for run #1 (V1) using Mass #1

V= Vf Vi

Where Vf is the final reading of buret with NaOH (fter the end of the titration) and Vi is the initial reading of buret with NaOH.

V1 = 35.98 0.58 = 35.4 mL

To calculate moles of NaOH = 35.410000.2061=0.007296 Mol Molar Mass MM1 = GramsMoles= 0.496890.007296=68.09 g/molSimilarly, Volume of NaOH solution used for the titration for run #2 (V2) using Mass #2

V2 = 35.68 0.48 = 35.2 mL

Moles of NaOH = = 35.210000.2061=0.007255 MolMolar Mass MM2 = GramsMoles= 0.484790.007255=66.81 g/mol Average Molecular Mass MMavg = MM1+ MM22=68.09+66.812=67.45 g/molAverage Deviation = MM1- MMavg+(MM2- MMavg)2= 68.09-67.45+( 66.81-67.45)2=0.64% Deviation = Average DeviationMean 100= 0.6467.45 100=0.95%

DISCUSSION

Determination of acetic acid

Mass % of acetic acid in vinegar sample was found to be 3.63 %. The expected value was 3.62% which is very close to our result, indicating that our measurement was precise. The content of acetic acid is less than 4%, thus it does not meet the federal standard.

Determination of the molar mass of an unknown acid

The % deviation for the molecular mass of the unknown acid was found to be 0.95% which is less than 1% indicating good accuracy of the measurements.

 

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