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Determination of Asa Content of Aspirin Essay

Aim: To determine the Molar Concentration of NaOH and HCl acid used in their Standardization processes and to determine the acetylsalicylic acid (ASA) content in Aspirin. Materials/Apparatus: materials used are the same as that outlined in the laboratory procedure prepared by the laboratory instructor. Procedure: The procedure used is the same as that outlined in the laboratory procedure prepared by the laboratory instructor. Abstract: Aspirin is said to be the oldest and generally most useful drug due to its analgesic and antipyretic properties. Aspirin contains acetylsalicylic acid (A.S.A) but not 100%, and as such its content is being determined by methods of titration in this experiment, using the indicator phenolpthalein. Titration was done using NaOH and after was hydrolyzed and back-titrated using HCl solution. The NaOH and HCl acids were standardized and its molar concentrations were found to be 0.09085molL-1 and 1.084×10-1molL-1 respectively. The average mass of ASA was found to be 0.437g or 437 mg. After performing the experiment, several calculations were done and the % average mass of ASA found in the brand of aspirin used was found to be 87.14%. Introduction: Acetylsalicylic acid was first obtained from the bark of willow trees and has been used for medicinal purposes for thousands of years. Today, it is sold as Aspirin and use as an analgesic (pain killer), antipyretic (fever reducing), and anti-inflammatory. Aspirin is also used for its anti-platelet (blood thinning) effects to prevent heart attacks.

In addition to containing acetylsalicylic acid, an aspirin tablet contains some additional compounds including starch, sodium cyclamate and sodium saccharine. . The objective of this experiment is to determine the ASA content in aspirin by hydrolysis using strong base, such as NaOH followed by back titration with HCl. The method used to determine the ASA content in aspirin is titration. This is the method of determining the concentration of a substance in a solution by adding to it a standard reagent of known concentration in carefully measured amount until reaction of define and known proportion is completed. This is indicated by a color change and then calculating the unknown concentration. . In this experiment the standard solution being used is sodium hydroxide. But first this has to be standardized with hydrogen phthalate to determine its “true” concentration. To determine the concentration of the sodium hydroxide solution, one must have especially pure acid (potassium biphthalate) to titrate against the strong base. A known quantity of a solid acid (potassium hydrogen phthalate) is dissolved in water in a flask, and phenolphthalein indicator is added. The NaOH solution is added from a burette into the flask containing the acid and the weak acid and the base react with one another according to the equation: 2 NaOH (s) + CO2 (g) Na2CO3(s) + H2O (l)

The volume of a known concentration of sodium hydroxide to titrate aspirin will be determined. The volume of sodium hydroxide can be used to determine the moles of sodium hydroxide reacted. From the moles of NaOH reacted, the mole of acetylsalicylic acid can be determined. The mass of the tablet divided by the moles of acetylsalicylic acid will provide an experimental value for the molar mass of acetylsalicylic acid. This can be compared to the reference molar mass (% error). The following balanced equation describes the double replacement reaction that will be observed in this experiment: HC9H7O4 (aq) +NaOH (aq) →NaC9H7O4 (aq) +H2O (l)

Pre-Lab Exercise:
Calculate the volume of 50% (by weight) NaOH solution necessary to prepare 1L of 0.1M NaOH. (The density of the 50% NaOH solution is 1.53± 0.01g/mL 50g NaOH100g soln×1 mol NaOH40.00g NaOH×1.53g solnmL soln×103 mL soln1L soln =19.125M

Dilute 19.125 to 0.1M
19.125M0.1M= Dilution Factor
Therefore, 1000mL191.25=5.23mL
Data Results:
Table 1: Showing the Standardization of NaOH
| Trial 1| Trial 2| Trial 3|
Mass of KHP/g| 0.5002| 0.5073| 0.5556|
Final vol of NaOH/ml| 27.33| 27.68| 30.36|
Initial vol of NaOH/ml| 0.29| 0.39| 0.45|
Titre vol of NaOH /ml| 27.04| 27.29| 29.91|

Table 2: Showing the Standardization of HCl
| Trial 1| Trial 2| Trial3|
Volume of HCl/ml| 25.00| 25.00| 25.00|
Final vol of NaOH/ml| 29.99| 29.49| 30.69|
Initial vol of NaOH/ml| 0.09| 0.13| 0.49|
Titre vol of NaOH/ml| 29.90| 29.36| 30.20|
Average Titre vol/ml| 29.82|

Table 3: Showing the Direct Titration of A.S.A content of Aspirin with NaOH | Trial 1| Trial 2|
Mass of aspirin/g| 0.5038| 0.4992|
Final vol of NaOH/ml| 28.23| 28.49|
Initial vol of NaOH/ml| 0.40| 0.09|
Titre vol of NaOH/ml| 27.83| 28.40|

Table 4: Showing the hydrolysis of the ASA content of aspirin with NaOH | Trial 1| Trial 2|
Mass of aspirin/g| 0.5038| 0.4992|
Final vol of NaOH/ml| 49.35| 47.29|
Initial vol of NaOH/ml| 6.52| 3.89|
Vol of NaOH added/ml| 42.83| 43.40|

Table 5: Showing the Back Titration of the ASA Content of Aspirin with HCl | Trial 1| Trial 2|
Mass of aspirin/g| 0.5038| 0.4992|
Final vol of HCl/ml| 19.29| 22.41|
Initial vol of HCl/ml| 3.48| 3.12|
Titre vol of HCl/ml| 15.81| 19.29|

Data Analysis:
1) From the titration reaction, calculate the mean molarity of the NaOH solution: Reaction that occurred in first titration with sodium hydroxide and potassium phthalate: KH5C8O4aq+NaOHaq→KH4C8O4Naaq+H2Ol

Average Mass of KHP used:
Avg.Mass=sample masses# of samples
Avg.Mass=0.5210 g
Therefore an average of 0.5210 g of KHP was diluted in 25mL. Number of moles of KHP in 0.5210g:
# of moles of KHP=given massmolar mass
Molar Mass of KHP=39.098+5×1.008+ 8×12.011+4×15.999gmol-1 Molar Mass of KHP=204.22gmol-1
# of moles of KHP=0.5210g204.22gmol-1
# of moles of KHP=2.551×10-3mol
From the reaction, NaOH and KHP reacted in a 1:1 mole ratio. # of moles of NaOH reacted=# of moles of KHP reacted
# of moles of KHP reacted= 2.551×10-3mol
# of moles of NaOH reacted=2.551×10-3mol
From the titration:
Avg.Volume of NaOH used to neutralise KHP=sample volumes# of samples Avg.Volume=27.04+27.29+29.91mL3
Therefore 28.08 mL of NaOH contains 2.551×10-3mol.
Molar Concentration of Sodium hydroxide:
Molar Concentration of NaOH=2.551×10-3mol28.08mL×1000mL1L

Molar Concentration of NaOH=0.09085molL-1

2) Use this molarity to calculate the concentration of the 0.1M HCl solution:

Reaction of second titration of sodium hydroxide and hydrochloric acid: NaOHaq+ HClaq → NaCl(aq) + H2O(l)
From the titration:
Avg.Volume of NaOH used to neutralise hydrochloric acid=sample volumes# of samples Avg.Volume=29.90+29.36+30.20mL3
# of mols of NaOH used to neutralise hydrochloric Acid=(2.982 x 10-2L)×0.09085molL-1 # of mols of NaOH used to neutralise hydrochloric Acid=2.709×10-3mol From the reaction, NaOH and hydrochloric acid reacted in a 1:1 mole ratio. # of moles of NaOH reacted=# of moles of hydrochloric acid reacted # of moles of NaOH reacted= 2.709×10-3mol

# of moles of hydrochloric acid reacted=2.709×10-3mol
Molar concentration of hydrochloric acid:
Molar Concentration of hydrochloric acid in 25 mL= 2.709×10-3mol25mL×1000mL1L Molar Concentration of hydrochloric acid in 25mL= 1.084×10-1molL-1

3) Using the results of the back titration and masses of sample used, calculate the mass of ASA in mg and % of ASA in the tablet:

Average mass of aspirin used:

Avg.Mass=sample masses# of samples
Avg.Mass=0.5015 g
Hence 0.4979g of Aspirin was dissolved in 25ml of ethanol.
Avg.Titre=sample volumes# of samples
Avg. Titre = (13.57 + 13.94)
= 13.76 mL
# of moles of HCl = Mean Titre vol of HCl x Molar Conc. Of HCl
= 1.376 x 10-2L x 1.084×10-1molL-1
=1.492 x 10-3 moles
Number of moles of A.S.A can be found by the difference between the mean number of moles of NaOH added for hydrolysis and the mean number of moles of HCl. Avg.vol=sample volumes# of samples
Avg. vol = (42.83 + 43.40)
= 43.12 mL
# of moles of NaOH = Mean vol of NaOH x Molar Conc. Of NaOH
= 4.312 x 10-2L ×0.09085molL-1
=3.917 x 10-3 moles
# of mol A.S.A = mean # of moles NaOH – mean # moles HCl
= 3.917 x 10-3 moles – 1.492 x 10-3 moles
= 2.425 x 10-3 moles
Mass of ASA = # of moles x RMM = 2.425 x 10-3 moles x 180.159gmol-1 = 0.437g x 1000 mg

= 437mg

Mass% of each sample = mass of A.S.A x 100
mass crushed tablet

mass of sample 1 = 437 mg x 100
503.8 mg
= 86.74 %

Mass of sample 2 = 437mg x 100
= 87.54%

Mean mass% of A.S.A= 86.74% +87.54%
= 87.14%

4) Determine what the average mass of ASA in a tablet be if the direct titration results were used. Avg.vol=sample volumes# of samples
Avg. vol = (27.83 + 28.40)
= 28.12 mL
# of moles of NaOH = Mean vol of NaOH x Molar Conc. Of NaOH
= 2.812 x 10-2L ×0.09085molL-1
=2.555 x 10-3 moles
moles of A.S.A = mean no. moles of NaOH – mean no. of molesof HCl = 2.555 x 10-3 moles – 1.492 x 10-3 moles
= 1.063 x 10-3 moles
Mass of ASA = # of moles of ASA x RMM
= 1.063 x 10-3 moles x 180.159 gmol-1
= 0.1915g
NB: Number of moles of HCl used within the titration, should be the same as the number of moles of excess NaOH. Therefore at some point in the experiment the end point was not reached or was overshot. Discussion: The mass of ASA calculated was 437 mg, while the yield given on the package of the Aspirin by the manufacturer is 324mg. The value that was calculated is way larger than what the manufacturer says is on the package, about 113mg larger. The percentage purity obtained from the experiment was 87.14%. The tablet’s active ingredient is A.S.A and the rest is made up of structural and supplementary substances. To make aspirin other substances are added to keep the tablet together and to keep it lubricated. Corn-starch and water is added to keep the tablet well bonded and to keep it from sticking to packaging. Diluents are also added to make the tablets easier to break down and taste better. These substances contribute to the percentage ASA purity of the aspirin tablets and can account for the other ingredients of the tablet. The manufactured mass of A.S.A was stated to be 324mg, as compared to the masses of the obtained which was 437mg. The indicator used, phenolphthalein, was used because the indicator needed to be soluble in ethanol. This is because the ASA is most soluble in cold ethanol so the indicator needed to be compatible with it. Since the ASA is most soluble in cold ethanol, the ethanol was brought to 150 C to increase the amount of dissolution occurring.

Due to the nature of the experiment there were certainly a few sources of error. The error that may have caused slight discrepancies could have been dissolving the powder in ethanol. Losing solution – too vigorous swirling can end in liquid splashing from the titration flask before the end point
had been reached. It may also happen that some titrant lands on the table instead of inside the flask. Some of the powder may have remained on the sides of the mortar and pestle, and thus reduced the quantity of aspirin in the experiment to cause less ASA present. There may also be errors in observing the colour change of the indicator at the end point. This is probably the most common one. Not only colour change is sometimes very delicate and slow, but different people have different sensitivity to colours. Misreading the volume at any moment will cause errors. This can be for example a parallax problem (when someone reads the volume looking at an angle), or error in counting unmarked graduation marks with regard to the burette’s uncertainty. When reading the volume on the burette scale it is not uncommon to read both upper and lower value in different lighting conditions, which can make a difference. Sometimes burettes leak slowly enough to allow titration, but will lose several tenths of millilitre if left for several minutes after titrant level has been set to zero and before titration started. Incomplete grinding and dissolution of the powder is a source of error affecting the final result of the titration of the ASA solution. The NaOH solution was prepared inside a sealed bottle hence decreasing the hydroscopic error of the experiment. Even though they were minimized as best as possible, errors due to the hygroscopic nature of NaOH may still have affected the experiment. Also, Instead of first weighing the tablet and then squash it. The experimenter should weigh the beaker before pouring in the powder and then weigh it after it has being poured, avoiding the source of error when leaving some powder in the paper or outside the beaker.

Conclusion: The molar concentration of NaOH was found to be 0.00985 mol/L and HCl 1.084 x 10-01 mol/L. The mean mass% off A.S.A found in the brand of aspirin used was found to be 87.14%.

Holler, J., & Skoog, F.(2004) Fundamentals of Analytical Chemistry (8th edition). Canada, Belmont
Harris, D.C. (1998) Quantitative Chemical Analysis (5th edition). New York, NY: W.H. Freeman and Company. Freeman, G. (2009) Standardization of NaOH.
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