Determination of Asa Content of Aspirin Essay

Purpose: To find the Molar Concentration of NaOH and HCl acid used in their Standardization procedures and to find the acetylsalicylic acid ( ASA ) content in Aspirin. Materials/Apparatus: stuffs used are the same as that outlined in the laboratory process prepared by the research lab teacher. Procedure: The process used is the same as that outlined in the laboratory process prepared by the research lab teacher. Abstraction: Aspirin is said to be the oldest and by and large most utile drug due to its analgetic and antipyretic belongingss. Aspirin contains acetylsalicylic acid ( A. S. A ) but non 100 % . and as such its content is being determined by methods of titration in this experiment. utilizing the index phenolpthalein. Titration was done utilizing NaOH and after was hydrolyzed and back-titrated utilizing 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 severally. The mean mass of ASA was found to be 0. 437g or 437 mg. After executing the experiment. several computations were done and the % mean mass of ASA found in the trade name of acetylsalicylic acid used was found to be 87. 14 % . Introduction: Acetylsalicylic acid was foremost obtained from the bark of willow trees and has been used for medicative intents for 1000s of old ages. Today. it is sold as Aspirin and usage as an analgetic ( trouble slayer ) . antipyretic ( fever cut downing ) . and anti-inflammatory. Aspirin is besides used for its anti-platelet ( blood thinning ) effects to forestall bosom onslaughts.

In add-on to incorporating acetylsalicylic acid. an aspirin tablet contains some extra compounds including amylum. Na cyclamate and Na saccharine. . The aim of this experiment is to find the ASA content in acetylsalicylic acid by hydrolysis utilizing strong base. such as NaOH followed by back titration with HCl. The method used to find the ASA content in acetylsalicylic acid is titration. This is the method of finding the concentration of a substance in a solution by adding to it a standard reagent of known concentration in carefully measured sum until reaction of define and known proportion is completed. This is indicated by a colour alteration and so ciphering the unknown concentration. . In this experiment the standard solution being used is sodium hydrated oxide. But first this has to be standardized with H phthalate to find its “true” concentration. To find the concentration of the Na hydroxide solution. one must hold particularly pure acid ( potassium biphthalate ) to titrate against the strong base. A known measure of a solid acid ( potassium H phthalate ) is dissolved in H2O in a flask. and phenolphthalein index is added. The NaOH solution is added from a burette into the flask incorporating the acid and the weak acid and the base react with one another harmonizing to the equation: 2 NaOH ( s ) + CO2 ( g ) Na2CO3 ( s ) + H2O ( cubic decimeter )

The volume of a known concentration of Na hydrated oxide to titrate acetylsalicylic acid will be determined. The volume of Na hydrated oxide can be used to find the moles of Na hydrated oxide 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 supply an experimental value for the molar mass of acetylsalicylic acid. This can be compared to the mention molar mass ( % mistake ) . The following balanced equation describes the dual replacing reaction that will be observed in this experiment: HC9H7O4 ( aq ) +NaOH ( aq ) >NaC9H7O4 ( aq ) +H2O ( cubic decimeter )

Pre-Lab Exercise:
Calculate the volume of 50 % ( by weight ) NaOH solution necessary to fix 1L of 0. 1M NaOH. ( The denseness of the 50 % NaOH solution is 1. 53± 0. 01g/mL 50g NaOH100g soln?1 mol NaOH40. 00g NaOH?1. 53g solnmL soln?103 milliliter soln1L soln =19. 125M

Dilute 19. 125 to 0. 1M
19. 125M0. 1M= Dilution Factor
=191. 25
Therefore. 1000mL191. 25=5. 23mL
Datas Consequences:
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 acetylsalicylic acid 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|



Datas Analysis:
1 ) From the titration reaction. cipher the average molar concentration of the NaOH solution: Chemical reaction that occurred in first titration with Na hydrated oxide and K phthalate: KH5C8O4aq+NaOHaq>KH4C8O4Naaq+H2Ol

Average Mass of KHP used:
Avg. Mass=sample masses # of samples
Avg. Mass=0. 5002+0. 5073+0. 5556g3
Avg. Mass=0. 5210 g
Therefore an norm 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 neutralize KHP=sample volumes # of samples Avg. Volume=27. 04+27. 29+29. 91mL3
Avg. Volume=28. 08mL
Therefore 28. 08 milliliter of NaOH contains 2. 551?10-3mol.
Molar Concentration of Sodium hydrated oxide:
Molar Concentration of NaOH=2. 551?10-3mol28. 08mL?1000mL1L
















Molar Concentration of NaOH=0. 09085molL-1

2 ) Use this molar concentration to cipher the concentration of the 0. 1M HCl solution:

Chemical reaction of 2nd titration of Na hydrated oxide and hydrochloric acid: NaOHaq+ HClaq > NaCl ( aq ) + H2O ( cubic decimeter )
From the titration:
Avg. Volume of NaOH used to neutralize hydrochloric acid=sample volumes # of samples Avg. Volume=29. 90+29. 36+30. 20mL3
Avg. Volume=29. 82mL
# of mols of NaOH used to neutralize hydrochloric Acid= ( 2. 982 ten 10-2L ) ?0. 09085molL-1 # of mols of NaOH used to neutralize 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 consequences of the back titration and multitudes of sample used. cipher the mass of ASA in milligram and % of ASA in the tablet:

Average mass of acetylsalicylic acid used:

Avg. Mass=sample masses # of samples
Avg. Mass=0. 5038+0. 4992g2
Avg. Mass=0. 5015 g
Therefore 0. 4979g of Aspirin was dissolved in 25ml of ethyl alcohol.
Avg. Titre=sample volumes # of samples
Avg. Titre = ( 13. 57 + 13. 94 )
2
= 13. 76 milliliter
# of moles of HCl = Mean Titre vol of HCl x Molar Conc. Of HCl
= 1. 376 ten 10-2L ten 1. 084?10-1molL-1
=1. 492 ten 10-3 moles
Number of moles of A. S. A can be found by the difference between the average figure of moles of NaOH added for hydrolysis and the average figure of moles of HCl. Avg. vol=sample volumes # of samples
Avg. vol = ( 42. 83 + 43. 40 )
2
= 43. 12 milliliter
# of moles of NaOH = Mean vol of NaOH x Molar Conc. Of NaOH
= 4. 312 ten 10-2L ?0. 09085molL-1
=3. 917 ten 10-3 moles
# of mol A. S. A = average # of moles NaOH – average # moles HCl
= 3. 917 ten 10-3 moles – 1. 492 ten 10-3 moles
= 2. 425 ten 10-3 moles
Mass of ASA = # of moles x RMM = 2. 425 ten 10-3 moles x 180. 159gmol-1 = 0. 437g ten 1000 milligram




















1g
= 437mg

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

mass of sample 1 = 437 milligram x 100
503. 8 milligram
= 86. 74 %

Mass of sample 2 = 437mg x 100
499. 2mg
= 87. 54 %

Average mass % of A. S. A= 86. 74 % +87. 54 %
2
= 87. 14 %

4 ) Determine what the mean mass of ASA in a tablet be if the direct titration consequences were used. Avg. vol=sample volumes # of samples
Avg. vol = ( 27. 83 + 28. 40 )
2
= 28. 12 milliliter
# of moles of NaOH = Mean vol of NaOH x Molar Conc. Of NaOH
= 2. 812 ten 10-2L ?0. 09085molL-1
=2. 555 ten 10-3 moles
moles of A. S. A = mean no. moles of NaOH – mean no. of molesof HCl = 2. 555 ten 10-3 moles – 1. 492 ten 10-3 moles
= 1. 063 ten 10-3 moles
Mass of ASA = # of moles of ASA x RMM
= 1. 063 ten 10-3 moles x 180. 159 gmol-1
= 0. 1915g
Niobium: Number of moles of HCl used within the titration. should be the same as the figure of moles of extra NaOH. Therefore at some point in the experiment the terminal point was non reached or was overshot. Discussion: The mass of ASA calculated was 437 mg. while the output given on the bundle of the Aspirin by the maker is 324mg. The value that was calculated is manner larger than what the maker says is on the bundle. about 113mg larger. The per centum pureness obtained from the experiment was 87. 14 % . The tablet’s active ingredient is A. S. A and the remainder is made up of structural and auxiliary substances. To do aspirin other substances are added to maintain the tablet together and to maintain it greased. Corn-starch and H2O is added to maintain the tablet good bonded and to maintain it from lodging to packaging. Dilutants are besides added to do the tablets easier to interrupt down and gustatory sensation better. These substances contribute to the per centum ASA pureness 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 multitudes of the obtained which was 437mg. The index used. phenolphthalein. was used because the index needed to be soluble in ethyl alcohol. This is because the ASA is most soluble in cold ethyl alcohol so the index needed to be compatible with it. Since the ASA is most soluble in cold ethyl alcohol. the ethyl alcohol was brought to 150 C to increase the sum of disintegration occurring.











Due to the nature of the experiment there were surely a few beginnings of mistake. The mistake that may hold caused little disagreements could hold been fade outing the pulverization in ethyl alcohol. Losing solution – excessively vigorous twirling can stop in liquid spatter from the titration flask before the terminal point
had been reached. It may besides go on that some titrant lands on the table alternatively of inside the flask. Some of the pulverization may hold remained on the sides of the howitzer and stamp. and therefore reduced the measure of acetylsalicylic acid in the experiment to do less ASA nowadays. There may besides be mistakes in detecting the color alteration of the index at the terminal point. This is likely the most common one. Not merely color alteration is sometimes really delicate and slow. but different people have different sensitiveness to color. Misreading the volume at any minute will do mistakes. This can be for illustration a parallax job ( when person reads the volume looking at an angle ) . or mistake in numbering unmarked graduation Markss with respect to the burette’s uncertainness. When reading the volume on the burette graduated table it is non uncommon to read both upper and lower value in different lighting conditions. which can do a difference. Sometimes burettes leak easy plenty to let titration. but will lose several ten percents of milliliter if left for several proceedingss after titrant degree has been set to zero and before titration started. Incomplete grinding and disintegration of the pulverization is a beginning of mistake impacting the concluding consequence of the titration of the ASA solution. The NaOH solution was prepared inside a certain bottle hence diminishing the hydroscopic mistake of the experiment. Even though they were minimized every bit best as possible. mistakes due to the hygroscopic nature of NaOH may still hold affected the experiment. Besides. Alternatively of first weighing the tablet and so crush it. The experimenter should weigh the beaker before pouring in the pulverization and so weigh it after it has being poured. avoiding the beginning of mistake when go forthing some pulverization in the paper or outside the beaker.

Decision: The molar concentration of NaOH was found to be 0. 00985 mol/L and HCl 1. 084 ten 10-01 mol/L. The average mass % off A. S. A found in the trade name of acetylsalicylic acid used was found to be 87. 14 % .

Mentions:
Holler. J. . & A ; Skoog. F. ( 2004 ) Fundamentalss of Analytic Chemistry ( 8th edition ) . Canada. Belmont
Harris. D. C. ( 1998 ) Quantitative Chemical Analysis ( 5th edition ) . New York. New york: W. H. Freeman and Company. Freeman. G. ( 2009 ) Standardization of NaOH.
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