anal chem 2b

8/12/2019 Anal Chem 2b

1/13

Experiment 2b

Determination of the aspirin content of Macs aspirin

Lab session: Monday 8-11am

Group Members: Laurenjay Henry 1001735

Camelia Samuels 0803467

Searena Treasure 0800417

Aim:To determine the amount of aspirin (acetylsalicylic acid) present in commercially available

aspirin tablets using a back titration method.

Abstract

A weighing dish was obtained and measured. Two aspirin tablets were weighed separately to

obtain the mass of the each tablet. The tablets were crushed in a mortar and pestle, and then three

0.5g samples were weighed. The weighing was transferred to a flask and 25ml of cooled

(15celcius) 95% ethanol was added to the flask and swirled to dissolve the aspirin. Two drops of

phenolphthalein was added and the sample was rapidly titrated with standard 0.1M NAOH to a

faint pink end point. The volume of NAOH used was recorded. The aspirin was hydrolyze by

adding additional NAOH from the burette. An extra 15ml more than what was required to reach

the initial endpoint was added. This volume of NAOH was added from the burette then the

burette reading was recorded again. The aspirin was hydrolyze by heating the mixture for about

15 minutes in a bath of boiling water in a 600ml.The flask was swirled occasionally. The flask

was cooled to room temperature with cooled tap water. The initial volume of standard HCL was

recorded and the excess base was back-titrated with HCL until the pink colour disappears. The

procedure was repeated with two samples.


2/13

The mass of aspirin was found to be 437.5mg and the average percentage of aspirin in sample

was found to be 85.11%.

Introduction

Aspirin is slightly acidic and reacts with bases in neutralization reactions. If the reaction is

followed with an indicator, a color change will occur when the acid is completely neutralized and

one drop of excess base has been added. The number of moles of base consumed and the

number of moles of acid in the sample can be calculated from the volume of the base needed to

obtain the color change. The relationship between these two mole values is determined form the

balanced chemical equation for the reaction.

In this experiment, an aspirin tablet with a known concentration of sodium hydroxide solution

was titrated. The end point will be determined with phenolphthalein indicator. The following

equation describes the neutralization reaction that will be observed in this experiment.

NaOH(aq) + C6H4OCOCH3COOH9(aq) C6H4OCOCH3COONa(aq)+ H2O(aq)

The mole ratio is 1:1, the number of moles of the base will equal the number of moles of acid at

the equivalence point of the titration. The molar mass of the solid aspirin can be approximated by

dividing the mass of the aspirin in the tablet by the moles of aspirin present. The mass of the

aspirin inside the tablet is usually near 325 mg. Inactive ingredients, such as binders, are added

to the aspirin during the manufacturing process. Therefore, the actual mass of the tablet exceeds

325 mg because it is not 100% pure aspirin.

Aspirin (acetylsalicylic acid, ASA) is a drug that hydrolyzes according to the chemical equation:


3/13

This hydrolysis is evident by the smell of acetic acid detected when a bottle of aspirin is opened.

The rate of hydrolysis is dependent on the temperature, pH and amount of moisture present. The

efficiency of any drug depends on its chemical stability. Hydrolysis of the drug can be a major

reason for the instability of drug solutions. Thus, when Aspirin undergoes hydrolysis, the

degradation products are salicylic acid and acetic acid.

Results:


4/13

Table 1: Values Obtained in Determining the Mass of KHP

Sample 1 Sample 2 Sample 3

Weight of beaker + KHP 34.073 34.116 34.360

Weight of beaker + Residue 33.495 33.608 33.845

Weight of KHP 0.578 0.508 0.521

Table 2: Titration Values Obtained in the Standardization of

Sodium Hydroxide (NaOH)and KHP

Sample 1 Sample 2 Sample 3

Mass of KHP 0.578 0.508 0.521

Final Volume (ml) 32.20 27.65 28.10

Initial Volume (ml) 1.70 0.40 1.30

Volume Used (ml) 30.5 27.25 26.80

Table 3: Values obtained for two (2) aspirin 325mg tablet

Weight of weighing dish 26.0252g

Aspirin tablets Weight of weighingdish + aspirin tablet

(g)

Weight of weighingdish + aspirin tablet

(g)

Actual mass (g)

Tablet 1 26.3646 --------- 0.3394

Tablet 2 ------------ 26.3532 0.3280

Average mass of aspirin tablets = 0.3394 + 0.3280

2

= 0.3337g


5/13


6/13

Pre-Lab Calculations

Calculate the volume of 50% (by weight) of NaOH solution necessary to prepare one litre of

0.1M NaOH. (The density of the 50% NaOH solution is 1.53 0.01 g/ml)

Molecular Mass of NaOH = 39.9972 = 39.9972g/mol

Diluting 50% to 0.1 M NaOH:

Therefore 19.13 mol/L is contained in a 50% solution of NaOH

Using C1V1= C2V2

19.13Mx V1 = 0.1Mx 1000ml

Therefore volume of solution V1 needed to make a 0.1 mol/L NaOH= 0.00523 L or 5.23 ml

This 5.23ml of NaOH was added to approximately 1L of previously boiled and cooled water.

To determine actual molar concentration of sodium hydroxide used (Standardization of

Sodium Hydroxide)

KH5C8O4(aq)+ NaOH(aq) KH4C8O4Na(aq)+ H2O(l ) (Equation 1)

Molar mass of KHP = (39.098 + (5 X 1.008) + (8 X 12.011) + (4 X 15.999) gmol-1

Molar mass of KHP = 204.22gmol-1

Sample 1:

x10

-3mol

From result Table 2, volume of NaOH used is 30.5 ml, thereforex 10-3

mol is contained in

30.5 ml of NaOH, therefore 1000 ml contain

= 0.09279 mol/L

Sample 2:


7/13

From result Table 2, volume of NaOH used is 27.25 ml, therefore mol is

contained in 27.25ml of NaOH, therefore 1000 ml contain

=

0.09130mol/L

Sample 3:

From result Table 2, volume of NaOH used is 26.8ml, therefore is

contained in 26.8 ml of NaOH, therefore 1000 ml contain

=

0.09519mol/L

Averaging all three (3) Molarities obtained =

= 0.09309 mol/L

From equation 1 NaOH and KHP reacted in a 1:1 mole ratio

# of moles of NaOH reacted = # of moles of KHP reacted

Therefore the molar concentration of sodium hydroxide used in the experiment was

0.09309mol/L

Finding the molar concentration of HCl solution

Determination of the actual molar concentration of HCl used (Standardization of Hydrochloric

acid):

Reaction that occurred between NaOH and HCl:

HCl(aq)+ NaOH(aq) NaCl(aq)+ H2O(l) (Equation 2)

From titration: Avg volume of NaOH used to neutralize HCl: (23.86+ 23.90+ 24.39) ml

3

= 24.05ml or 0.02405L

From equation 2 NaOH and HCl reacted in a 1:1 mole ratio

Moles = Avg Volume of HCl used x Molarity of NaOH


8/13

= 0.02405L X 0.09309 mol/L

= 2.240 X 10-3

mol NaOH

Since 1 mole NaOH reacts with 1 mole of HCl (see equation 2) then the number of moles of

NaOH reacted with 2.240 X 10

-3

mol so therefore the number of HCl reacted = 2.240 X 10

-3

moles

Molar concentration of HCl: 2.240 X 10-3

mol

0.02500L

= 0.08960M of HCl

To determine the amount of ASA

NaOH(aq) + C6H4OCOCH3COOH9(aq)C6H4OCOCH3COONa(aq)+ H2O(aq) (Equation 3)

Molar mass of ASA is 180.1g

Sample 1: From table 4: - 0.5033g of aspirin tablet powder

From table 6: - 27.34ml of 0. 0.09309 mol/L NaOH used for neutralization and excess

NaOH was back titrated with 17.14ml of 0.08960M of HCl.

27.34ml NaOH was used to neutralize all the material of an acidic nature. However for

hydrolysis additional 44.48ml was added.

# of moles NaOH used for hydrolysis is: 44.48ml x 0.08960M = 3.985 X 10-3

moles

1000ml

Moles of HCl is equal to moles of excess NaOH

therefore = 0.08960M X 17.14ml = 1.536 X 10-3

moles

1000ml

The difference between the moles of NaOH added for hydrolysis and that consumed by

hydrolysis equals the moles of NaOH used for hydrolysis.

3.985 X 10-3

moles - 1.536 X 10-3

moles = 2.449 x 10-3

moles

Since the mole ratio of NaOH and ASA is 1:1 (see equation 3) then there were 2.449 x 10-3

moles

of ASA in the sample.

2.449 x 10-3

moles x 180.1g of ASA = 0.4412g or 441.2mg of ASA

Sample 2: From table 4: - 0.4979g of aspirin


9/13






moles

1000ml


therefore = 0.08960M X 16.50ml = 1.478 X 10-3

moles

1000ml



3.950 X 10-3

moles - 1.478 X 10-3

moles = 2.472x 10-3

moles


moles


2.472 x 10-3


Sample 3: From table 4: - 0.5262g of aspirin






moles

1000ml


therefore = 0.08960M X 15.26ml = 1.367 X 10-3

moles

1000ml




10/13

3.796 X 10-3

moles - 1.367 X 10-3

moles = 2.429 x 10-3

moles


moles


2.429 x 10

-3


Percentage weight of ASA in the sample = mass of ASA X 100

Weight of sample

% weight of ASA in the sample 1 = 0.4412g x 100 = 82.78%

0.5033g

% weight of ASA in the sample 2= 0.4452g x 100 = 89.42%

0.4979g

% weight of ASA in the sample 3 = 0.4375g x 100 = 83.14%

0.5262g

Average weight of ASA in tablet = (82.78+ 89.42+ 83.14)% = 85.11%

3

Discussion


11/13

Sodium hydroxide was standardized using KHP and the molarity was found to be 0.09309M.

The average volume of NaOH from table 2 used in the neutralization of the KHP was 28.18ml.

This process was done as NaOH is not a highly pure solid and so was titrated against KPH which

was the primary standard. The hydrochloric acid used in the back-titration was also standardized

against the 0.09309M NaOH solution. The average volume of NaOH used in the standardization

of HCl was 24.05ml and the molarity was calculated to be 0.0896M.

Back titration was the method used in the determining the amount of aspirin (macs) in the

sample. The indicator of choice was phenolthalein as the pH at the equivalence point of this

titration is basic and the end point would therefore be observable, which is pink/ fuchsia. The

reverse of this would be seen since the solution was pink with the excess NaOH. At the end point

of the back titration the solution became colourless. From table 6, the volumes used to achieve

the end point of the reaction were 27.34ml, 28.54ml, 27.11ml, ie the amount of base needed to

neutralize all the material of acidic nature. The aspirin content in the flask was hydrolyzed with

addition NaOH of 44.48ml, 44.08ml, and 42,37ml respectively. This was done to obtain acetic

acid and salicylic acid, to have complete reaction of all the acidic content present.

The amount of excess NaOH that was present in the sample was reacted with the standard

solution of 0.0896M HCl. This was done as the analyte may be speed up in the presence of

excess NaOH. From this reaction, the amount of mole of the sample that the HCL reacted with

can be determined. The average amount of mole of NaOH used/ added for the Hydrolysis was

found to be 4.059x10-3

mole and that which was consumed in the hydrolysis process was

1.209x10-3

mole, the difference of both values was the number of moles of aspirin present/reacted

in the sample. The average mass of aspirin present in the sample was found to be 0.4413g.


12/13

Post Lab Question

1. What would be the average mass of aspirin in a tablet if the results from the directtitration was used?

Average: From table 4: -average mass of aspirin tablet powder is 0.5091g

From table 6: -average volume 27.66ml of 0. 0.09309 mol/L NaOH used for

neutralization of ASA using direct titration.

Number of moles of NaOH used in the neutralization= molarity * volume

=

* 27.66ml

= 2.575x10-3

moles

Since Sodium hydroxide and aspirin is a 1: 1 ratio, ASA= 2.575x10-3

moles

Mass of ASA= # of moles * molecular mass = 2.575x10-3

moles * 180.1g= 0.4637g

Average weight of one tablet=0.3337g

If 0.4637g of ASA is contained in 0.5091g of sample aspirin tablet powder

Then x is contained in 0.3337; x= 0.3039g

Therefore 0.3039g/303.9mg of aspirin is present in one tablet of Macs aspirin.


13/13

anal chem 2b

Documents