Abstract:
For this experiment, back titration is a technique used to determine the actual aspirin content in an aspirin pill. The titration of an aspirin pill is a two step reaction involving one fast acid-base neutralization in addition to a slow hydrolysis step. Because of this two step reaction, the molar ratio of our NaOH standard obtained last week will be required to titrate the acid-base neutralization, and excess of the NaOH standard will be added to ensure that the end point is passed. It is then possible to heat the solution to drive the reaction to completion. Upon doing this, we were able to back titrate the excess NaOH by using Hcl. In this report, calculations involving the amounts of NaOH and Hcl are used to determine the concentration of aspirin in an aspirin tablet.

Experimental:
The preparation of the roughly 0.1M Hcl solution is the first step in the experiment. 100 mL of distilled water are placed in a bottle along with a pre-calculated amount of Hcl (2.08 grams). After mixing the solution, about 150 mL of water were added to the solution. 
	A buret was used to add 35, 40, and 45 mL of  0.1 M Hcl to 3 flasks containing 50 mL of distilled water and 3 drops of phenolphthalein indicator. Upon doing this, the 0.1M standardized NaOH solution was used to back titrate the 3 flasks with the NaOH.
	Three aspirin tablets were weighed, crushed to a fine power, and re-weighed by difference and added to 3 flasks. About 20 mL of ethanol – aids in dissolving- were added to the flask in addition to phenolphthalein indicator. All of these flasks were titrated with NaOH using a buret. Because the aspirin molecule reacts with two hydroxide ions, it is necessary to multiply the volume of NaOH used by 2 and add 10 mL to overshoot. For this instance, about 40 mL of NaOH were added to each flask.
	Boiling chips were added to the flask and the flask then heated in a water bath for 15 minutes to speed up the reaction. The samples were then allowed to cool for 5 minutes. The flasks were then back titrated with the 0.1M Hcl. 

Safety Issues and Chemical Hazard Information:

Chemical Name:                                      Physical Hazards                 Health Hazards
Hydrochloric acid
Water-reactive
Toxic, irritant
Phenolphthalein
none
irritant
Aspirin
none
Toxicant, irritant, sensitizer
Sodium Hydroxide
Water-reactive, corrosive
Toxic, irritant
Ethanol
Flammable
Irritant

Results:

Comparison of Hcl to standardized NaOH
Flask #
Starting Volume (mL)
Ending Volume (mL)
Total Volume (mL)
1
0.52 ± 0.02 mL
35.56 ± 0.02 mL
35.04 ± 0.03 mL
2
3.06 ± 0.02 mL
43.02 ± 0.02 mL
39.96 ± 0.03 mL
3
1.32 ± 0.02 mL
46.32 ± 0.02 mL
45.00 ± 0.03 mL

Titration with added NaOH
Flask #
Starting Volume (mL)
Ending Volume (mL)
Total Volume (mL)
1
1.02 ± 0.02 mL
30.60 ± 0.02 mL
29.58 ± 0.03 mL
2
1.96 ± 0.02 mL 
35.64 ± 0.02 mL
33.68 ± 0.03 mL
3
2.98 ± 0.02 mL
40.72 ± 0.02 mL
37.74 ± 0.03 mL





Sample Preparation:
Flask #
Weight by difference
Ending Volume
1
21.9879  ± 0.0001 g
0.3159 ± 0.0001 g
2
21.6719  ± 0.0001 g
0.3160 ± 0.0001 g
3
21.3806  ± 0.0001 g
0.2913 ± 0.0001 g

Aspirin titration with base:
Flask #
Starting Volume (mL)
Ending Volume (mL)
Total Volume (mL)
1
1.10 ± 0.02 mL
15.58 ± 0.02 mL
14.48 ± 0.03 mL
2
0.38 ± 0.02 mL
16.02 ± 0.02 mL
15.64 ± 0.03 mL
3
1.42 ± 0.02 mL
17.02 ± 0.02 mL
15.60 ± 0.03 mL

Back titration with acid: 
Flask #
Starting Volume (mL)
Ending Volume (mL)
Total Volume (mL)
1
0.14 ± 0.02 mL
12.14 ± 0.02 mL
12.00 ± 0.03 mL
2
12.14 ± 0.02 mL
23.02 ± 0.02 mL
10.88  ± 0.03 mL
3
23.02 ± 0.02 mL
34.08 ± 0.02 mL
11.06 ± 0.03 mL

Data Analysis:
1)
NaOH molarity (found last week) = 0.1012 ± 0.001 M
For the 3 titrations:
	M1V1 = M2V2
	M1V1/V2 = M2
	[NaOH] * (mL NaOH) / (mL Hcl) = [Hcl]
	
	First titration:
	0.1012 ±  0.0003 * (29.58 ± 0.03 mL NaOH) / ( 35.04 ± 0.03 mL Hcl) = 0.088  ± 0.003  M Hcl
	Second titration:
	0.1012 ±  0.0003 * (33.68 ± 0.03 mL NaOH) / (39.96 ± 0.03 mL Hcl) = 0.085  ± 0.003 M Hcl
	Third titration: 
	0.1012 ±  0.0003 * (37.74 ± 0.03 mL NaOH) / (45.00 ± 0.03 mL Hcl) = 0.085  ± 0.003  M Hcl

Standardized HCl concentration (average of the three values above):

x = (value 1 + value 2 + value 3) / 3 total measurements
x = (0.088 ± 0.003) + (0.085 ± 0.003) + (0.085 ± 0.003)
			      3
x = 0.086  ± 0.005 M
                                                                                                                                                      
s = √(measurement 1- 0.086)2 + ( measurement 2 – 0.086)2  + (measurement 3 - 0.086)2 / (n-1)
s = √(0.088 – 0.086)2 + ( 0.085 – 0.086)2 + (0.085 – 0.086)2)/(2)       
s = 0.002 M
             
%RSD = (100 x (standard deviation / mean))
%RSD = (100 x (0.002M / 0.086))
%RSD = 2.73 %
90% confidence interval:
μ = x ± (t*s/√n)                       
μ = 0.086 ±  (2.920*0.002)/√2
μ = 0.086 ± (0.004)
2)
NaOH molarity (found last week) = 0.1012 ± 0.001 M
Total number of moles added to each of the aspirin samples:
Flask #
Volume of NaOH added (L)
Molarity*Volume
 Moles NaOH
1
0.01448 ± 0.00003 L
0.1012 * 0.01448
0.00146
2
0.01564 ± 0.00003 L
0.1012 * 0.01564
0.00158
3
0.01560 ± 0.00003 L
0.1012 * 0.01560
0.00158
 

Discussion Questions:
1)The first reaction step of the aspirin with the hydroxide ions in the acid-base neutralization step is complete when a colour change occurs from clear to pink.
2)In this experiment, it does not seem to be possible to determine when the second hydrolysis step of the reaction has begun.
3)It is necessary to use a buret and not a graduated cylinder to add the excess NaOH so a precise and accurate amount can be added and measured. It is important to know the amount added so that the amount of Hcl added subsequently can be calculated
4)It is important that the analyte solution remain pink throughout the heating stage of the experiment because it ensures that the reaction is going to completion. The slow hydrolysis reaction is occurring during this step.
5)Ethanol was used to help dissolve the acetylsalicylic acid in the aspirin tablets. Ethanol is slightly acidic and reacts with NaOH, because of this, the final determination of the mass of aspirin per tablet would be too low because the amount of Hcl reacting with the NaOH would be less. Because we use Hcl in subsequent calculations to determine the mass of aspirin per tablet, the smaller amount of Hcl would cause the amount of aspirin per tablet to be less
6)A blank correction experiment could be done by titrating an NaOH solution/standard with ethanol. We could then use the results of the blank correction experiment by replacing the amount of ethanol that would have theoretically reacted with the NaOH with HCL, as the ethanol takes the place of Hcl in reacting with the NaOH.
7)____

Results: