Question

Aspirin Determination using Back Titration

Began with 0.1 M HCl solution and 0.1 M NaOH solution.

1:2 aspirin to NaOH mole ratio, therefore, mol HCl (back titration) = excess mol NaOH AND total mol NaOH used - excess mol NaOH=required mol NaOH AND required mol NaOH (1 mol aspirin / 2 mol NaOH) = mol aspirin.

Each step builds off the last

1) Determination of HCl Concentration. This was the first titration. Concentrated HCl was diluted to .1M before titrating. n is number of moles:

Trial	Starting Vol HCl	M NaOH	Vol NaOH used in titration	n NaOH	n HCl	M HCl after titration
1	35mL	.1 M	59.3mL	5.93x10^-3	5.93x10^-3	0.169 M
2	40mL	.1 M	66.5mL	6.65x10^-3	6.65x10^-3	0.166 M
3	45mL	.1 M	74.9mL	7.49x10^-3	7.49x10^-3	0.166 M

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2) Moles of NaOH added to aspirin for back titration. More NaOH was added to the flasks, giving the total vol NaOH. The flasks were gently heating to speed up the process. I would appreciate if you checked these answers for n:

trial	M NaOH	Vol NaOH	Add. Vol NaOH	Total Vol NaOH	Total n NaOH
1	0.1 M	25.9mL	35.9mL	61.8mL	6.18x10^-3
2	0.1 M	25.0mL	35.0mL	60.0mL	6.00x10^-3
3	0.1 M	24.2mL	34.2mL	58.4mL	5.84x10^-3

3) Moles of HCl used in back titration. The heated solution was titrated again:

Trial	M HCl	Vol HCl	n HCl	n NaOH excess
1	0.1 M	9.1mL	9.1x10^-4	9.1x10^-4
2	0.1 M	8.6mL	8.6x10^-4	8.6x10^-4
3	0.1 M	7.0 mL	7.0x10^-4	7.0x10^-4

We started with 5 tablets and the total weight was 1.8721g. The tablets were crushed and 0.5 g powder was added to each flask and then mixed with 20mL ethanol to help with solubility This is what I got for the first table but the second table I am hopeless. Please help (ignore the empty two columns in this table):

Trial	Total n NaOH added	n NaOH excess	n NaOH reacted			n Aspirin
1	6.18x10^-3	9.1x10^-4	.00527			2.63x10^-3
2	6.00x10^-3	8.6x10^-4	.00514			2.57x10^-3
3	5.84x10^-3	7.0x10^-4	.00514			2.57x10^-3

This is the last table HELP HERE. I don't know what the mass of sample means or how to find the % weight or mg per tablet:

Trial	n Aspirin	Mass Aspirin in sample (g)	Mass of sample (g)	% weight Aspirin	Average mass of  5 tablets (g)	mg of Aspirin per tablet (mg)
1	2.63x10^-3	.5000g			1.8721g
2	2.57x10^-3	.5003g			1.8721g
3	2.57x10^-3	.5001g			1.8721g

I was given that HCl has a density of 1.188g/mL and is 37 wt % (g). 2 moles of NaOH are per 1 mole of aspirin.

Answer 1

In the last table , in the 3rd column mass of aspirin is not 0.5 gm. It can be found out using formula

. wt of substance = Molecular wt of sample *no of moles

(Molecular weight of aspirin is 180 g/mol)

So the third column becomes

1. 2.63 x 10^-3 x 180 = 0.4734 gm

2. 2.57 x 10^-3 x 180 = 0.4626 gm

3. 2.57 x 10^-3  x 180 = 0.4626 gm

In the fourth column mass of sample is

1.0.5gm

2.0.5003 gm

3.0.5001 gm

In the fifth column percentage of aspirin is given by

% weight of aspirin = (wt of aspirin / wt of sample) x 100

1. (0.4734/0.5) x100 = 94.68 %

2 (0.4626/0.5003) x 100 = 92.46 %

3 (0.4626/0.5001)x 100 = 92.5%

We have avg of 5 tablets is 1.8721, then avg mass of each tablet = 1.8721/5 = 0.374 gm = 374.42 mg

Column 7 becomes

1. 94.68/100 = 0.9468 mg aspirin per mg of tablet

2. 92.46/100 = 0.9246 mg aspirin per mg of tablet

3. 92.5/100 = 0.925 mg of aspirin per mg of tablet