Analysing aspirin tablets.

Method: A known amount of standard sodium hydroxide solution is used in excess to hydrolyse a known mass of aspirin tablets. The remaining sodium hydroxide acid is then titrated with standard solution. The amount of alkali required can be calculated from the equation above and furthermore, the amount (in moles) of acetylsalicylic acid which has been hydrolysed can be found.

Procedure:

We were provided with 0.1 mol dm-3 hydrochloric acid (HCl) and 1 mol dm-3 sodium hydroxide (NaOH). We worked in pairs.

• Partner No.1

1. Using a safety filler, I pipetted exactly 25 cm3 of the approx. 1.0 mol dm-3 NaOH solution into a 250 cm3 standard flask and made up to the mark.
2. I titrated 25 cm3 of this solution against 0.10 mol dm-3 HCl using phenolphthalein indicator.
3. I repeated it 3 times to get more accurate results and put the data in a table.

• Partner No.2

4. Meanwhile Partner No.2 weighted accurately 1.8 g of the aspirin tablets (3 pieces) into a conical flask.
5. Using a safety filler, she pipetted exactly 25 cm3 of the approx. 1.0 mol dm-3 NaOH on to the tablets, followed by the same volume of lime water.
6. She simmered the mixture gently on a tripod and gauzed over a bunsen for 10 minutes to hydrolyse the acetylsalicylic acid.
7. The mixture was then cooled and transferred with washings to a 250 cm3 standard flask and made up to the mark with lime water.

• Both Partners

8. We pipetted the hydrolysed solution into a conical flask. We titrated this against 0.10 mol dm-3 HCl using phenolphthalein indicator.
9. Finally, we estimated the quantity of unused NaOH after the hydrolysis.

Data Collection

Data Processing and Presentation

• Standardisation of the approx. 1.0 mol dm-3 NaOH

During titration the reaction below takes place:

HCl(aq) + NaOH(aq) → NaCl(aq) + H20(l)

That is why the solution changes its colour from pink into transparent.

The ratio between HCl and NaOH is 1/1, so one mol of HCl reacts with one mol of NaOH.

n = c ∙ V

NaOH

VNaOH = 25 cm3 = 0.025 dm3

cNaOH = ?

nNaOH = nHCl

HCl

VHCl = 23.8 cm3 = 0.0238 dm3

cNaOH = 0.10 mol dm-3

nNaOH = 0.10 mol dm-3 ∙ 0.0238 dm3 = 0.00238 mol

 

1. From the balanced chemical equation find the mole ratio
         NaOH:HCl
         1:1
2. Find moles HCl
         NaOH: HCl is 1:1
         So n(NaOH) = n(HCl) = 3 x 10-3 moles at the equivalence point
3. Calculate concentration of HCl: M = n ÷ V
         n = 3 x 10-3 mol,       V = 25.0 x 10-3L
         M(HCl) = 3 x 10-3 ÷ 25.0 x 10-3 = 0.12M or 0.12 mol L-1 

The titration formula is:

concentration of NaOH ∙ mean amount of NaOH / concentration of HCl ∙ amount of HCl = 1/1

cNaOH ∙ VNaOH / cHCl ∙ VHCl  = 1/1

When substituting the values received in the first trial into the titration formula:

(0.100 ∙ 5.26) / (cHCl ∙ 25.0) = 1/1

cHCl = (0.100 ∙ 5.26) / 25.0 = 0.0210 mol dm-3

To calculate the total amount of excess HCl we have to multiply this by the volume of the calibrated flask:

250 cm3 = 0.25 dm3

0.25 dm3 ∙ 0.0210 mol dm-3 = 0.0053 mol

Now, to calculate the amount of calcium carbonate in egg shell we have to find out the amount of acid added to the shells:

0.040 dm3 ∙ 1.20 mol dm-3 = 0.0480 mol

And subtract the total amount of excess HCl from it:

0.0480 mol - 0.0053 mol = 0.0427 mol

The equation for the reaction of the acid with egg shells:

CaCO3(s) + 2HCl(aq) →CaCl2(aq) + CO2(g) + H2O(l)

As we can see from the ratio, the amount of moles of CaCO3 in 1.511g egg shells is half the amount in moles of HCl that reacted = 0.0214 mol.

The molar mass of CaCO3 is 100 g mol-1, so the mass of CaCO3 in egg shells:

0.0214 mol ∙ 100 g mol-1 = 2.14 g

Finally, the percentage CaCO3 in egg shells is:

2.14 g / 1.51 g ∙ 100% = 141.72 %

Which obviously states that the experiment was done wrongly.

• Hydrolysis

During hydrolysis the reaction below takes place:

CH3COOC6H4COOH + 2OH- → CH3COO- + HOC6H4COO- + H2O

As a result of above reaction, aspirin is hydrolysed to ethanoate ions and 2-hydroxybenzoate ions.

Errors and uncertainty

1. Balance: 0.06 %

0.001g / 1.511g * 100% = 0.0662 %

2. Pipette: 0.16 %

V = 25 cm3 → 0.04 cm3

0.04 cm3 / 25 cm3 * 100% = 0.16%

3. burette: 0.5%
4. calibrated flask: 0.2%

Total: 0.92 %

Conclusion and Evaluation

As we can see from the results above, the prediction made at the very beginning of this lab was correct. Neither type of acid or base nor the concentration of acid does not have influence on the enthalpy of neutralisation.

Hence we may assume that the enthalpy of neutralisation is equal to the enthalpy change for

H+ + OH-→ H2O.

The enthalpy change for this reaction, however, is –57.9 kJ mol-1. The differences between my results and the theoretical value may come from the fact that the measurements were not very accurate. The temperatures of the acids, bases and mixtures might have been influenced by cool beakers. Therefore the temperatures were a bit lower than they should have been. If the ∆T was higher by 3oC, the enthalpy of neutralisation would be almost the same as in the sources.

I do not know how to improve the experiment so that data gathered will be similar to theoretical values. I reckon in classroom conditions such mistake is not a serious one.

Sources:

1. Green J, Damji S. 2001. Chemistry. Second edition. IBID Press, Victioria, Australia.
2. http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_neutralisation

http://en.wikipedia.org/wiki/Aspirin

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