24/7 writing help on your phone
Aspirin1, or acetylsalicylic acid is a drug in the family of salicylates, often used as an analgesic (to relieve minor aches and pains), antipyretic (to reduce fever), and as an anti-inflammatory. It also has an antiplatelet (“blood-thinning”) effect and is used in long-term, low doses to prevent heart attacks and cancer.
Aspirin tablets consist mainly of 2-ethanoylhydroxybenzoic acid (acetylsalicylic acid, CH3COOC6H4COOH, figure 3).
Although the acidic conditions found in stomach do not affect aspirin, the alkaline juices in the intestines hydrolyse it to ethanoate (acetate) ions and 2-hydroxybenzoate ions.
CH3COOC6H4COOH + 2OH- ï¿½ CH3COO- + HOC6H4COO- + H2O
Aim: To determine the percentage of CH3COOC6H4COOH in aspirin tablets.
Hypothesis: When a known amount of standard sodium hydroxide solution is used in excess to hydrolyse a known mass of aspirin tablets, we may determine the percentage of acetylsalicylic acid in those tablets.
Prediction: The quantity of acetylsalicylic acid will be smaller than the one given on the packaging and in the leaflet enclosed.
– 3 aspirin tablets (Acidium acetylsalicylium 500 mg)
– pipette (25 cm3)
– safety filler
– burette (25 cm3) and stand
– small beaker
– 2 standard flasks (250 cm3)
– 3 conical flasks (250 cm3)
– small funnel
– lime water
– Bunsen burner
– Tripod and gauze
– 1.0 mol dm-3 sodium hydroxide solution (30 cm3)
– 0.10 mol dm-3 hydrochloric acid (150 cm3)
– phenolphthalein as indicator
– eye protection
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.
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.
Mass of 3 aspirin tablets/ g
Mean mass of one tablet/ g
Final burette reading/ cm3
Initial burette reading/ cm3
Volume of 0.10 mol dm-3 HCl added/ cm3
Mean volume of HCl added
25 cm3 = 0.025 dm3
23.8 cm3 = 0.023.8 dm3
0.10 mol dm-3
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) ï¿½ 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 ï¿½ V
VNaOH = 25 cm3 = 0.025 dm3
cNaOH = ?
nNaOH = nHCl
VHCl = 23.8 cm3 = 0.0238 dm3
cNaOH = 0.10 mol dm-3
nNaOH = 0.10 mol dm-3 ï¿½ 0.0238 dm3 = 0.00238 mol
1. From the balanced chemical equation find the mole ratio
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 ï¿½ V
n = 3 x 10-3 mol, V = 25.0 x 10-3L
M(HCl) = 3 x 10-3 ï¿½ 25.0 x 10-3 = 0.12M or 0.12 mol L-1
The titration formula is:
concentration of NaOH ï¿½ mean amount of NaOH / concentration of HCl ï¿½ amount of HCl = 1/1
cNaOH ï¿½ VNaOH / cHCl ï¿½ VHCl = 1/1
When substituting the values received in the first trial into the titration formula:
(0.100 ï¿½ 5.26) / (cHCl ï¿½ 25.0) = 1/1
cHCl = (0.100 ï¿½ 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 ï¿½ 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 ï¿½ 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) ï¿½CaCl2(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 ï¿½ 100 g mol-1 = 2.14 g
Finally, the percentage CaCO3 in egg shells is:
2.14 g / 1.51 g ï¿½ 100% = 141.72 %
Which obviously states that the experiment was done wrongly.
During hydrolysis the reaction below takes place:
CH3COOC6H4COOH + 2OH- ï¿½ 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 ï¿½ 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-ï¿½ 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.
👋 Hi! I’m your smart assistant Amy!
Don’t know where to start? Type your requirements and I’ll connect you to an academic expert within 3 minutes.get help with your assignment