dDetermination of the partition coefficient of ethanoic acid between water and butan-2-ol.

25cm3 of aqueous ethanoic acid and 15cm3 of butan-2-ol

7. For each experiment, the ratio of the concentration of ethanoic acid in the aqueous layer to that in the butan-2-ol layer was calculated.

Result

Room temperature: 29℃

Volume of butan-2-ol: 15 cm3 

Conclusion

The partition coefficient of ethanoic acid between water and butan-2-ol is :

=

=0.786

Discussion

1. Shaking is necessary in step (2) because it made it faster to attain equilibrium state.
2. When temperature increases, the solubility of the two solvents increase. But the rate of the increase in solubility are not the same, it is expected that the partition coefficient varies with temperature.
3. The aim of titration is to find the concentration of the solvent, but not the total number of mole in the solvent. Therefore, the volumes of the aqueous and alcohol solution used in the titration must be known as accurately as possible in order to find accurately concentration. The aim of adding aqueous ethanoic acid and 2-methypropan-1-ol is only to leave the mixture to equilibrium and provide enough solvent for the titration. Therefore the amounts of aqueous ethanoic acid and 2-methypropan-1-ol need not be measured out accurately.
4. The following assumptions are made:

1. The temperature of the mixture remained constant throughout the experiment. This assumption was valid as it was felt (by hand) that the temperature of the separating funnel did not changed throughout the experiment.
2. Ethanoic acid, water and butan-2-ol are non-volatile and do not evaporate slowly. This assumption is not valid because there is a smell of alcohol over the separating funnel. That means that there are particles coming out from the mixture in the separating funnel.

5. Solvent extraction is more efficient if the extraction solvent is added in small portions several times instead of all at once. Therefore it is more efficient to extract a solute with two 25cm3 portions of solvent rather than with a single 50cm3 extraction.
6. The applications of the partition law:

1. By partition law, the KD can be found experimentally. The amount of the solute that can be extracted using solvent extraction can be predicted, instead of using other complex method.
2. By partition law, we know that the amount of solute extracted is more when the extracting solvent is added in several small portions instead of all at once.

7. Butan-2-ol is much lighter than water. Therefore butan-2-ol is at the top of the mixture while water is at the bottom.

Result

Titration Result

The first try encountered error, and only the result of the second try was used.

Calculation

I2+2S2O32- → 2I-+S4O62-

2Mn(OH)3(s) + 2I-(aq) + 6H+(aq) → 2Mn2+(aq) + I2(aq) + 6H2O(l)

8OH-(aq) + 4Mn2+(aq) + O2(aq) + 2H2O(l) → 4Mn(OH)3(s)

No. of mole of sodium thiosulphate used: 6 × 10-6 × 0.0125 = 7.5 × 10-8

No. of mole of I2 in 100cm3 sample: 7.5 × 10-8 ÷2 = 3.75 × 10-8

No. of mole of Mn(OH)3 in 100cm3 sample: 3.75 × 10-8 × 2 = 7.5 × 10-8

No. of mole of dissolved O2 in 100cm3 sample: 7.5 × 10-8 ÷4 = 1.875 × 10-8

No. of mole of dissolved O2 per dm3 : 1.875 × 10-8 ÷100 × 106 = 1.875 × 10-4

Mass of dissolved O2 per dm3 : 1.875 × 10-4 × 16 × 2 = 0.006g

DO value of the water sample is: 0.006 × 103 = 6mg dm-3

Conclusion

The Do value of the water sample is 6mg dm-3.

Discussion

1. No air bubble should be trapped as the oxygen in the bubbles may dissolve in the water sample and affect the result.
2. During titration, starch solution is added as indicator. Starch solution should not be added too early as it may react with the iodine.