1. You are given two clear solutions and told that one is lauric acid solution and the other is oleic acid. State and clearly outline the test you would use to differentiate one from the other. Hint: Name of test, Procedure of test, and Results.
2. Write the chemical equations for the reactions taking place in the test above (question 1).
1. To differentiate between a lauric acid solution and an oleic acid solution, you can perform a solubility test using the sodium hydroxide (NaOH) test.
Procedure:
1. Take a small amount of each sample (lauric acid solution and oleic acid solution) in separate test tubes.
2. Add a few drops of sodium hydroxide (NaOH) solution to each test tube.
3. Observe the results.
Results:
- Lauric acid solution: Lauric acid is a saturated fatty acid and is soluble in NaOH. Upon adding NaOH, you would observe a clear solution, and the lauric acid would dissolve.
- Oleic acid solution: Oleic acid is an unsaturated fatty acid. It does not dissolve readily in NaOH. Upon adding NaOH, you would observe a precipitation or the formation of a milky or cloudy solution.
2. The chemical equations for the reactions taking place in the solubility test are as follows:
For lauric acid:
Lauric acid + Sodium hydroxide → Sodium laurate + Water
C12H24O2 + NaOH → C12H23O2Na + H2O
For oleic acid:
Oleic acid + Sodium hydroxide → No reaction (Insoluble)
In the case of lauric acid, it reacts with sodium hydroxide to form sodium laurate and water. However, oleic acid is insoluble in sodium hydroxide and would not undergo any reaction in this test.