Lab 1 - Lab 1: Investigating Solubility and Acid-Base...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Lab 1: Investigating Solubility and Acid-Base Reactions In part I, our solvent was Hexane, a non-polar compound. First, we added Diethyl Ether in it, a non-polar compound and it dissolved completely, from this we can conclude that like dissolve like. However, all our other compounds were polar: Ethyl Acetate, Dichloromethane, Acetone and Ethanol, and they all showed miscibility with Hexane, probably as a result of the intermolecular forces (dipole-induced dipole) taking place within the mixture. For our last test tube, we added water (a very polar compound) to Hexane (non-polar) and the result, as expected was non-miscibility, forming two layers in the test tube. In part II, our solvent was Water, a polar compound. We had two solutes, Diethyl Ether and Toluene, which are non-polar and we could see this since neither of them dissolved but instead formed two layers in the test tube. The other four compounds are polar and our predictions would be that they would dissolve; however, only two: Acetone and Ethanol dissolved because of their polar conformation. Dichloromethane, although polar, has many non- polar qualities that explain its immiscibility. Finally, Ethyl Acetate also formed two layers but probably because we went over the limit of its solubility in water.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/25/2009 for the course CHEM Chemistry taught by Professor Nitche during the Fall '08 term at University of California, Berkeley.

Page1 / 3

Lab 1 - Lab 1: Investigating Solubility and Acid-Base...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online