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Compare The Solubility Of Two Different Solutes

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Various solids and liquids were immersed in solvents, ranging from water to TTE, in order to observe the solubility of the given solutes in each solvent. Each solute was placed in 3 different test tubes filled with water, ethanol and TTE. In part one of the lab, 2 small crystals of iodine, nonpolar solutes, were added to 5 ml of water, ethanol and TTE. Upon our observation iodine did not dissolve in all three solvents and instead produced a dark reddish-orange colour and magenta colour in ethanol and TTE respectively. The same procedure done to the 2 small iodine crystals was repeated with the 2 small ammonium chloride crystals, polar solutes. Unlike iodine, ammonium chloride dissolved in almost every solute completely. The only exception was that it did not dissolve fully in ethanol and left behind a powdery …show more content…

Our last experiment involved investigating the solubility of an unknown solid. The unknown substance did not dissolve in most of the solvents with the exception of water. The solubility of the unknown substance leads me to conclude that the substance is polar, however it is as polar as water thus indicating that it is more polar than ethanol and TTE. With our noted observations, we came to the general conclusion that the solubility of a substance is determined by how similar the polarities of the solute and the solvent it is immersed in. For instance, iodine remained in its solid form in water and dissolved slightly in TTE. This indicates that TTE has a polarity more similar to iodine than that of water hence why the iodine’s stronger pull by the molecules of TTE allowed it to dissolve slightly better. Similarly, ammonium chloride was able to dissolve easier in water than TTE. Water is a more polar substance than TTE and created a stronger attraction between the polar molecules of ammonium

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