Thermodynamics of the Dissolution of Borax
* Study a system of salt and water solution.
* Determining a variety of important thermodynamics quantities from the solubility information at various temperatures.
The salt and water solution in this experiment has relatively simple solubility equilibrium of borax in water.
Na2B4O7 . 10 H2O 2Na + + B4O5(OH)42- + 8H2O
This reaction is an equilibrium process and 8 water molecules from the hydrated salt are lost to the reaction medium. The equilibrium constant expression for this reaction is:
K = [Na+]2 [B4O5(OH)42-] [H2O]8
[Na2B4O7 . 10 H2O]
In this experiment we will always make sure there is some solid borax remaining in the sample mixture before removing some of it to analyze it. Therefore, we can assume that the concentration of solid borax is constant. In addition, the water molecules which were originally part of the borax’s crystalline matrix is lost to the sample mixture; however, it doesn’t significantly affect the concentration of the water. The equilibrium constant expression can now be simplified to become:
K = [Na+]2 [B4O5(OH)42-]
The first equilibrium expression and the balanced solubility equilibrium reaction allow us to express either borate ion or sodium ion in terms of the other. So, it is possible to determine the constant in terms of either ion. After substituting borax ion in place of the sodium ion, ([Na+] = 2 [B4O5(OH)42-] , K = (2 [B4O5(OH)42-] )2 * [B4O5(OH)42-] ) we get:
K = 4 [B4O5(OH)42-]3 .
Finding the concentration of borate ion in any sample at any given temperature gives us the solubility product at that temperature.
* 5mL pipet
* Test tubes
* Hot plate
* 250 mL beaker
* Electric balance...
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