Chemical Equibrium in Solution #3 Essay
Experimental Method 2
Chemical Equilibrium in Solution
Ginger Rimestad 10 December 2005
Abstract The experiment, Chemical Equilibrium in Solution, makes use of a titration of a heterogeneous solution. This is done in order to find the distribution of molecular Iodine, I2, as the solute between two immiscible liquid phases, water and a hexane solution. The average values obtained for (I2) = 6.11E-06 M, (I-) = 0.1097, (I3-) = 2.82E-04. The results that were found in this experiment show an inaccuracy. This may have been due to the third run in …show more content…
2S2O32- + I3- → S4O6 2- + 3I-
Iodine in hexanes, (I2)h, is found by taking the molarity of S2O32- and multiplying it by the amount of the KI titrated into the hexane layer. Then the balanced equation was used to understand that this amount is then divided by 2. This volume of S2O32- titrated is documented in Table 3. For the calculations, the average of the two titrations is used. Iodine in water, (I2)w, is found in the same manner and results of volume of S2O32- titrated is found in Table 2. Using the results found for the two layers of I2 we solved for distribution constant. This is defined by using Eq.2. A representation of the calculations from runs 1-3 are found on Graph 1. Note: each run was done two times. Therefore Tables 2 & 3 show two set of 3 runs.
A trend line was added to show the variation of the concentration. This trend line is interpolated thru the y-axis. This will give the distribution constant needed to find (I2)w when KI is present. Use the same procedure in finding for (I2)h runs 4-6 as was done for runs 1-3. This data can be found on Table 5.