Tri-Iodide Ion Essay

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Experiment 2.15. The stability Constant of the Tri-Iodide Ion. Nicole Murray Analytical Science Abstract: The aims of this experiment are: * To determine the stability constant of the tri-iodide ion. * To improve titration skills. * To improve mathematical skills. The stability constants of the tri-iodide ion was found to be Introduction: Non-polar octamethylcyclorasiloxame, volasil 244, and polar potassium iodide are mixed together and the iodide distribution between them was examined. Volasil 244 is a source of I2, potassium iodide is a source of I- and in equilibrium they are I2 + I- I3-. The equilibrium or stability constant is denoted by [I3-]/[I2][I-]. I2 is the only species that is undergoing distribution. KI3 and KI are insoluble in the non-polar solvent as they are ionic. The concentration of I2 is related to its KD, or characteristic distribution coefficient. The concentration of I2 in the aqueous layer can be determined by measuring the I2 concentration in the organic layer as long as there is knowledge of KD. the total titratable iodide that was present is greater than the concentration of the I2 in the aqueous phase as some is being bound as I3-. This is how the stability constant is calculated. Method: As per lab manual page 89. Results: Organic Layer A: Moles of I2: 0.25/1000*0.95 = 2.375*10^-4/2 = 1.1875*10^-4 KD = [I2]H2O/[I2]Volasil 244 = 0.0118 I2 1.1875*10^-4*0.0118 = 2.2125*10^-6 Titrated S2O3 = 2(I2 + I3-)H2O 2.375*10^-4 = 2(2.2125*10^-6 + I3) I3 = 1.1654*10^-4 Iint = (0.2M * volume)/1000 Iint = 0.2/1000*15 = 3.0*10^-3M Iint = (I- + I3-)H2O 3*10^-3 = I- + 1.1654*10^-4 I- = 0.192M I2 in M/L : 1.1875*10^-4/15 * 1000 = 0.0125M/L I3= 1.1654*10^-4/15 *1000 = 0.007769M/L K= [I3]/[I2][i-] K= 0.007769/(0.0125)(0.192) K= 3.237 Organic Layer B: 0.25/1000*1.12= 2.8*10^-4/2 = 1.4*10^-4 moles of

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