Final Exam Physical Chemistry Laboratory! Fall 2017 4. (20 pts) Here we wish to

Question

Final Exam Physical Chemistry Laboratory! Fall 2017 4. (20 pts) Here we wish to investigate the equilibrium of the sparingly soluble iodate, AglO,, in water at 25'C. The reaction is: And the equilibrium constant for this reaction can be writen as (with assumptions): We would like to monitor the concentration of I0, spectroscopically, h owever, IO, has ho absorbance that is suitable. So, the IO, concentration must be determined indirectly, by running a second reaction: The li ion is orange and its concentration can be determined by measuring the absorbance at 565 the concentration as a function of absorbance. nm. As this absorbance can be compared to a Beer's Law plot, giving Here is the procedure: -A saturated solution ofAglos is maintained at 250. - An aliquot of 5mL of the solution is drawn and put into a flask. - 5mL of HCI and 5mL of KI are mixed with the 5mL aliquot. - Some time is allowed to pass for rxn 2 to complete (it turns orange). -3 mL of the resulting solution is drawn into a cuvette and placed in the colorimeter. . The absorbance of the solution at 565 nm is measured to be 0.125%. Before this procedure was carried out, the result of a Beer's Law calibration plot gavea slope of 0.040 (M/%) a. (7 pts) Determine the concentration of I; in the solution and determine the number of b. (5 pts) Determine the number of moles of 10s and Ag' ions. moles of 13 present. c. (5 pts) Determine the e d. (3 pts) Determine the equilibrium constant for this dissociation at 250.

Explanation / Answer

From the given experiment and the set of data given

Absorbance of solution = 0.125 x 10^-3

from graph molar absorptivity = 0.04

a. concentration of [I3-] in 15 ml solution = 0.125 x 10^-3/0.04 = 0.03125 M

moles I3- present = 0.03125 M x 15 ml = 0.00047 mol

b. moles of IO3- in 5 ml original solution = 0.00047 x 15/3 x 5 = 0.00047 mol

c. concentration of [IO3-] = [Ag+] = 0.00047 mol/0.005 L = 0.094 M

d. equilibrium constant for dissociation = (0.094)(0.094) = 8.8125 x 10^-3

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