Question

Hi I need help with finding the triiodide in diluted soultion ( i think you use M1V1=M2V2) but I'm not sure with what values. I also need it for the excinction coefficient value. b for the equation is 1.3, please help!

Procedure termination of the absorption spectrum oftrilodide. Please review the discussion n spectroscopy in Experiment 15. Iodine and triodide solutions are both colored and of absorption absorb light in the visible region of the spectrum. Iodine has nm in the blue green region but fortunately for this experiment, the absorption tails energy to a value (extinction coefficient of about 16) that is low enough in the near ultravio nm) that it causes an insignificant error at the wavelength used Triiodide has two peaks in the ultraviolet (288 nm and one to be determined near 350 nm) are very strong absorptions. The absorption then tails off at lower energy an absorption maximum around 460 off at higher let (<400 for study in this experiment. and both into the visible giving rise to its yellowish color. Use of the peak in the near ultraviolet (~350 nm) avoids the use of expensive quartz cells needed at 288 nm (Pyrex glass does not transmit below about 300 nm) and it is just within reach of inexpensive spectrometers such as the Spectronic 20 (capable of reaching to 340 nm). The first step in this experiment will be to determine the optimum wavelength for the study by determining the absorption spectrum of triiodide. Alternatively, the instructor can run the this spectrum on a recording spectrophotometer or provide the optimum wavelength. However, if is done, the absorption of a solution of triiodide of known concentration should still be determine at the absorption maximum so that the results do not depend on a literature value of the extinction coefficient. Because round tubes will probably be used and other variables are difficult to exactly duplicate, extinction coefficients should generally be determined with the equipment and conditions at hand or an error will be introduced. A solution of triiodide with an easily measured absorption throughout the range of interest (340 nm- 420 nm) needs to be prepared. Because the solution will be prepared by mixing solutions ofiodine and iodide, a very high iodide concentration will be prepared to make sure the equilibrium strongly favors triiodide. In addition to providing sufficient triiodide, this technique makes the iodine concentration very low. As a result, the absorption of iodine in the region of interest is kept to an insignificant level. Into a dry 125 mL Erlenmeyer, pipet 5.00 mL of a stock solution of 1.00x10-4 M aqueous iodine and 10.00 mL ofa 0.5 M KI solution. Review the instructions for your spectrometer and determine the absorption spectrum of the solution every 10 nm from 340 nm to 420 nm using water as a blank. Plot the data and determine the wavelength and extinction coefficient at the wavelength of maximum absorption. Set the spectrometer on that wavelength

Answer 1

5 mL of 1.00 * 10^-4 M of aqueous I₂ is taken. Moles of I₂ present in the solution = Volume in Liter * Molarity = 0.005 L * 1.00 * 10^-4 M = 5.0 * 10^-7 moles

10 mL of 0.5 M KI solution is taken. Moles of KI = Volume in Liters * Molarity = 0.01 L * 0.5 M = 0.005 moles

The balanced reaction is:

KI + I₂ = KI₃

So, here the mole ratio of I₂ : KI = 1:1

Here moles of I₂= 5.0 * 10^-7 moles, moles of KI = 0.005 moles.

So, the limiting reagent is I₂.

5.0 * 10^-7 moles of I₂ produces 5.0 * 10^-7 moles of KI₃

Volume of resulting soution = 5 + 10 mL = 15 mL = 0.015 L

So, concentration of triiodide = concentration of KI3 ( potassium triiodide) = 5.0 * 10^-7 moles/ 0.015 L= 3.33 * 10^-5 M

Calculation of molar absorbptivity coefficient:

A= E * b * c

1.01 = E * 1.3 cm * 3.33 * 10^-5 M

E = 23308 M^-1cm^-1