To determine the equilibrium constant for the reaction I_2 + 2SCN I(SCN)^-_2 + I

Question

To determine the equilibrium constant for the reaction I_2 + 2SCN I(SCN)^-_2 + I^- 25.0 mL of a 0.0100 M aqueous solution of I_2 were extracted with 10.0 mL of CHCIl_3. After extraction, spectrophotometric measurements revealed that the I_2 concentration of the aqueous layer was 1.12 times 10^-4 M. An aqueous solution that was 0.0100 M in I_2 and 0.100 M KSCN was then prepared. After extraction of 25.0 mL of this solution with 10.0 mL of CHCl_3, the concentration of I_2 in the CHCl_3 layer was found from spectrophotometric measurement to be 1.02 times 10^-3 M. What is the partition coefficient for I_2 between CHCl_3 and water? What is the formation constant for I(SCN)_2?

Explanation / Answer

a. partition coeffcient for I2 between CHCl3 and water = K

The non–polar iodine is much more soluble in the non–polar organic solvent than in the highly polar water solvent.

K = concentration of I2 in water/concentration of I2 in CHCl3

k= (1.12 x 10-4) / (1.02 x 10-3)

k=0.109

b. formation constant of I(SCN)2 =

= {[I(SCN)2] [I]}/[I2] [SCN-]2

= [(1.02 X 10-3M) x (1.12 X 10-4M)] / [(0.0100M) x (0.100M)2]

=1.142 X 10 -3

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