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F-1
EXPERIMENT F: DETERMINATION OF THE
IODINE-TRIIODIDE EQUILIBRIUM CONSTANT
OBJECTIVES
The numerical value for the equilibrium constant will be determined for the reaction:
I
2(aq)
+ I
-
(aq)
I3
-
(aq)
(1)
In the process you will be introduced to the concept of a distribution equilibrium as demonstrated by
the partitioning of iodine (I2) between two immiscible liquid phases: water and dichloromethane (CH2Cl2).
You will also be introduced to the analytical procedure of titration, which involves the quantitative
determination of the amount of an unknown substance in a solution. This procedure will be used to
determine the equilibrium concentrations of iodine and the triiodide ions.
THEORETICAL CONSIDERATIONS
Iodine is only slightly soluble in water at 25
C. Its solubility can be increased by the adding iodide (I
-
)
ion which reacts with the iodine molecule (I2) to form the triiodide ion (I3
-
(aq)
), thereby pulling more I2 into
solution. If the concentration of either of the reactants (I
2(aq)
or I
-
(aq)
) in equation (1) is increased,
Le Châtelier's Principle predicts that the position of the equilibrium is shifted to the right to form more I3
-
(aq)
. Conversely, removal of iodine by some means causes a shift of the equilibrium position to the left by
dissociation of I3
-
(aq)
ions. The equilibrium expression for reaction (1) is:
K
eq
=
[I3
-
(aq)
]
[I2
(aq)
][I
-
(aq)
]
(2)
In this experiment you will determine the numerical value for this equilibrium constant. To accomplish this
you will need to know [I
2(aq)
], [I
(aq)
-
], and [I
3(aq)
-
] at equilibrium.
However, there is no direct chemical method for distinguishing between iodine molecules
(I2 ) and triiodide (I3
-
) ions in aqueous solution. Thus, if you add 1 mole of iodine to one liter of aqueous
potassium iodide, you cannot say that the iodine concentration is 1.0 M, because most of it will convert to
the triiodide ion (I3
-
).
