(3) A. Equilibrium Position -- A + B = C -- Effect of Dilution Iodine is sparing

ID: 570651 • Letter: #

Question

(3) A. Equilibrium Position -- A + B = C -- Effect of Dilution

Iodine is sparingly soluble in pure water. However, it does `dissolve' in solutions containing excess iodide ion because of the following reaction:

For each of the following cases calculate the equilibrium ratio of [I3-] to [I2].

a) 8.00×10-2 mol of I2 is added to 1.00 L of 8.00×10-1 M KI solution.
5.11×102(this is correct)

b) The solution above is diluted to 14.50 L.

B. Equilibrium Constant Relationships

Given the following equilibrium constants at 427°C:

What would be the value of the equilibrium constant for each of the following reactions, at 427°C?

I-(aq) + I2(aq) I3-(aq) K = 710.

If you dilute the solution to 14.5 Liters, the concentrations will change

New I2 concentration will be:

0.08 / 14.5 = 0.00552 M

For the KI you have a concentration of 0.8 mol in 1 L

new concentration will be = 0.8 / 14.5 = 0.0552 M

ICE chart will be

I- (aq) + I2 (aq) = I3- (aq) ,

I 0.0552 0.00552 0

C -X -X X

C 0.0552-X 0.00552-X X

K = x / ([0.0552-x]*[0.00552-x])

710 * [0.0552-x]*[0.00552-x] = x

If you solve this equation (i am using a spreadsheet) you will get a value of

x = 0.005368 M

so I3 / I2 will be

0.005368 / ( 0.00552- 0.005368)

0.005368 / (0.000151) = 35.38

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