Question

Solubility of I₂(s) in water at 25°C K(I₂) = 0.0013, and iodine can with iodine ions create tri iodine ions:
I₂(aq)+I^-(aq) = I₃^-(aq) K = 700

Calculate the solubility of I₂(s) in a KI(aq) solution with a concentration of 0.10 M.

Answer 1

As the reaction, I₂ (aq) + I^- (aq) <-----------> I₃^-

K = [ I₃^-] / [I₂][ I^-] = 700 ------------------------- (eqn 1)

Given, I^- = 0.10 M

I₂ = 0.0013 M

ICE table for above equilibrium can be prepared for reaction of x moles of [I₂] as:

	[I2]	[I-]	[I3-]
Initial (M)	0.0013	0.10	0
Change (M)	-x	-x	+x
Equilibrium (M)	0.0013-x	0.10-x	+x

Now, substituting the equilibrium concentrations in eqn 1,

x/[(0.0013-x)(0.10-x)] = 700

x = 700 (0.0013-x)(0.10-x)

x = 0.091 - 70x -0.91 x + 700x²

700x² -71.91x + 0.091 = 0

On solving the quadratic equation for x,

x = 0.00128 M

So, at equilibrium, concentration of I₂ = 0.0013-0.00128 = 0.00002 M

As the concentration of I₂ is decreased, it means it gets converted to I₃^-, so, its solubility increases.