Solving Solubility Problem: I_2\text{(s)} in KI(aq)

[broegger]

Hi. I'm preparing for a chemistry exam and I have trouble with this former exam problem:

The solubility of $$I_2\text{(s)}$$ in water at 25ºC is 0.0013 M.

Calculate the solubility of $$I_2\text{(s)}$$ in a 0.1 M solution of KI(aq) by considering this equilibrium:

$$I_2\text{(aq)}+I^-\text{(aq)} \leftrightharpoons I_3^-\text{(aq)}, \quad K = 700 M^{-1}$$​

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I assume that I2(s) just dissolves like I2(s) <-> I2(aq). I have found the solubility to be 0.0489 M, but I lack a systematic way of solving this. Any hints?

 

[PPonte]

$$I_2 (s) \leftrightharpoons I_2 (aq)$$

That's the solubility reaction of $I_2$.

Now, consider that you add a a 0.1 M solution of $KI$. One of the ions of this salt, $I^-$, reacts with $I_2 (aq)$. Acoording to Le Châtelier's Principle what will happen to the equilibrium above since the concentration of $I_2 (aq)$ got lower.

 

[broegger]

It will shift to the right, so the solubility will go up (consistent with my result.) I should compute the I3- concentration and add it to the I2(aq) concentration to find the new solubility, right?

 

[PPonte]

I assume that I2(s) just dissolves like I2(s) <-> I2(aq). I have found the solubility to be 0.0489 M, but I lack a systematic way of solving this. Any hints?

How did you found that solubility?

 

[broegger]

Hmm, not sure. But I think I know how to do it now. Thanks.