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Solubility problem

  1. Jul 28, 2006 #1
    Hi. I'm preparing for a chemistry exam and I have trouble with this former exam problem:

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

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

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

    ---

    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?
     
  2. jcsd
  3. Jul 28, 2006 #2
    [tex]I_2 (s) \leftrightharpoons I_2 (aq)[/tex]

    That's the solubility reaction of [itex]I_2[/itex].

    Now, consider that you add a a 0.1 M solution of [itex]KI[/itex]. One of the ions of this salt, [itex]I^-[/itex], reacts with [itex]I_2 (aq)[/itex]. Acoording to Le Châtelier's Principle what will happen to the equilibrium above since the concentration of [itex]I_2 (aq)[/itex] got lower.
     
  4. Jul 28, 2006 #3
    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?
     
  5. Jul 28, 2006 #4
    How did you found that solubility?
     
    Last edited by a moderator: Jul 28, 2006
  6. Jul 30, 2006 #5
    Hmm, not sure. But I think I know how to do it now. Thanks.
     
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