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Equilibrium Position - Effect of Dilution

  1. Nov 9, 2007 #1
    Iodine is sparingly soluble in pure water. However, it does `dissolve' in solutions containing excess iodide ion because of the following reaction:

    I-(aq) + I2(aq) I3-(aq) K = 710 L/mol
    For each of the following cases calculate the equilbrium ratio of [I3-] to [I2]:

    2.00×10-2 mol of I2 is added to 1.00 L of 2.00×10-1 M KI solution.

    I did this question using the ICE table and got the RIGHT answer of


    but now the SECOND part asked.. same thing but

    The solution above is diluted to 5.50 L.

    How do i approach this?
  2. jcsd
  3. Nov 9, 2007 #2
    nevermind i got it :)
  4. Nov 9, 2007 #3
    did you re-use the value you found?
  5. Nov 9, 2007 #4
    not really, i just made did the following

    (2.00×10-1 M)(1.00 L) = C2(5.50L)

    solved for c2 and then

    2.00×10-2 mol of I2 is added to 5.50 L of C2 KI solution.

    then again did ICE table and solved at equilibrium.. then did the same thing as part a to find ratio of I3 to I2 at equilibrium
  6. Nov 10, 2007 #5
    Same question

    I have the same question but with different numbers
    can you show me how to solve it

    8.00×10-2 mol of I2 is added to 1.00 L of 8.00×10-1 M KI solution.

    The solution above is diluted to 13.00 L.
  7. Nov 10, 2007 #6
    [tex]M_{1} \times V_{1} = M_{2} \times V_{2}[/tex]

    solve for final Molarity
  8. Nov 10, 2007 #7
    yea just solve for molarity first

    (8.00x10^-1)(1.00L) = (c2)(13.00)

    solve for c2 now ur new question is

    8.00×10-2 mol of I2 is added to 13.00 L of c2 M KI solution.

    make Ice table now so u have

    --------------------- = k(watever k value ur given)
    ((0.08/13)-x)((c2) - x)

    solve for ur x using quadratic formula or if ur lazy find one online and just type in a,b and c values to find x, then plug back in to get

    -- = --------------
    Last edited: Nov 10, 2007
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