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Equlibrium in saturated I2 solution find Keq help please

  1. Sep 11, 2015 #1
    1. The problem statement, all variables and given/known data
    A saturated solution of iodine in water contains 0.330g of I2/L. More than this can dissolve in a KI solution because of the following equilibrium

    I2(aq) + I- (aq) <=> I3-
    A 0.1M KI solution actually dissolves 12.5g of iodine/L, mmost of which is converted to I3-. Assuming that the concentration of I2 in all saturated solutions is same, calculate the equilibrium constant for the above reaction.



    This ICE table isn't like most I know. I know to convert to moles for I2 in water and also I2 in KI. But where do I put them it doesn't make sense to me after that point. I also know that the 12.5g will give the molarity of the I3- also and that I need to subtract that from the free I2 conc. I am still fuzzy on why please help. This is a tough problem.



    2. Relevant equations
    Keq= [A]/[C][D]
    Moles= Mass/Molar mass



    3. The attempt at a solution
    0.33 g I2* mol I2/254g= 1.3*10-3mol (this is the amount of I2 saturated in water.)

    12.5g I2 * mol I2/254g = 0.0492 mol (this is the amount of I2 saturated in I- solution)

    0.0492 is also the amount of I3-(aq) initially as problem states I2 all converts.

    0.1 is the amount of I- (or KI).

    I know that when I3- dissociates it"loses" the 1.3 *10-3 (at equilibrium we have 0.0479 I3-) but what Im confused about is why do we care if this is strictly the amount in water. I know all of this is occuring in water but is there an implication here that both water and I- act to be saturated? I'm feel like im missing something here.

    I also know that the equilibrium of I3-is the "change" amount of I-(aq). I know this sounds funny but why is that? I have a ICE table but it confuses me. Or do we need it at all? We are left with 0.0521M

    Thanks and again like my other post sorry if im being tedious. First time poster here! grin.gif


    So to summarize

    Initial Amounts
    I- 0.1M
    I3- 0.0492M

    Change Amounts
    I2 0.0013M
    I- 0.0479M
    I-3 -0.0013M

    Equilibrium amounts
    I2 .0013M
    I- 0.0521M
    I3- 0.0479M
     
  2. jcsd
  3. Sep 11, 2015 #2

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    Can you write Keq in terms of concentrations of specific chemical species?
     
  4. Sep 11, 2015 #3
    I wanted to start over because Im confused about how we get some and use these numbers. I know that KI is 0.1M and that I2 is 0.0013M But to get I3- we subtract the two? I don't really know why.
     
  5. Sep 12, 2015 #4

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    It is implicit (no guarantees) in the problem statement that there is excess solid I2 maintaining the saturation of the solution, and that the properties of that solution are not changed significantly enough by the addition of KI to affect the solubility of I2 in the resulting solution.

    Get you started now?
     
  6. Sep 12, 2015 #5

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    Saturated I2 in water, with xs solid I2; plus KI (very water soluble) in the same solution as a second solute. I2 in solution reacts with I- to form I3- as governed by the equilibrium constant, and dissolves whatever xs solid I2 is necessary to maintain the I2 concentration at 0.0013M.
     
  7. Sep 12, 2015 #6
    ok thought so. So the full amount of I3 is offset by 0.0013 moles of I2? (we're in the wrong thread :D ) I.e. subtract 0.0013 from the amount of I3- . THAT equil amount is subtracted from 0.1M of I-(im confused as to why though.)
     
  8. Sep 12, 2015 #7

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    The quantity of I- ion available is fixed by the KI concentration specified, 0.1 M (?), and the I3- formation reaction is reducing it until equilibrium is reached.
     
  9. Sep 12, 2015 #8
    So the thing being saturated I- is the thing that's also in excess. It's being used up by the amount of I3- being formed.
     
  10. Sep 12, 2015 #9

    epenguin

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    Answering question of #2 would start way out of confusion.

    It doesn't.

    It states
    A saturated solution of iodine in water contains 0.330g of I2/L.

    It also gives you this indication: Assuming that the concentration of I2 in all saturated solutions is same,

    This is a tough problem.
    Not when you use the information you are given.
     
    Last edited: Sep 12, 2015
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