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Moose100
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Homework Statement
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.
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Homework Equations
Keq= [A]/[C][D]
Moles= Mass/Molar mass
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 I am confused about is why do we care if this is strictly the amount in water. I know all of this is occurring in water but is there an implication here that both water and I- act to be saturated? I'm feel like I am 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 I am being tedious. First time poster here!
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[/B]