[Moose100]

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)  I^3- (aq)   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.

 I₂(aq) +I^-(aq)  I^3- (aq)

 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 I^3- also and that I need to subtract that from the free I2 conc. I am still fuzzy on why please *delete me* This is a tough problem.

[Borek]

What are the initial concentrations of I₂, I^- and I₃^-?

What is the equilibrium concentration of I₂? If so, how much I₂ reacted? With what?

If so, what is the equilibrium concentration of I₃^-? I^-?

[Moose100]

Ok here's what I have.


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!


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

[Moose100]

Ok here's what I have.


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!


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

And again I think my issue is setting up the ICE table and how to know to use which numbers this isnt a typical problem.

[Borek]

Assuming the reaction is

I₂(aq) +I^-(aq)  I^3- (aq)

initial concentration of I₃^- is 0. Think about it as if we started we a solution of KI to which we add enough I₂ to make the solution saturated.

Please remember ICE table is just a way of organizing data around the reaction stoichiometry. Sometimes it can't be applied directly, but the stoichiometry of the reaction (and the way amounts of substances are linked to each other by the stoichiometry) still holds.

I asked my questions is a very specific order for a reason - that's how you can follow the stoichiometry.

[Moose100]

I understand that part about I₃^-. Also the I₂ concentration at equlibrium has to be 0.0013? This is also the bit that doesn't saturate and is in water instead?


If so that makes the difference between the two(0.0492) the change concentration of I₃^-?

[Borek]

Also the I₂ concentration at equlibrium has to be 0.0013?

Yes, that's what you are told.

This is also the bit that doesn't saturate and is in water instead?

Not sure what you mean by that. 0.0013 M is the equilibrium concentration of free I₂ in the 0.1M solution of I^- saturated with iodine.

If so that makes the difference between the two(0.0492) the change concentration of I₃^-?

0.0492 M is not the difference between anything, if anything it is a sum. It is the total amount of iodine in both forms (I₂ and I₃^-).

[Moose100]

Look Im not trying to frustrate you or anything. I asked that because I subtracted two numbers to get that.

[Borek]

Look Im not trying to frustrate you or anything. I asked that because I subtracted two numbers to get that.

I am far from being frustrated

0.0492 is not a result of subtraction - that's what you got just from known amount of iodine (total) present in the solution. Solution contains 12.5 g of iodine in two forms - I₂ and I₃^-. That was given in the problem, and from this number you calculated (total) concentration of both forms to be 0.0492 M.

[Moose100]

Hmm ok. I thought you were. Well let me get my notes out I am at work.

[Moose100]

Oops I meant 0.0479 is.

0.0492-0.0013=0.0497.

Like why do this? Is it because some of the I2 complexes with water?

[Borek]

Oops I meant 0.0479 is.

0.0492-0.0013=0.0497.

Like why do this? Is it because some of the I2 complexes with water?

This is just following the stoichiometry. You know the solution contains 0.0492 M of iodine (in total). Of that 0.0013 M is in the form of free iodine. That's like saying you took a 1 L of 0.1 M KI solution, added 12.5 g and left it till it got to equilibrium. At equilibrium you found the free I₂ concentration to be 0.0013 M. What happened to the rest of I₂? And how much was "the rest" of iodine, if the initial amount was 0.0492 M and what is left is 0.0013 M?

Take a look at the reaction equation -

[Moose100]

So the rest is why we subtract 0.0492 from 0.0013M to get 0.0479M. SInce we know the amount of I3- at equilibrium it has to be that I- depletes by that amount.  We get 0.0492 for I3- because we know that the I2 saturates to form I3-. 

Thats what I meant about the iodine complexing in water that is was FREE Iodine. I said it wrong probably due to confusion. So that makes sense.   I wish I could find a trove of problems like this to practice. The idea that we put MORE of something in is a little different.

[Moose100]

Also I think I probalby shouldnt have put for I3- 0.0493(initial) and -0.0013M(change). THese dont go there the difference in these two concentrations are the EQUILIBRIUM Molarity. Like I don't "start" with 0.0492 of I3-. 0.0479 is produced based on what is left over. Which is the same amount that I- is used up.

Right?

[Borek]

Also I think I probalby shouldnt have put for I3- 0.0493(initial)

That was the very first thing I told you (in the post #5) commenting on your approach. I have not commented on every other things that was wrong, as they were just consequences of this mistake.