Equlibrium in saturated I2 solution find Keq help please

[Moose100]

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 I^3- 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.
[/B]

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!

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[/B]

 

[Bystander]

calculate the equilibrium constant for the above reaction.

Keq= [A]/[C][D]

Can you write K_eq in terms of concentrations of specific chemical species?

 

[Moose100]

I wanted to start over because I am 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.

 

[Bystander]

also "know" that I2 is 0.0013M but why use this at all?

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?

 

[Bystander]

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.

 

[Moose100]

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.)

 

[Bystander]

THAT equil amount is subtracted from 0.1M of I-(im confused as to why though.)

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.

 

[Moose100]

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.

 

[epenguin]

I wanted to start over because I am confused ... I don't really know why.

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

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

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.