Equlibrium in saturated I2 solution find Keq help please

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Discussion Overview

The discussion revolves around calculating the equilibrium constant (Keq) for the reaction between iodine (I2) and iodide ions (I-) to form triiodide ions (I3-) in a saturated iodine solution. Participants explore the implications of concentrations in both water and KI solutions, addressing the complexities of the ICE table and the relationships between the species involved.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents the equilibrium reaction I2(aq) + I-(aq) <=> I3- and provides initial concentrations based on the problem statement.
  • Another participant questions how to derive the concentration of I3- from the given amounts of I2 and I- and expresses confusion about the use of these values.
  • Some participants discuss the assumption that the concentration of I2 remains constant due to the presence of excess solid I2 in the solution.
  • There is a suggestion that the equilibrium concentration of I3- should account for the initial concentration of I2, leading to a subtraction from the I- concentration.
  • One participant emphasizes that the concentration of I- is fixed by the KI concentration and that the formation of I3- reduces the available I- until equilibrium is reached.
  • Another participant expresses uncertainty about the implications of the saturation of I2 and how it relates to the overall equilibrium.
  • There is a clarification that the problem does not explicitly state that all I2 converts to I3-, which adds to the confusion regarding the calculations.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the calculations and the relationships between the species involved. There is no consensus on the correct approach to calculating Keq, and multiple viewpoints on how to interpret the problem persist.

Contextual Notes

Participants highlight the complexity of the ICE table in this context and the potential implications of assuming constant concentrations in saturated solutions. There is also uncertainty regarding the initial conditions and how they affect the equilibrium calculations.

Who May Find This Useful

Students working on equilibrium problems in chemistry, particularly those involving saturated solutions and complex ion formation, may find this discussion relevant.

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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.
[/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!
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[/B]
 
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Moose100 said:
calculate the equilibrium constant for the above reaction.

Moose100 said:
Keq= [A]/[C][D]
Can you write Keq in terms of concentrations of specific chemical species?
 
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.
 
Moose100 said:
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?
 
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.
 
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.)
 
Moose100 said:
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.
 
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.
 
Moose100 said:
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.

Moose100 said:
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.
 
Last edited:

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