How Is the Equilibrium Constant Calculated for I2 and I- in KI Solution?

AI Thread Summary
The discussion focuses on calculating the equilibrium constant for the reaction between iodine (I2) and iodide ions (I-) in a potassium iodide (KI) solution. A saturated iodine solution contains 0.330g of I2/L, while a 0.1M KI solution can dissolve 12.5g of iodine/L, primarily converting it to I3-. Participants clarify that the total concentration of iodine in the KI solution includes both free I2 and I3-, and they emphasize the importance of understanding the stoichiometry involved. The equilibrium constant expression is established as K=[I3-]/[I2][I-], leading to a clearer understanding of the concentrations at equilibrium. The conversation concludes with a participant expressing gratitude for the clarification on the problem.
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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- (aq) where 2 and 3 are subscripts and - is negative charge

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.

Homework Equations


The Attempt at a Solution



0.33g/L is 0.0013 mole and 12.5g/L is 0.0492 mole
I don't know how to proceed
 
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Start writing expression for equilibrium constant. Then think which concentrations you are given, and which you can calculate from the simple stoichiometry.
 
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K=[I3-]/[I2][I-]

Actually, I did not understand the problem. A saturated solution of Iodine in water is given to us.
What is the meaning of 'A 0.1M KI solution actually dissolves 12.5g of iodine/L'
how does this info help us?
 
Why does the iodides solution contain more iodine than solution in pure water?
 
It is so because KI is added to it.
0.1M KI dissolves 0.0492 mole I2
but we have only 0.0013 mole I2 so all of it dissolves and converts into I3-
moles of KI left = 0.1-0.0013
Is this approach correct?
 
No, there was excess iodine present. It was not dissolved, but it was added as a solid, thus there is much more I3- present.

How much? That's where the mass of dissolved iodine comes handy.
 
excess iodine present in KI solution or saturated solution of Iodine in water?
 
No idea what you are asking about. SOLID iodine. Solutions are in contact with the solid. Concentrations given are for the SOLUTIONS.
 
I would be thankful to you if you write the concentrations at equilibrium as this question is very vague to me.
 
  • #10
Concentration of free iodine is identical with the concentration in pure water. Total concentration of iodine dissolved is sum of free iodine and I3-. Simple subtraction will give you concentration of I3-, then simple stoichiometry will tell you how much I- was left.
 
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  • #11
Thanks alot. I understood it clearly now.
 
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