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Question

A saturated solution of I2 in water contains 0.33 g litre1 of I2 more than this can be dissolved in a KI solution because of the following equilibrium:
I2+II3. A 0.1 M KI solution (0.1M I) actually dissolves 12.5g/litre I2, most of which is converted to Assuming that the concentration of I2 in all saturated solution is the same, calculate the equilibrium constant (Kc) for the above reaction. What is the effect of adding water to clear saturated solution of I2 in KI solution.

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Solution

I2+II3
Keq=[I3][I2][I]
Saturated solution contains I2 at concentration of 0.330g/L0.330gL(1254)=1.3×103molL
Now, in this solution of KI tyhe concentration of I2 is 12.5g/L or 1.25gI2/L×(1254)=0.049molI2L
We have 0.0013M as Free I2
[I3]=0.0477M
KI solution originally had [I]=101M but 0.477×101M of it is used up to form I[I]0.04770.0523M
Keq=[I3][I2][I]=0.0477(0.0013)(0.0523)=701

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