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Standard XII

Chemistry

Common Ion Effect

A saturated s...

Question

A saturated solution of $I_{2}$ in water contains $0.33 g l i t r e^{- 1}$ of $I_{2}$ more than this can be dissolved in a $K I$ solution because of the following equilibrium:
$I_{2} + I^{-} ⇌ I_{3}^{-}$ . A $0.1 M K I$ solution $(0.1 M I^{-})$ actually dissolves $12.5 g / l i t r e I_{2}$ , most of which is converted to Assuming that the concentration of $I_{2}$ in all saturated solution is the same, calculate the equilibrium constant $(K_{c})$ for the above reaction. What is the effect of adding water to clear saturated solution of $I_{2}$ in $K I$ solution.

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Solution

$I_{2} + I^{-} ⇋ I_{3}^{-}$
$K_{e q} = \frac{[I_{3}]}{[I_{2}] [I^{-}]}$
Saturated solution contains $I_{2}$ at concentration of $0.330 g / L ⟶ 0.330 \frac{g}{L} (\frac{1}{254}) = 1.3 \times 10^{- 3} \frac{m o l}{L}$
Now, in this solution of $K I$ tyhe concentration of $I_{2}$ is $12.5 g / L$ or $1.25 g I_{2} / L \times (\frac{1}{254}) = 0.049 \frac{m o l I_{2}}{L}$
We have $0.0013 M$ as Free $I_{2}$
$[I_{3}^{-}] = 0.0477 M$
$K I$ solution originally had $[I^{-}] = 10^{- 1} M$ but $0.477 \times 10^{- 1} M$ of it is used up to form $I^{-} ⟹ [I^{-}] \equiv 0.0477 \equiv 0.0523 M$
$K_{e q} = \frac{[I_{3}^{-}]}{[I_{2}] [I^{-}]} = \frac{0.0477}{(0.0013) (0.0523)} = 701$

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