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Derivation of Kp and Kc

A saturated s...

Question

A saturated solution of iodine in water contains $0.33$ g iodine/L. More than this can dissolve in a $K I$ solution because of the following equilibrium: $I_{2} (a q) + I^{-} \to I_{3} - (a q)$ . A $0.1 M$ $K I$ solution $(0.1 M$ $I^{-})$ actually dissolves $12.5$ g of Iodine/L, most of which is converted to $I_{3}^{-} .$ Assuming that the concentration of $I_{2}$ in all saturated solutions is the same, calculate the equilibrium constant for the above reaction: (Mol. wt. of $I$ = $258 g m o l^{- 1}$ )

A

709.3

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B

465.6

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C

287.4

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D

687.4

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Solution

The correct option is A $709.3$
The molar mass of Iodine is $253.8$ g/mol.
$0.33$ g of Iodine corresponds to $0.0013$ moles. Thus, $12.5$ g of Iodine corresponds to $0.04925$ moles.
Thus, when $0.04925$ moles of Iodine are added, $0.0013$ moles will remain at equilibrium and the remaining ( $0.04795$ moles) will react with equal number of moles of KI to form $0.04795$ moles of iodate ion.
The number of moles of iodide ions remaining at equilibrium will be $0.10 - 0.04795 = 0.0520$ mol.
The value for the equilibrium constant will be $\frac{0.04795}{0.0013 \times 0.0520} = 709.3$ .

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