A saturated solution of iodine in water contains 0.254 g of in 1 L. More than this can dissolve in a — Ionic Equilibrium Chemistry Question
Question
A saturated solution of iodine in water contains 0.254 g of $\text{I}_2$ in 1 L. More than this can dissolve in a KI solution because of the following equilibrium:<br><br>$\text{I}_2\text{(aq)} + \text{I}^-\text{(aq)} \rightleftharpoons \text{I}_3^-\text{(aq)}$<br><br>A 0.1 M – KI solution actually dissolved 12.7 g of iodine per litre, most of which is converted to $\text{I}_3^-$. Assuming that the concentration of $\text{I}_2$ in all saturated solutions is the same, calculate the equilibrium constant for the above reaction. (Take: $9.6 \times 5.1 = 49$, Atomic mass of iodine = 127)
💡 Solution & Explanation
Molar mass of $\text{I}_2 = 254\text{ g/mol}$. Free saturated $[\text{I}_2] = 0.254 / 254 = 10^{-3}\text{ M}$. In the 0.1 M KI solution, total dissolved $\text{I}_2 = 12.7 / 254 = 0.05\text{ M}$. The complexed iodine $[\text{I}_3^-] = \text{Total} - \text{Free} = 0.05 - 0.001 = 0.049\text{ M}$. The initial KI concentration is $0.1\text{ M}$. The remaining free $[\text{I}^-] = 0.1 - 0.049 = 0.051\text{ M}$. The equilibrium constant $K = \frac{[\text{I}_3^-]}{[\text{I}_2][\text{I}^-]} = \frac{0.049}{(10^{-3})(0.051)} = \frac{49}{0.051} \approx 960.78$. Rounding to the nearest specified format integer gives 0960.