# How do we formulate "iodide ion" in aqueous solution?

Apr 20, 2017

Iodide anion is treated as ${I}^{-}$ in aqueous solution..........

#### Explanation:

Elemental iodine is binuclear, i.e. we treat it as ${I}_{2}$, but this species can be reduced to give water-soluble ions, ${I}^{-}$........ This species can react with (some) heavy metal ions to give insoluble precipitates.

And thus we interrogate the equilibrium:

$A g I \left(s\right) r i g h t \le f t h a r p \infty n s A {g}^{+} + {I}^{-}$

This is an equilibrium reaction, and typically we write the $\text{solubility product}$, where ${K}_{s p} = \left[A {g}^{+}\right] \left[{I}^{-}\right] = 8.1 \times {10}^{-} 17$ under standard conditions..........

And so, let us call the solubility of $A g I$, a yellow salt, $S$, and thus............

${K}_{\text{sp}} = {S}^{2}$, and thus we can make a quantitative measurement of solubility, $S = \sqrt{{K}_{\text{sp}}} = \sqrt{8.1 \times {10}^{-} 17} = 9 \times {10}^{-} 9 \cdot m o l \cdot {L}^{-} 1$.

So, what do you have to learn? Iodide, like other halides, exists in aqueous solution as the ion, ${I}^{-}$, $C {l}^{-}$, $B {r}^{-}$, ${F}^{-}$ etc. The addition of solid iodine to solutions of iodide CAN produce triiodide species such as ${I}_{3}^{-}$?