What is the solubility in mol/L of silver iodide, "AgI" ?

Its ${K}_{s p}$ value is $8.3 \cdot {10}^{-} 17$.

Feb 17, 2017

$9.11 \cdot {10}^{- 9} \text{M}$

Explanation:

Silver iodide, $\text{AgI}$, is an insoluble ionic compound, which basically means that when this compound is added to water, an equilibrium is established between the undissolved solid and the dissolved ions.

${\text{AgI"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "I}}_{\left(a q\right)}^{-}$

Notice that every mole of silver iodide that dissolves produces $1$ mole of silver cations and $1$ mole of iodie anions.

The expression for the solubility product constant, ${K}_{s p}$, of silver iodide looks like this

${K}_{s p} = \left[{\text{Ag"^(+)] * ["I}}^{-}\right]$

If you take $s$ to be the molar solubility of silver iodide, you can say that you will have

${K}_{s p} = s \cdot s = {s}^{2}$

This means that the molar solubility of this salt will be equal to

$s = \sqrt{{K}_{s p}} = \sqrt{8.3 \cdot {10}^{- 17}} = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{9.11 \cdot {10}^{- 9} \text{M}}}}$

In other words, you can only hope to dissolve $9.11 \cdot {10}^{- 9}$ moles of silver iodide for every liter of water at room temperature.