91Ó°ÊÓ

In polar liquids, the ionic material dissociates into its ions in ionic equilibrium. The ions generated in the solution are always in equilibrium with the undissociated solute.

Consider the given information,

$\text{c}\left({\text{NH}}_{\text{3}}\right)\text{= 2.5M}$

We can start by writing the dissolution equation for $\mathrm{Agl}$:

$$\begin{gathered} \text{AgI(s)}⇌{\text{Ag}}^{\text{+}}{\text{(aq) +I}}^{\text{-}}\text{(aq)}        \text{(1)} \\ \text{Ksp =}\left[{\text{Ag}}^{\text{+}}\right]\left[{\text{I}}^{\text{-}}\right]{\text{= 8.3×10}}^{\text{- 17}}       \left(2\right) \end{gathered}$$

We can now write the forming equation $\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}^{\text{+}}$:

$$\begin{gathered} {\text{Ag}}^{\text{+}}{\text{(aq) + 2NH}}_{\text{3}}\text{(aq)}⇌\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}^{\text{+}}\left(\text{aq}\right)     \left(3\right) \\ {\text{K}}_{\text{f}}\text{=}\frac{{\displaystyle \left[\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}^{\text{+}}\right]}}{{\displaystyle \left[{\text{Ag}}^{\text{+}}\right]\text{×}{\left[{\text{NH}}_{\text{3}}\right]}^{\text{2}}}}{\text{=1.7×10}}^{\text{7}}                           \left(4\right) \end{gathered}$$

When we add the equations 1 and 3 together, we get the equation for the solubility of Agl in ${\text{NH}}_{\text{3}}$

${\text{AgI(s) + 2NH}}_{\text{3}}\text{(aq)}⇌\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}^{\text{+}}{\text{(aq) +I}}^{\text{-}}\text{(aq)}         \text{(5)}$

$\text{K =}\frac{{\displaystyle \left[\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}^{\text{+}}\right]\text{×}\left[{\text{I}}^{\text{-}}\right]}}{{\displaystyle {\left[{\text{NH}}_{\text{3}}\right]}^{\text{2}}}}            \left(6\right)$

Since, equation 5 is the sum of equation 1 and 3, the equilibrium constant would be the product of the equilibrium constants of equation 1 and 3.

${\text{K =K}}_{\text{sp}}{\text{×°­}}_{\text{f}}             \left(7\right)$

Preparing the ICE table :

Substituting the values in equation (6):

$$\begin{gathered} \text{K =}\frac{{\displaystyle \left[\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}^{\text{+}}\right]\text{×}\left[{\text{I}}^{\text{-}}\right]}}{{\displaystyle {\left[{\text{NH}}_{\text{3}}\right]}^{\text{2}}}} \\ {\text{K}}_{\text{sp}}{\text{K}}_{\text{f}}=\frac{{\displaystyle \left[\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}^{\text{+}}\right]\text{×}\left[{\text{I}}^{\text{-}}\right]}}{{\displaystyle {\left[{\text{NH}}_{\text{3}}\right]}^{\text{2}}}} \\ \left(1.7×{\text{10}}^{\text{7}}\right)×\left(8.3×{\text{10}}^{\text{-17}}\right)=\frac{\mathrm{x}×\mathrm{x}}{{\displaystyle \text{(2.5 - 2x)}}} \\ \mathrm{x}=5.9×{\text{10}}^{\text{- 5}}\text{M} \end{gathered}$$

Therefore, the solubility of Agl is$5.9×{10}^{-5}\mathrm{M}$