Find the solubility of $\mathbf{Agl}$in ${\mathbf{\text{2.5 M NH}}}_{\mathbf{\text{3}}}$[${\mathbf{\text{K}}}_{\mathbf{\text{sp}}}$of $\mathbf{\text{Agl}}\mathbf{ }\mathbf{\text{=}}\mathbf{ }{\mathbf{\text{8.3×10}}}^{\mathbf{\text{- 17}}}$; ${\mathbf{\text{K}}}_{\mathbf{\text{f}}}$of${\mathbf{\text{Ag(NH}}}_{\mathbf{\text{3}}}{{\mathbf{\text{)}}}_{\mathbf{\text{2}}}}^{\mathbf{\text{+}}}{\mathbf{\text{= 1.7×10}}}^{\mathbf{\text{7}}}$].

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 $\mathrm{Agl}$:

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

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

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

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

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

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

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.

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

Preparing the ICE table :

Substituting the values in equation (6):

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

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