Question

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${\mathbf{\text{Ag(NH}}}_{\mathbf{\text{3}}}{{\mathbf{\text{)}}}_{\mathbf{\text{2}}}}^{\mathbf{\text{+}}}{\mathbf{\text{= 1.7×10}}}^{\mathbf{\text{7}}}$].

Short answer

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

Step-by-step solution

Concept Introduction

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.

Solubility of Agl

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 :

	${\text{NH}}_{\text{3}}$	$\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}^{\text{+}}$	${\text{I}}^{\text{-}}$
Initial	2.5	0	0
Change	-2x	x	x
equilibrium	2.5-2x	x	x

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$5.9\times {10}^{-5}\mathrm{M}$