Q19.102P - Chemistry: Molecular Nature Of Matter And Change

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

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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 $5.9\times {10}^{-5}\mathrm{M}$