Question

a. Calculate the molar solubility of AgI in pure water. ${\mathit{K}}_{\mathbf{s}\mathbf{p}}$for AgI is$\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{16}}$ .

b. Calculate the molar solubility of AgI in $\mathbf{3}\mathbf{.}\mathbf{0}\mathit{M}\mathit{N}{\mathit{H}}_{\mathbf{3}}$. The overall formation constant for is $\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}$.

c. Compare the calculated solubilities from part a and b. Explain any differences.

Short answer

a. The molar solubility of AgI in pure water is $1.2\times {10}^{-8}M$.

b. The molar solubility of AgI in $3.0MN{H}_3$is $1.53\times {10}^{-4}M$.

The solubility of AgI increases in the presence of $N{H}_3$. Therefore, part (b) has greater solubility of AgI and the concentration of $A{g}^{+}$ ions

Step-by-step solution

Subpart (a) The molar solubility of AgI in pure water.

The number of moles of the solute that can be dissolved per liter of solution before the solution becomes saturatedis defined as molar solubility.

$\begin{array}{l}Agl\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+l^{-}\left(aq\right)\\ \text{initial\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}--\\ {\text{equil}}^{\text{m}}xx\end{array}$

Hence,

$\begin{array}{l}{K}_{sp}=\left[A{g}^{+}\right]\left[I^{-}\right]\\ 1.5\times {10}^{-16}=x^2\\ x=1.2\times {10}^{-8}M\end{array}$

Subpart (b) The molar solubility of AgI in3.0M NH3  .

$\begin{array}{l}Agl\left(s\right)+2N{H}_3\left(aq\right)\rightleftharpoons \text{Invalid <msub> element}A{g}^{+}\left(aq\right)+l^{-}\left(aq\right)\\ \text{initial\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3.0M--\\ {\text{equil}}^{\text{m}}(3.0-2x)xx\end{array}$

$\begin{array}{c}K=\frac{\left[Ag{{\left(N{H}_3\right)}_2}^{+}\right]\left[I^{-}\right]}{{\left[N{H}_3\right]}_2}\\ 2.6\times {10}^{-9}=\frac{x^2}{{(3.0-2x)}^2}\end{array}$

And,

$5.1\times {10}^{-5}=\frac{x}{3.0-2x}$

$\begin{array}{l}1.53\times {10}^{-4}-1.02\times {10}^{-4}x=x\\ x=1.53\times {10}^{-4}M\end{array}$

Subpart (c) Comparison of the calculated solubilities from part a and b.

Whenis added, $A{g}^{+}$present in the solution forms a stable complex as$Ag{{\left(N{H}_3\right)}_2}^{+}$. The concentration ofions decrease.

The solubility equilibrium of AgI shifts towards right forming more$A{g}^{+}$ ions.

Hence, the solubility of AgI increases in the presence of (part b).