Question

Consider the following equilibria in aqueous solution:

(1)${\mathbf{Ag}}^{\mathbf{+}}\mathbf{+}{\mathbf{Cl}}^{\mathbf{-}}\mathbf{\rightleftharpoons }\mathbf{AgCl}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\hspace{1em}}\mathbf{K}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}$

(2)$\mathbf{AgCl}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{Cl}}^{\mathbf{-}}\mathbf{\rightleftharpoons }{\mathbf{AgCl}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{}\mathbf{K}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{1}}$

(3)$\mathbf{AgCl}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\rightleftharpoons }{\mathbf{Ag}}^{\mathbf{+}}\mathbf{+}{\mathbf{Cl}}^{\mathbf{-}}\mathbf{\hspace{1em}}\mathbf{K}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{)}$

(a) Calculate the numerical value of the equilibrium constant for the reaction$\mathbf{AgCl}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{AgCl}\mathbf{\left(}\mathbf{aq}\right)$.

(b) Calculate the concentration of$\mathbf{AgCl}\mathbf{\left(}\mathbf{aq}\right)$in equilibrium with excess undissolved solid AgCl.

(c) Find the numerical value of$\mathbf{K}$for the reaction${\mathbf{AgCl}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{=}\mathbf{AgCl}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{Cl}}^{\mathbf{-}}$

Short answer

a) The equilibrium constant (K) for given reaction is $=3.6\times {10}^{-7}$

b) The concentration of aqueous $\left\{\mathrm{AgCI}\right\}$is role="math" localid="1663566428457" $=3.6\times {10}^{-7}\mathrm{M}$

c) The equilibrium constant $k$for given reaction is $=3.6\times {10}^4$

Step-by-step solution

equilibrium constant

For given equilibria,

The equilibrium constant ($\mathrm{K}$) for given reaction must be calculated.

Equilibrium constant (K):

In the equilibrium reaction, the ratio of concentration of the reactant and concentration of the product. If value of $\mathbf{K}$is small than $\mathbf{1}$the reaction should be move to the left and $\mathit{k}$is more than $\mathbf{1}$the reaction should be move to right.

equilibrium constant and chemical equation for the given reaction

Given

$$\begin{gathered} {\mathrm{Ag}}^{+}+\mathrm{Cl}\xcancel{}{\mathrm{AgCl}}_{\left(\mathrm{aq}\right)}\mathrm{K}\hspace{0.33em}=\hspace{0.33em}2.0\times {10}^3 \\ {\mathrm{AgCl}}_{\left(\mathrm{s}\right)}\xcancel{}{\mathrm{Ag}}^{+}+{\mathrm{Cl}}^{-}\mathrm{K}\hspace{0.33em}=\hspace{0.33em}1.8\times {10}^{-10} \end{gathered}$$

Equilibrium constant and chemical equation can be obtained by adding these two equilibria and multiplying their equilibrium constants.

$$\begin{gathered} {\mathrm{Ag}}^{+}+\mathrm{Cl}\xcancel{}{\mathrm{AgCl}}_{\left(\mathrm{aq}\right)}{\mathrm{K}}_1\hspace{0.33em}=\hspace{0.33em}2.0\times {10}^3 \\ {\mathrm{AgCl}}_{\left(\mathrm{s}\right)}\xcancel{}{\mathrm{Ag}}^{+}+{\mathrm{Cl}}^{-}{\mathrm{K}}_2\hspace{0.33em}=\hspace{0.33em}1.8\times {10}^{-10} \\ {\mathrm{AgCl}}_{\left(\mathrm{s}\right)}\xcancel{}{\mathrm{AgCl}}_{\left(\mathrm{aq}\right)}{\mathrm{K}}_3\hspace{0.33em}=\hspace{0.33em}3.6\times {10}^{-7} \end{gathered}$$

Concept for concentration of aqueous 

For given equilibria,

The concentration of aqueous$\mathbf{AgCl}$must be calculated.role="math" localid="1663567674612" ${\mathbf{AgCl}}_{\left(\mathrm{aq}\right)}$is equilibrium with excess undissolved${\mathbf{AgCl}}_{\left(s\right)}$.

concentration of aqueous 

Given

$$\begin{gathered} {\mathrm{Ag}}^{+}+\mathrm{Cl}\xcancel{}{\mathrm{AgCl}}_{\left(\mathrm{aq}\right)}\mathrm{K}\hspace{0.33em}=\hspace{0.33em}2.0\times {10}^3 \\ {\mathrm{AgCl}}_{\left(\mathrm{s}\right)}\xcancel{}{\mathrm{Ag}}^{+}+{\mathrm{Cl}}^{-}\mathrm{K}\hspace{0.33em}=\hspace{0.33em}1.8\times {10}^{-10} \end{gathered}$$

By adding these two equilibria and multiply these equilibrium constants we can get equilibrium constant and given chemical equation.

role="math" localid="1663568680415" $$\begin{gathered} {\mathrm{Ag}}^{+}+\mathrm{Cl}\xcancel{}{\mathrm{AgCl}}_{\left(\mathrm{aq}\right)}\&{\mathrm{K}}_1\hspace{0.33em}=\hspace{0.33em}2.0\times {10}^3 \\ {\mathrm{AgCl}}_{\left(\mathrm{s}\right)}\xcancel{}{\mathrm{Ag}}^{+}+{\mathrm{Cl}}^{-}\&{\mathrm{K}}_2\hspace{0.33em}=\hspace{0.33em}1.8\times {10}^{-10} \\ {\mathrm{AgCl}}_{\left(\mathrm{s}\right)}\xcancel{}{\mathrm{AgCl}}_{\left(\mathrm{aq}\right)}\&{\mathrm{K}}_3\hspace{0.33em}=\hspace{0.33em}3.6\times {10}^{-7} \end{gathered}$$

When a reaction reaches equilibrium, the equilibrium constant defines the relationship between reactant and product concentrations. As a result, the concentration will be equal to the equilibrium constant. Thus, the concentration of aqueous $\mathrm{AgCI}$is found to be $\hspace{0.33em}3.6\times {10}^{-7}\mathrm{M}$.

equilibrium constant

Given

$$\begin{gathered} {\mathrm{AgCl}}_2^{-}\xcancel{}{\mathrm{AgCl}}_{\left(\mathrm{aq}\right)}+{\mathrm{Cl}}^{-}\mathrm{K}=9.3\times {10}^1 \\ {\mathrm{Ag}}^{+}+\mathrm{Cl}\xcancel{}{\mathrm{AgCl}}_{\left(\mathrm{s}\right)}\mathrm{K}=1.8\times 10-{10}^{-10} \\ {\mathrm{AgCl}}_{\left(\mathrm{aq}\right)}\xcancel{}{\mathrm{AgCl}}_{\left(\mathrm{s}\right)}+{\mathrm{Cl}}^{-}\mathrm{K}=2.0\times {10}^3 \end{gathered}$$

Equilibrium constant and chemical equation can be obtained by adding these two equilibria and multiplying their equilibrium constants.

If the reaction is reversed,$\mathrm{K}=\frac{1}{\mathrm{K}}$
localid="1663569203523" $$\begin{gathered} {\mathrm{AgCl}}_2^{-}\xcancel{}{\mathrm{AgCl}}_{\left(\mathrm{aq}\right)}+{\mathrm{Cl}}^{-}1/\mathrm{K}=9.3\times {10}^1 \\ {\mathrm{Ag}}^{+}+\mathrm{Cl}\xcancel{}{\mathrm{AgCl}}_{\left(\mathrm{s}\right)}1/\mathrm{K}=1.8\times {10}^{-10} \\ {\mathrm{AgCl}}_{\left(\mathrm{aq}\right)}\xcancel{}{\mathrm{AgCl}}_{\left(\mathrm{s}\right)}+{\mathrm{Cl}}^{-}1/\mathrm{K}=2.0\times {10}^3 \\ {\mathrm{AgCl}}_{\left(2\right)}\xcancel{}{\mathrm{AgCl}}_{\left(\mathrm{s}\right)}+{\mathrm{Cl}}^{-}{\mathrm{K}}_3=3.0\times {10}^4 \end{gathered}$$