Question

Calculate the ${\mathbf{K}}_{\mathbf{f}}$of$\mathbf{Ag}\mathbf{(}\mathbf{N}{\mathbf{H}}_{\mathbf{3}}{{\mathbf{)}}_{\mathbf{2}}}^{\mathbf{+}}$from:

role="math" localid="1663934413634" $$\begin{gathered} {\mathbf{Ag}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{e}}^{\mathbf{-}}\mathbf{\rightleftharpoons }\mathbf{Ag}{\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{}{\mathbf{E}}^{\mathbf{o}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{80}\mathbf{V} \\ }^{\mathbf{+}} \\ \mathbf{Ag}\mathbf{(}\mathbf{N}{\mathbf{H}}_{\mathbf{3}}{\mathbf{)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{e}}^{\mathbf{-}}\mathbf{\rightleftharpoons }\mathbf{Ag}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{N}{\mathbf{H}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{,}\mathbf{}{\mathbf{E}}^{\mathbf{o}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{37}\mathbf{V} \end{gathered}$$

Short answer

The ${\mathrm{K}}_{\mathrm{f}}$of $\mathrm{Ag}(\mathrm{N}{\mathrm{H}}_3{{)}_2}^{+}$is obtained as:$1.88\times {10}^7$ .

Step-by-step solution

Define chemical reaction

Chemical synthesis or, alternatively, chemical breakdown into two or more separate chemicals occurs when one component interacts with another to generate a new material. These processes are known as chemical reactions, and they are generally irreversible until followed by other chemical reactions.

Evaluate the value of Kf

Silver is oxidised during the production of $\mathrm{Ag}(\mathrm{N}{\mathrm{H}}_3{{)}_2}^{+}$.

$\begin{array}{c}\mathrm{Oxidation}:{\mathrm{Ag}}_{\left(\mathrm{s}\right)}+2{\mathrm{NH}}_{3\left(\mathrm{aq}\right)}\to \mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_{2\left(\mathrm{aq}\right)}^{+}+{\mathrm{e}}^{-}\\ {\mathrm{Reduction}:\mathrm{Ag}}_{\left(\mathrm{aq}\right)}^{+}+{\mathrm{e}}^{-}\to {\mathrm{Ag}}_{\left(\mathrm{s}\right)}\\ {\mathrm{Reaction}:\mathrm{Ag}}_{\left(\mathrm{aq}\right)}^{+}+2{\mathrm{NH}}_{3\left(\mathrm{aq}\right)}\to \mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_{2\left(\mathrm{aq}\right)}^{+}\end{array}$

$\begin{array}{c}{\mathrm{E}}_{\mathrm{oxidation}}^{\mathrm{o}}=0.37\mathrm{V}\\ {\mathrm{E}}_{\mathrm{reduction}}^{\mathrm{o}}=0.80\mathrm{V}\\ {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}=0.80\mathrm{V}-0.37\mathrm{V}\\ {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}=0.43\mathrm{V}\end{array}$

$\begin{array}{c}{\mathrm{\Delta G}}^{\mathrm{o}}={-\mathrm{nFE}}_{\mathrm{cell}}^{\mathrm{o}}\\ =-1{\mathrm{e}}^{-}\times 96,500\frac{\mathrm{C}}{{\mathrm{mole}}^{-}}\times 0.43\mathrm{V}\\ =-41,495\mathrm{J}\end{array}$

${\mathrm{\Delta G}}^{\mathrm{o}}=-\mathrm{RTlnK}$in Action the for may be calculated as:

$\begin{array}{c}-41,495=-\mathrm{RTlnK}\\ \mathrm{lnK}=\frac{41,495\mathrm{J}}{\mathrm{RT}}\\ {\mathrm{K}}_{\mathrm{sp}}=\mathrm{e}\frac{41,495\mathrm{J}}{\mathrm{RT}}\\ {\mathrm{K}}_{\mathrm{sp}}=\mathrm{e}\frac{41,495\mathrm{J}}{298\mathrm{K}\times 8.314\frac{\mathrm{J}}{\mathrm{molK}}}\\ {\mathrm{K}}_{\mathrm{sp}}=1.88\times {10}^7\end{array}$

Therefore, the value is$1.88\times {10}^7$.