Use Appendix B to determine the ${\mathbf{\text{K}}}_{\mathbf{\text{sp}}}$of${\mathbf{\text{Ag}}}_{\mathbf{\text{2}}}\mathbf{\text{S}}$.

Solubility: The maximum amount of solute that can be dissolved in the solvent at equilibrium is defined as solubility.

Constant of soluble product: For equilibrium between solids and their respective ions in a solution, the solubility product constant is defined. In general, the term "solubility product" refers to water-equilibrium insoluble or slightly soluble ionic substances.

Considering the given Chemical reaction:

The following is the equation for the solubility of $\left({\text{Ag}}_{\text{2}}\text{S}\right)$:

${\text{Ag}}_{\text{2}}\text{S(s)}\rightleftharpoons {\text{2Ag}}^{\text{+}}{\text{(aq)+S}}^{\text{2-}}\text{(aq)}$

Calculate ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$ from ${\text{ΔG}}_{\text{f}}^{\text{o}}$values of products and reactants,

The Gibbs free energy equation is as follows:

${\text{ΔG}}_{\text{rxn}}^{\text{°}}\text{=}\sum \text{m}{\text{ΔG}}_{\text{f}}^{\text{°}}\text{(Products)-}\sum \text{n}{{\text{ΔG}}_{\text{f}}}^{\text{°}}\text{(Reactants)}$

The reaction's free energy change is calculated as follows:

$\begin{array}{c}{\text{ΔG}}_{\text{rrn}}^{\text{°}}\text{=}[\left({\text{2molAg}}^{\text{2+}}\right)({\text{ΔG}}_{\text{f}}^{\text{o}}\text{of}{\text{Ag}}^{\text{2+}})\text{+}\left({\text{1mol}}^{\text{2-}}\right)({\text{ΔG}}_{\text{f}}^{\text{o}}\text{of}{\text{S}}^{\text{2-}})][(\text{1}\text{mol}{\text{Ag}}_{\text{2}}\text{S})({\text{ΔG}}_{\text{f}}^{\text{o}}\text{of}{\text{Ag}}_{\text{2}}\text{S})]\\ {\text{ΔG}}_{\text{rxnn}}^{\text{0}}\text{=}\left[\left({\text{1molAg}}^{\text{2+}}\right)\text{(77.111kJ/mol)+}\left({\text{1molS}}^{\text{2-}}\right)\text{(83.7kJ/mol)}\right]\left[\left({\text{1molAg}}_{\text{2}}\text{S}\right)\text{(-40.3kJ/mol)}\right]\\ {\text{ΔG}}_{\text{rxn}}^{\text{°}}\text{=278.222kJ}\end{array}$

The value of ${\text{ΔG}}_{\text{rxn}}^{\text{°}}$is $\underset{\_}{\text{278.222kJ}}$and these value of ${\text{ΔG}}_{\text{f}}^{\text{o}}$are referred from the Appendix B.

K, the equilibrium constant, is calculated.

We are aware of the equilibrium equation.

${\text{ΔG=ΔG}}^{\text{o}}\text{+RTln(K)}$

Rearrange the equation above,

$\begin{array}{l}\text{lnK=-}\frac{{\text{ΔG}}^{\text{o}}}{\text{RT}}\text{=}\left(\frac{\text{278.222kJ/mol}}{\text{-(8.314J/mol-K)(298K)}}\right)\left(\frac{{\text{10}}^{\text{3}}\text{J}}{\text{1kJ}}\right)\\ \text{InK=-112.296232}\end{array}$

Hence,

$\begin{array}{l}{\text{K=e}}^{\text{-112.296232}}\\ {\text{K=1.69967585×10}}^{\text{-49}}\text{(or)}\\ {\text{K=1.70×10}}^{\text{-49}}\end{array}$

Therefore, the required solubility product${\text{K}}_{\text{sp}}$ of Silver sulfide$\left({\text{Ag}}_{\text{2}}\text{S}\right)$is$\text{K=}\underset{\_}{\text{1.70}}{\text{×10}}^{\text{-49}}$