72P - Chemistry: Molecular Nature Of Matter And Change

Question

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

Step-by-Step Solution

Answer

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

Step 1: Definition of Concept

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.

Step 2: Find the solubility product K sp of Silver sulfide ( Ag 2 S )

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:

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

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,

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

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 $\text{K=}\underset{\_}{\text{1.70}}{\text{×10}}^{\text{-49}}$