Chapter 23, Problem 23.72P

(a)

Interpretation Introduction

Interpretation:

The solubility-product constant of $AgCl\left(s\right)$ has to be calculated.

Concept Introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by $\text{ΔG}$.

Value of $\Delta G_{rxn}^o$ can be calculated by the equation:

  $\Delta G_{rxn}^o\text{= }G°\left[\text{products}\right]-G°\left[\text{reactants}\right]$

Equilibrium constant of a reaction can be determined from the value of $\Delta G_{rxn}^o$ using the given formula,

  ${\text{ΔG}}_{rxn}^o=-\text{RT}\text{ln K}$

Solubility product constant ${\text{K}}_{\text{sp}}$ refers to the level in which the solute gets dissolved in a solution.  It explains the equilibrium of a solid with its corresponding ions in a solution.

It is the product of concentration of ions that are dissolved raised to power of their corresponding coefficient of stoichiometry.

  $\begin{array}{l}{\text{A}}_{\text{c}}{\text{B}}_{\text{d}}\rightleftharpoons {\text{cA}}^{\text{+}}\text{+}{\text{dB}}^{-}\\ \\ {\text{K}}_{\text{sp}}\text{=}{\left[{\text{A}}^{\text{+}}\right]}^{\text{c}}{\left[{\text{B}}^{-}\right]}^{\text{d}}\end{array}$

Answer to Problem 23.72P

Value of solubility-product constant is $\text{2}\times {\text{10}}^{10}$.

Explanation of Solution

The given reaction is shown below,

  $AgCl\left(s\right)\underset{}{\to }A{g}^{+}\left(aq\right)+C{l}^{-}\left(aq\right)$

• Calculate the value of $\Delta G_{rxn}^o$:

Standard free energy of formation values is given below,

  $\begin{array}{c}\text{Δ}G°\left[AgCl\left(s\right)\right]\text{=}-\text{109}\text{.8 kJ/mol}\\ \\ \text{Δ}G°\left[A{g}^{+}\left(aq\right)\right]\text{=}77.1\text{ kJ/mol}\\ \\ \text{Δ}G°\left[C{l}^{-}\left(aq\right)\right]\text{=}-131.2\text{ kJ/mol}\end{array}$

Value of $\Delta G_{rxn}^o$ can be calculated by the equation:

  $\Delta G_{rxn}^o\text{= }G°\left[\text{products}\right]-G°\left[\text{reactants}\right]$

Substitute the values as follows,

  $\begin{array}{c}\Delta G_{rxn}^o\text{= }\left[\left(1\right)\left(77.1\text{ kJ/mol}\right)+\left(1\right)\left(-131.2\text{ kJ/mol}\right)\right]-\left[\left(1\right)\left(-\text{109}\text{.8 kJ/mol}\right)\right]\\ \\ =55.7\text{ kJ/mol}\end{array}$

Therefore, the value of $\Delta G_{rxn}^o$ is $55.7\text{ kJ/mol}$.

• Calculate the value of the solubility-product constant:

For the above reaction, equilibrium constant can be written as below,

  $K\text{ = }\frac{\left[A{g}^{+}\left(aq\right)\right]\left[C{l}^{-}\left(aq\right)\right]}{\left[AgCl\left(s\right)\right]}$

However, $AgCl$ is a solid and therefore,

  $K\text{ = }\left[A{g}^{+}\left(aq\right)\right]\left[C{l}^{-}\left(aq\right)\right]{\text{ = K}}_{sp}$

Value of the solubility-product constant is determined as follows,

  $\begin{array}{c}T\text{ = 25}°\text{C}\text{= }\left(25+273\right)\text{ = 298 K}\\ \\ {\text{ΔG}}_{rxn}^o=-\text{RT}\text{ln K}\\ \\ 55.7\text{ kJ/mol =}-\left(\text{8}\text{.314}\times {\text{10}}^{-3}\text{ kJ}{\text{.K}}^{-1}.mo{l}^{-1}\right)\left(\text{298 K}\right)\text{ln }{K}_{sp}\\ \\ \text{ln }{K}_{sp}=-\frac{55.7\text{ kJ/mol}}{\left(\text{8}\text{.314}\times {\text{10}}^{-3}\text{ kJ}{\text{.K}}^{-1}.mo{l}^{-1}\right)\left(\text{298 K}\right)}\text{ =}-\text{22}\text{.482}\\ \\ {\text{K}}_{sp}\text{ = 2}\times {\text{10}}^{10}\end{array}$

Therefore, the value of the solubility-product constant is $2\times 10^{10}$.

(b)

Interpretation Introduction

Interpretation:

The solubility-product constant of $AgBr\left(s\right)$ has to be calculated.

Concept Introduction:

Refer to (a).

Answer to Problem 23.72P

Value of solubility-product constant is $\text{5}\text{.4}\times {\text{10}}^{-13}$.

Explanation of Solution

The given reaction is shown below,

  $AgBr\left(s\right)\underset{}{\to }A{g}^{+}\left(aq\right)+B{r}^{-}\left(aq\right)$

• Calculate the value of $\Delta G_{rxn}^o$:

Standard free energy of formation values is given below,

  $\begin{array}{c}\text{Δ}G°\left[AgBr\left(s\right)\right]\text{=}-\text{96}\text{.9 kJ/mol}\\ \\ \text{Δ}G°\left[A{g}^{+}\left(aq\right)\right]\text{=}77.1\text{ kJ/mol}\\ \\ \text{Δ}G°\left[B{r}^{-}\left(aq\right)\right]\text{=}-\text{104}\text{.0 kJ/mol}\end{array}$

Value of $\Delta G_{rxn}^o$ can be calculated by the equation:

  $\Delta G_{rxn}^o\text{= }G°\left[\text{products}\right]-G°\left[\text{reactants}\right]$

Substitute the values as follows,

  $\begin{array}{c}\Delta G_{rxn}^o\text{= }\left[\left(1\right)\left(77.1\text{ kJ/mol}\right)+\left(1\right)\left(-\text{104}\text{.0 kJ/mol}\right)\right]-\left[\left(1\right)\left(-\text{96}\text{.9 kJ/mol}\right)\right]\\ \\ =70\text{ kJ/mol}\end{array}$

Therefore, the value of $\Delta G_{rxn}^o$ is $70\text{ kJ/mol}$.

• Calculate the value of the solubility-product constant:

For the above reaction, equilibrium constant can be written as below,

  $K\text{ = }\frac{\left[A{g}^{+}\left(aq\right)\right]\left[B{r}^{-}\left(aq\right)\right]}{\left[AgBr\left(s\right)\right]}$

However, $AgCl$ is a solid and therefore,

  $K\text{ = }\left[A{g}^{+}\left(aq\right)\right]\left[B{r}^{-}\left(aq\right)\right]{\text{ = K}}_{sp}$

Value of the solubility-product constant is determined as follows,

  $\begin{array}{c}T\text{ = 25}°\text{C}\text{= }\left(25+273\right)\text{ = 298 K}\\ \\ {\text{ΔG}}_{rxn}^o=-\text{RT}\text{ln K}\\ \\ 70\text{ kJ/mol =}-\left(\text{8}\text{.314}\times {\text{10}}^{-3}\text{ kJ}{\text{.K}}^{-1}.mo{l}^{-1}\right)\left(\text{298 K}\right)\text{ln }{K}_{sp}\\ \\ \text{ln }{K}_{sp}=-\frac{70\text{ kJ/mol}}{\left(\text{8}\text{.314}\times {\text{10}}^{-3}\text{ kJ}{\text{.K}}^{-1}.mo{l}^{-1}\right)\left(\text{298 K}\right)}\text{ =}-\text{28}\text{.25347}\\ \\ {\text{K}}_{sp}\text{ = 5}\text{.4}\times {\text{10}}^{-13}\end{array}$

Therefore, the value of the solubility-product constant is $5.4\times 10^{-13}$.