Chapter 16, Problem 35PE

(a)

Interpretation Introduction

Interpretation:

Solubility product constant for ${\text{BaSO}}_{\text{4}}$ has to be calculated.

Concept Introduction:

The equilibrium constant used for the partially soluble salt in water is termed as solubility product constant $K_{\text{sp}}$. Consider solubility of $\text{AB}$ in water and equilibrium equation of $\text{AB}$ is as follows:

  $\text{AB}\left(s\right)\rightleftharpoons {\text{A}}^{+}\left(aq\right)+{\text{B}}^{-}\left(aq\right)$

The expression for $K_{\text{sp}}$ is as follows:

  $K_{\text{sp}}=\frac{\left[{\text{A}}^{+}\right]\left[{\text{B}}^{-}\right]}{\left[\text{AB}\left(s\right)\right]}$

Generally the concentration of solid is taken as constant. Therefore the expression for $K_{\text{sp}}$ of $\text{AB}$ is as follows:

  $K_{\text{sp}}=\left[{\text{A}}^{+}\right]\left[{\text{B}}^{-}\right]$

Explanation of Solution

The equilibrium reaction for ${\text{BaSO}}_{\text{4}}$ is as follows:

  ${\text{BaSO}}_{\text{4}}\left(s\right)\rightleftharpoons {\text{Ba}}^{+2}\left(aq\right)+{\text{SO}}_4^{2-}\left(aq\right)$

In expression for $K_{\text{sp}}$ molar concentration of solid is considered constant. Therefore, expression for ${\text{BaSO}}_{\text{4}}$ is as follows:

  $K_{\text{sp}}=\left[{\text{Ba}}^{+2}\right]\left[{\text{SO}}_4^{2-}\right]$        (1)

Ionization of solid ${\text{BaSO}}_{\text{4}}$ produces ${\text{Ba}}^{+2}$ and ${\text{SO}}_4^{2-}$ in equal amounts that is has equal molar concentrations.

Substitute $\text{3}\text{.9}\times {\text{10}}^{-5}$ for $\left[{\text{Ba}}^{+2}\right]$ and $\text{3}\text{.9}\times {\text{10}}^{-5}$ for $\left[{\text{SO}}_4^{2-}\right]$ in equation (1).

  $\begin{array}{c}{K}_{\text{sp}}=\left(\text{3}\text{.9}\times {\text{10}}^{-5}\right)\left(\text{3}\text{.9}\times {\text{10}}^{-5}\right)\\ =1.5\times {\text{10}}^{-9}\end{array}$

Hence, $K_{\text{sp}}$ of ${\text{BaSO}}_{\text{4}}$ is $1.5\times {\text{10}}^{-9}$.

(b)

Interpretation Introduction

Interpretation:

Solubility product constant for ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The equilibrium reaction for ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is as follows:

  ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}\left(s\right)\rightleftharpoons 2{\text{Ag}}^{+}\left(aq\right)+{\text{CrO}}_4^{2-}\left(aq\right)$

In expression for $K_{\text{sp}}$ molar concentration of solid is considered constant. Therefore, expression for ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is as follows:

  $K_{\text{sp}}={\left[{\text{Ag}}^{+}\right]}^2\left[{\text{CrO}}_4^{2-}\right]$        (2)

Ionization of solid ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$ produces two moles of ${\text{Ag}}^{+}$ and one mole of ${\text{CrO}}_4^{-}$ and therefore concentration of ${\text{Ag}}^{+}$ is $\text{15}\text{.6}\times {\text{10}}^{-5}$.

Substitute $\text{15}\text{.6}\times {\text{10}}^{-5}$ for $\left[{\text{Ag}}^{+}\right]$ and $\text{7}\text{.8}\times {\text{10}}^{-5}$ for $\left[{\text{CrO}}_4^{-}\right]$ in equation (2).

  $\begin{array}{c}{K}_{\text{sp}}={\left(\text{15}\text{.6}\times {\text{10}}^{-5}\right)}^2\left(\text{7}\text{.8}\times {\text{10}}^{-5}\right)\\ =1.825\times {\text{10}}^{-12}\end{array}$

Hence, $K_{\text{sp}}$ of ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is $1.825\times {\text{10}}^{-12}$.

(c)

Interpretation Introduction

Interpretation:

Solubility product constant for ${\text{CaSO}}_4$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The expression to convert solubility of ${\text{CaSO}}_4$ from $\text{g/L}$ to $\text{mol/L}$ is as follows:

  $\text{Solubility in mol/L}=\left[\begin{array}{c}\left(\frac{\text{solubilty of compound}\left(\text{g}\right)}{\text{1 L}}\right)\\ \left(\frac{\text{1}}{{\text{molar mass of CaSO}}_4\left(\text{g/mol}\right)}\right)\end{array}\right]$        (3)

Substitute $0.67\text{ g}$ for solubility of compound and $136.14\text{ g/mol}$ for molar mass of ${\text{CaSO}}_4$ in equation (3).

  $\begin{array}{c}\text{Solubility in mol/L}=\left(\frac{0.67\text{ g}}{\text{1 L}}\right)\left(\frac{\text{1}}{136.14\text{ g/mol}}\right)\\ =0.005\text{ mol}/\text{L}\end{array}$

The equilibrium reaction for ${\text{CaSO}}_4$ is as follows:

  ${\text{CaSO}}_4\left(s\right)\rightleftharpoons {\text{Ca}}^{+2}\left(aq\right)+{\text{SO}}_4^{2-}\left(aq\right)$

In expression for $K_{\text{sp}}$ molar concentration of solid is considered constant. Therefore, expression for ${\text{CaSO}}_4$ is as follows:

  $K_{\text{sp}}=\left[{\text{Ca}}^{+2}\right]\left[{\text{SO}}_4^{2-}\right]$        (4)

Ionization of solid ${\text{CaSO}}_4$ produces ${\text{Ca}}^{+2}$ and ${\text{SO}}_4^{2-}$ in equal amounts that is has equal molar concentrations.

Substitute $0.005$ for $\left[{\text{Ca}}^{+2}\right]$ and $0.005$ for $\left[{\text{SO}}_4^{2-}\right]$ in equation (4).

  $\begin{array}{c}{K}_{\text{sp}}=\left(0.005\right)\left(0.005\right)\\ =2.5\times {\text{10}}^{-5}\end{array}$

Hence, $K_{\text{sp}}$ of ${\text{CaSO}}_4$ is $2.5\times {\text{10}}^{-5}$.

(d)

Interpretation Introduction

Interpretation:

Solubility product constant for $\text{AgCl}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The expression to convert solubility of $\text{AgCl}$ from $\text{g/L}$ to $\text{mol/L}$ is as follows:

  $\text{Solubility in mol/L}=\left[\begin{array}{c}\left(\frac{\text{solubilty of compound}\left(\text{g}\right)}{\text{1 L}}\right)\\ \left(\frac{\text{1}}{\text{molar mass of AgCl}\left(\text{g}\right)}\right)\end{array}\right]$        (5)

Substitute $0.0019\text{ g}$ for solubility of compound and $143.32\text{g/mol}$ for molar mass of ${\text{CaSO}}_4$

 in equation (3).

  $\begin{array}{c}\text{Solubility in mol/L}=\left(\frac{0.0019\text{ g}}{\text{1 L}}\right)\left(\frac{\text{l}}{143.32\text{g/mol}}\right)\\ =1.33\times {10}^{-5}\text{ mol}/\text{L}\end{array}$

The equilibrium reaction for $\text{AgCl}$ is as follows:

  $\text{AgCl}\left(s\right)\rightleftharpoons {\text{Ag}}^{+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$

In expression for $K_{\text{sp}}$ molar concentration of solid is considered constant. Therefore, expression for $\text{AgCl}$ is as follows:

  $K_{\text{sp}}=\left[{\text{Ag}}^{+}\right]\left[{\text{Cl}}^{-}\right]$        (1)

Ionization of solid $\text{AgCl}$ produces ${\text{Ag}}^{+}$ and ${\text{Cl}}^{-}$ in equal amounts that is has equal molar concentrations.

Substitute $1.33\times {10}^{-5}$ for $\left[{\text{Ag}}^{+}\right]$ and $1.33\times {10}^{-5}$ for $\left[{\text{Cl}}^{-}\right]$ in equation (1).

  $\begin{array}{c}{K}_{\text{sp}}=\left(1.33\times {10}^{-5}\right)\left(1.33\times {10}^{-5}\right)\\ =1.77\times {\text{10}}^{-10}\end{array}$

Hence, $K_{\text{sp}}$ of $\text{AgCl}$ is $1.77\times {\text{10}}^{-10}$.