Chapter 16, Problem 36PE

(a)

Interpretation Introduction

Interpretation:

Solubility product constant for $\text{ZnS}$ 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 $\text{ZnS}$ is as follows:

  $\text{ZnS}\left(s\right)\rightleftharpoons {\text{Zn}}^{+2}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)$

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

  $K_{\text{sp}}=\left[{\text{Zn}}^2{}^{+}\right]\left[{\text{S}}^{2-}\right]$        (1)

Ionization of solid $\text{ZnS}$ produces ${\text{Zn}}^2{}^{+}$ and ${\text{S}}^{2-}$ in equal amounts that is has equal molar concentrations.

Substitute $\text{3}\text{.5}\times {\text{10}}^{-12}$ for $\left[{\text{Zn}}^2{}^{+}\right]$ and $\text{3}\text{.5}\times {\text{10}}^{-12}$ for $\left[{\text{S}}^{2-}\right]$ in equation (1).

  $\begin{array}{c}{K}_{\text{sp}}=\left(\text{3}\text{.5}\times {\text{10}}^{-12}\right)\left(\text{3}\text{.5}\times {\text{10}}^{-12}\right)\\ =1.2\times {\text{10}}^{-23}\end{array}$

Hence, $K_{\text{sp}}$ of $\text{ZnS}$ is $1.2\times {\text{10}}^{-23}$.

(b)

Interpretation Introduction

Interpretation:

Solubility product constant for $\text{Pb}{\left({\text{IO}}_{\text{3}}\right)}_2$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The equilibrium reaction for $\text{Pb}{\left({\text{IO}}_{\text{3}}\right)}_2$ is as follows:

  $\text{Pb}{\left({\text{IO}}_{\text{3}}\right)}_2\left(s\right)\rightleftharpoons {\text{Pb}}^{+2}\left(aq\right)+2{\text{IO}}_3^{-}\left(aq\right)$

In expression for $K_{\text{sp}}$ molar concentration of solid is considered constant. Therefore, expression for $\text{Pb}{\left({\text{IO}}_{\text{3}}\right)}_2$ is as follows:

  $K_{\text{sp}}=\left[{\text{Pb}}^{+2}\right]{\left[{\text{IO}}_3^{-}\right]}^2$        (2)

Ionization of solid $\text{Pb}{\left({\text{IO}}_{\text{3}}\right)}_2$ produces one mole of ${\text{Pb}}^{+2}$ and two moles of ${\text{IO}}_3^{-}$. Therefore concentration of ${\text{IO}}_3^{-}$ is two times of concentration of $\text{Pb}{\left({\text{IO}}_{\text{3}}\right)}_2$ and that is $8\times {\text{10}}^{-5}$.

Substitute $\text{4}\text{.0}\times {\text{10}}^{-5}$ for $\left[{\text{Pb}}^{+2}\right]$ and $8\times {\text{10}}^{-5}$ for $\left[{\text{IO}}_3^{-}\right]$ in equation (2).

  $\begin{array}{c}{K}_{\text{sp}}=\left(\text{4}\text{.0}\times {\text{10}}^{-5}\right){\left(8\times {\text{10}}^{-5}\right)}^2\\ =2.56\times {\text{10}}^{-13}\end{array}$

Hence, $K_{\text{sp}}$ of $\text{Pb}{\left({\text{IO}}_{\text{3}}\right)}_2$ is $2.56\times {\text{10}}^{-13}$.

(c)

Interpretation Introduction

Interpretation:

Solubility product constant for ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The expression to convert solubility of ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$ from $\text{g/L}$ to $\text{mol/L}$ is as follows:

  $\text{Solubility in mol/L}=\left[\begin{array}{c}\left(\frac{\text{solubilty of compound}\left(\text{g}\right)}{\text{1 L}}\right)\\ \left(\frac{1}{{\text{molar mass of Ag}}_{\text{3}}{\text{PO}}_{\text{4}}\left(\text{g/mol}\right)}\right)\end{array}\right]$        (3)

Substitute $\text{6}\text{.73}\times {\text{10}}^{-3}\text{g}$ for solubility of compound and $481.58\text{g/mol}$ for molar mass of ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$ in equation (3).

  $\begin{array}{c}\text{Solubility in mol/L}=\left(\frac{\text{6}\text{.73}\times {\text{10}}^{-3}\text{ g}}{\text{1 L}}\right)\left(\frac{\text{1}}{418.58\text{ g/mol}}\right)\\ =1.60\times {\text{10}}^{-5}\text{ mol}/\text{L}\end{array}$

The equilibrium reaction for ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$ is as follows:

  ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}\left(s\right)\rightleftharpoons 3{\text{Ag}}^{+}\left(aq\right)+{\text{PO}}_4^{3-}\left(aq\right)$

In expression for $K_{\text{sp}}$ molar concentration of solid is considered constant. Therefore, expression for ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$ is as follows:

  $K_{\text{sp}}={\left[{\text{Ag}}^{+}\right]}^3\left[{\text{PO}}_4^{3-}\right]$        (4)

Ionization of solid ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$ produces three moles of ${\text{Ag}}^{+}$ and one mole of ${\text{PO}}_4^{3-}$. Therefore concentration of ${\text{Ag}}^{+}$ is three times of concentration of ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$ and that is $4.8\times {\text{10}}^{-5}$.

Substitute $4.8\times {\text{10}}^{-5}$ for $\left[{\text{Ag}}^{+}\right]$ and $1.60\times {\text{10}}^{-5}$ for $\left[{\text{PO}}_4^{3-}\right]$ in equation (4).

  $\begin{array}{c}{K}_{\text{sp}}={\left(4.8\times {\text{10}}^{-5}\right)}^3\left(1.60\times {\text{10}}^{-5}\right)\\ =1.8\times {\text{10}}^{-18}\end{array}$

Hence, $K_{\text{sp}}$ of ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$ is $1.8\times {\text{10}}^{-18}$.

(d)

Interpretation Introduction

Interpretation:

Solubility product constant for $\text{Zn}{\left(\text{OH}\right)}_{\text{2}}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The expression to convert solubility of $\text{Zn}{\left(\text{OH}\right)}_{\text{2}}$ from $\text{g/L}$ to $\text{mol/L}$ is as follows:

  $\text{Solubility in mol/L}=\left[\begin{array}{c}\left(\frac{\text{solubilty of compound}\left(\text{g}\right)}{\text{1 L}}\right)\\ \left(\frac{1}{{\text{molar mass of Ag}}_{\text{3}}{\text{PO}}_{\text{4}}\left(\text{g/mol}\right)}\right)\end{array}\right]$        (5)

Substitute $\text{2}\text{.33}\times {\text{10}}^{-4}\text{g}$ for solubility of compound and $99.39\text{g/mol}$ for molar mass of $\text{Zn}{\left(\text{OH}\right)}_{\text{2}}$

 in equation (5).

  $\begin{array}{c}\text{Solubility in mol/L}=\left(\frac{\text{2}\text{.33}\times {\text{10}}^{-4}\text{ g}}{\text{1 L}}\right)\left(\frac{\text{1}}{99.39\text{g/mol}}\right)\\ =2.34\times {\text{10}}^{-6}\text{ mol}/\text{L}\end{array}$

The equilibrium reaction for $\text{Zn}{\left(\text{OH}\right)}_{\text{2}}$ is as follows:

  $\text{Zn}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)\rightleftharpoons {\text{Zn}}^{+2}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)$

In expression for $K_{\text{sp}}$ molar concentration of solid is considered constant. Therefore, expression for $\text{Zn}{\left(\text{OH}\right)}_{\text{2}}$ is as follows:

  $K_{\text{sp}}=\left[{\text{Zn}}^2{}^{+}\right]{\left[{\text{OH}}^{-}\right]}^2$        (6)

Ionization of solid $\text{Zn}{\left(\text{OH}\right)}_{\text{2}}$ produces one mole of ${\text{Zn}}^2{}^{+}$ and two moles of ${\text{OH}}^{-}$. Therefore concentration of ${\text{OH}}^{-}$ is two times of concentration of $\text{Zn}{\left(\text{OH}\right)}_{\text{2}}$ and that is $4.68\times {\text{10}}^{-6}$.

Substitute $2.34\times {\text{10}}^{-6}$ for $\left[{\text{Zn}}^2{}^{+}\right]$ and $4.68\times {\text{10}}^{-6}$ for $\left[{\text{OH}}^{-}\right]$ in equation (6).

  $\begin{array}{c}{K}_{\text{sp}}=\left(2.34\times {\text{10}}^{-6}\right){\left(4.68\times {\text{10}}^{-6}\right)}^2\\ =5.127\times {\text{10}}^{-17}\end{array}$

Hence, $K_{\text{sp}}$ of $\text{Zn}{\left(\text{OH}\right)}_{\text{2}}$ is $5.127\times {\text{10}}^{-17}$.