Chapter 16, Problem 22RQ

(a)

Interpretation Introduction

Interpretation:

Among $\text{Mn}{\left(\text{OH}\right)}_2$ and ${\text{Ag}}_2{\text{CrO}}_4$ that has higher molar solubility of $\text{Mn}{\left(\text{OH}\right)}_2$ and ${\text{Ag}}_2{\text{CrO}}_4$ has to be determined.

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 $\text{Mn}{\left(\text{OH}\right)}_2$ is as follows:

  $\text{Mn}{\left(\text{OH}\right)}_2\left(s\right)\rightleftharpoons {\text{Mn}}^{2+}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)$

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

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

Ionization of solid $\text{Mn}{\left(\text{OH}\right)}_2$ produces one mole of ${\text{Mn}}^{2+}$ and two moles of ${\text{OH}}^{-}$. Therefore concentration of ${\text{OH}}^{-}$ is two times of concentration of ${\text{Mn}}^{2+}$ and that is concentration of ${\text{Mn}}^{2+}$ is $s$ and concentration of ${\text{OH}}^{-}$ is $2s$.

Substitute $s$ for $\left[{\text{Mn}}^{2+}\right]$ and $2s$ for $\left[{\text{OH}}^{-}\right]$ in the equation (1).

  $\begin{array}{c}{K}_{\text{sp}}=\left(s\right){\left(2s\right)}^2\\ \sqrt[3]{\frac{K_{\text{sp}}}{4}}=s\end{array}$

Substitute $2\times {10}^{-13}$ for $K_{\text{sp}}$ in above equation.

  $\begin{array}{c}s=\sqrt[3]{\frac{2\times {10}^{-13}}{4}}\\ =3.68\times {10}^{-5}\end{array}$

The equilibrium reaction for ${\text{Ag}}_2{\text{CrO}}_4$ is as follows:

  ${\text{Ag}}_2{\text{CrO}}_4\left(s\right)\rightleftharpoons 2{\text{Ag}}^{+}\left(aq\right)+{\text{CrO}}_4^{2-}\left(aq\right)$

In expression for $K_{\text{sp}}$ molar concentration of solid is considered constant. Therefore, expression for ${\text{Ag}}_2{\text{CrO}}_4$ is as follows:

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

Ionization of solid ${\text{Ag}}_2{\text{CrO}}_4$ produces one mole of ${\text{CrO}}_4^{2-}$ and two moles of ${\text{Ag}}^{+}$. Therefore concentration of ${\text{Ag}}^{+}$ is two times of concentration of ${\text{CrO}}_4^{2-}$ and that is concentration of ${\text{CrO}}_4^{2-}$ is $s$ and concentration of ${\text{Ag}}^{+}$ is $2s$.

Substitute $2s$ for $\left[{\text{Ag}}^{+}\right]$ and $s$ for $\left[{\text{CrO}}_4^{2-}\right]$ in the equation (2).

  $\begin{array}{c}{K}_{\text{sp}}={\left(2s\right)}^2\left(s\right)\\ \sqrt[3]{\frac{K_{\text{sp}}}{4}}=s\end{array}$

Substitute $1.9\times {10}^{-12}$ for $K_{\text{sp}}$ in above equation.

  $\begin{array}{c}s=\sqrt[3]{\frac{1.9\times {10}^{-12}}{4}}\\ =7.80\times {10}^{-5}\end{array}$

Value of molar solubility is greater in ${\text{Ag}}_2{\text{CrO}}_4$ than $\text{Mn}{\left(\text{OH}\right)}_2$.

(b)

Interpretation Introduction

Interpretation:

Among ${\text{BaCrO}}_4$ and ${\text{Ag}}_2{\text{CrO}}_4$ that has higher molar solubility of $\text{Mn}{\left(\text{OH}\right)}_2$ and ${\text{Ag}}_2{\text{CrO}}_4$ has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

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

  ${\text{BaCrO}}_4\left(s\right)\rightleftharpoons {\text{Ba}}^{2+}\left(aq\right)+{\text{CrO}}_4^{2-}\left(aq\right)$

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

  $K_{\text{sp}}=\left[{\text{Ba}}^{2+}\right]\left[{\text{CrO}}_4^{2-}\right]$        (3)

Ionization of solid ${\text{BaCrO}}_4$ produces ${\text{Ba}}^{2+}$ and ${\text{CrO}}_4^{2-}$ in equal amounts that is has equal molar concentrations. Value of concentration of ${\text{Ba}}^{2+}$ and ${\text{CrO}}_4^{2-}$ is $s$.

Substitute $s$ for $\left[{\text{Ba}}^{2+}\right]$ and $s$ for $\left[{\text{CrO}}_4^{2-}\right]$ in the equation (1).

  $\begin{array}{c}{K}_{\text{sp}}=\left(s\right)\left(s\right)\\ \sqrt{K_{\text{sp}}}=s^2\end{array}$

Substitute $8.5\times {10}^{-11}$ for $K_{\text{sp}}$ in above equation.

  $\begin{array}{c}s=\sqrt{8.5\times {10}^{-11}}\\ =9.21\times {10}^{-6}\end{array}$

The equilibrium reaction for ${\text{Ag}}_2{\text{CrO}}_4$ is as follows:

  ${\text{Ag}}_2{\text{CrO}}_4\left(s\right)\rightleftharpoons 2{\text{Ag}}^{+}\left(aq\right)+{\text{CrO}}_4^{2-}\left(aq\right)$

In expression for $K_{\text{sp}}$ molar concentration of solid is considered constant. Therefore, expression for ${\text{Ag}}_2{\text{CrO}}_4$ is as follows:

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

Ionization of solid ${\text{Ag}}_2{\text{CrO}}_4$ produces one mole of ${\text{CrO}}_4^{2-}$ and two moles of ${\text{Ag}}^{+}$. Therefore concentration of ${\text{Ag}}^{+}$ is two times of concentration of ${\text{CrO}}_4^{2-}$ and that is concentration of ${\text{CrO}}_4^{2-}$ is $s$ and concentration of ${\text{Ag}}^{+}$ is $2s$.

Substitute $2s$ for $\left[{\text{Ag}}^{+}\right]$ and $s$ for $\left[{\text{CrO}}_4^{2-}\right]$ in the equation (2).

  $\begin{array}{c}{K}_{\text{sp}}={\left(2s\right)}^2\left(s\right)\\ \sqrt[3]{\frac{K_{\text{sp}}}{4}}=s\end{array}$

Substitute $1.9\times {10}^{-12}$ for $K_{\text{sp}}$ in above equation.

  $\begin{array}{c}s=\sqrt[3]{\frac{1.9\times {10}^{-12}}{4}}\\ =7.80\times {10}^{-5}\end{array}$

Value of molar solubility is greater in ${\text{Ag}}_2{\text{CrO}}_4$ than ${\text{BaCrO}}_4$.