Chapter 14, Problem 14.69QE

(a)

Interpretation Introduction

Interpretation:

The solubility product constant for ${\text{BaCrO}}_{\text{4}}$ has t o be determined.

Concept Introduction:

Solubility product is the equilibrium constant for the reaction that occurs when an ionic compound is dissolved to produce its constituent ions. It is represented by $K_{\text{sp}}$. Consider ${\text{A}}_x{\text{B}}_y$ to be an ionic compound. Its dissociation occurs as follows:

  ${\text{A}}_x{\text{B}}_y\rightleftharpoons x{\text{A}}^{y+}+y{\text{A}}^{x-}$

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

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

Answer to Problem 14.69QE

The solubility product constant of ${\text{BaCrO}}_{\text{4}}$ is $1.2\times {10}^{-10}$.

Explanation of Solution

The dissociation reaction of ${\text{BaCrO}}_{\text{4}}$ is as follows:

  ${\text{BaCrO}}_{\text{4}}\rightleftharpoons {\text{Ba}}^{2+}+{\text{CrO}}_4^{2-}$

The solubility of ${\text{BaCrO}}_{\text{4}}$ is $1.1\times {10}^{-5}M$. Therefore the concentration of both ${\text{Ag}}^{+}$ and ${\text{CrO}}_4^{2-}$ is $1.1\times {10}^{-5}M$.

The formula to calculate the molar solubility of ${\text{BaCrO}}_{\text{4}}$ is as follows:

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

Substitute $1.1\times {10}^{-5}M$ for $\left[{\text{Ba}}^{2+}\right]$ and $1.1\times {10}^{-5}M$ for $\left[{\text{CrO}}_4^{2-}\right]$ in equation (1).

  $\begin{array}{c}{K}_{\text{sp}}=\left(1.1\times {10}^{-5}\right)\left(1.1\times {10}^{-5}\right)\\ =1.2\times {10}^{-10}\end{array}$

Therefore the solubility product constant of ${\text{BaCrO}}_{\text{4}}$ is $1.2\times {10}^{-10}$.

(b)

Interpretation Introduction

Interpretation:

The solubility product constant for ${\text{CsMnO}}_{\text{4}}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 14.69QE

The solubility product constant of ${\text{CsMnO}}_{\text{4}}$ is $76\times {10}^{-18}$.

Explanation of Solution

The dissociation reaction of ${\text{CsMnO}}_{\text{4}}$ is as follows:

  ${\text{CsMnO}}_{\text{4}}\rightleftharpoons {\text{Cs}}^{+}+{\text{MnO}}_4^{-}$

The formula to calculate the molar solubility of ${\text{CsMnO}}_{\text{4}}$ is as follows:

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

The solubility of ${\text{CsMnO}}_{\text{4}}$ can be calculated as follows:

  $\begin{array}{c}{\text{Solubility of CsMnO}}_{\text{4}}=\left(\frac{0.22\text{ g}}{100\text{ mL}}\right)\left(\frac{1\text{ mL}}{{10}^{-3}\text{L}}\right)\\ =0.0000022\text{ g/L}\end{array}$

The formula to convert solubility of ${\text{CsMnO}}_{\text{4}}$ to moles per litre is as follows:

  $\text{Molar solubility}\left(\text{mol/L}\right)=\frac{\text{Solubility}\left(\text{g/L}\right)}{{\text{Molar mass of CsMnO}}_{\text{4}}}$        (3)

Substitute $0.0000022\text{ g/L}$ for solubility of ${\text{CsMnO}}_{\text{4}}$ and $251.841\text{ g/mol}$ for the molar mass of ${\text{CsMnO}}_{\text{4}}$ in equation (3).

  $\begin{array}{c}\text{Molar solubility}\left(\text{mol/L}\right)=\left(\frac{0.0000022\text{ g}}{1\text{ L}}\right)\left(\frac{1\text{ mol}}{\text{251}\text{.841 g}}\right)\\ =8.7\times {10}^{-9}M\end{array}$

The solubility of ${\text{CsMnO}}_{\text{4}}$ is $8.7\times {10}^{-9}M$. Therefore the concentration of both ${\text{Cs}}^{+}$ and ${\text{MnO}}_4^{-}$ is $8.7\times {10}^{-9}M$.

Substitute $8.7\times {10}^{-9}M$ for $\left[{\text{Cs}}^{+}\right]$ and $8.7\times {10}^{-9}M$ for $\left[{\text{MnO}}_4^{-}\right]$ in equation (2).

  $\begin{array}{c}{K}_{\text{sp}}=\left(8.7\times {10}^{-9}\right)\left(8.7\times {10}^{-9}\right)\\ =76\times {10}^{-18}\end{array}$

Therefore the solubility product constant of ${\text{CsMnO}}_{\text{4}}$ is $76\times {10}^{-18}$.

(c)

Interpretation Introduction

Interpretation:

The solubility product constant for ${\text{Ag}}_3{\text{PO}}_{\text{4}}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 14.69QE

The solubility product constant of ${\text{Ag}}_3{\text{PO}}_{\text{4}}$ is $9.67\times {10}^{-17}$.

Explanation of Solution

The dissociation reaction of ${\text{Ag}}_3{\text{PO}}_{\text{4}}$ is as follows:

  ${\text{Ag}}_3{\text{PO}}_{\text{4}}\rightleftharpoons 3{\text{Ag}}^{+}+{\text{PO}}_4^{3-}$

The ICE table for the above reaction is as follows:

$\begin{array}{cccccc}\text{Equation}& {\text{Ag}}_3{\text{PO}}_{\text{4}}& \rightleftharpoons & {\text{3Ag}}^{+}& +& {\text{PO}}_4^{3-}\\ \text{Initial}\left(M\right)& \text{Solid}& & 0& & 0\\ \text{Change}\left(M\right)& -4.4\times {10}^{-5}& & 3\left(4.4\times {10}^{-5}\right)& & 4.4\times {10}^{-5}\\ \text{Equilibrium}\left(M\right)& \text{Solid}& & 13\times {10}^{-5}& & 4.4\times {10}^{-5}\end{array}$

The formula to calculate the molar solubility of ${\text{Ag}}_3{\text{PO}}_{\text{4}}$ is as follows:

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

Substitute $13\times {10}^{-5}M$ for $\left[{\text{Ag}}^{+}\right]$ and $4.4\times {10}^{-5}M$ for $\left[{\text{PO}}_4^{3-}\right]$ in equation (4).

  $\begin{array}{c}{K}_{\text{sp}}={\left(13\times {10}^{-5}\right)}^3\left(4.4\times {10}^{-5}M\right)\\ =9.67\times {10}^{-17}\end{array}$

Therefore the solubility product constant of ${\text{Ag}}_3{\text{PO}}_{\text{4}}$ is $9.67\times {10}^{-17}$.