Chapter 15, Problem 35E

(a)

Interpretation Introduction

Interpretation: The solubility product of each pair of solids is given. The solid that has the smallest molar solubility is to be identified from each given pairs.

Concept introduction: Solubility is defined as the maximum amount of solute that can dissolve at a certain amount of solvent at certain temperature. The solubility product, $K_{\text{sp}}$ is the equilibrium constant that is applied when salt partially dissolve in a solvent. The solubility product of dissociation of ${\text{A}}_x{\text{B}}_y$ is calculated as,

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

Answer to Problem 35E

Answer

The compound ${\text{CaF}}_{\text{2}}$ has the smallest molar solubility as compared to ${\text{BaF}}_{\text{2}}$.

Explanation of Solution

Explanation

To identify: The solid that has the smallest molar solubility from ${\text{CaF}}_{\text{2}}$ and ${\text{BaF}}_{\text{2}}$.

The solubility of ${\text{CaF}}_{\text{2}}$ is $\underset{\_}{2.1\times {10}^{-4}mol/L}$.

Given

Solubility product of ${\text{CaF}}_{\text{2}}$ is $\text{4}\text{.0}\times {\text{10}}^{-\text{11}}$.

Since, solid ${\text{CaF}}_{\text{2}}$ is placed in contact with water. Therefore, compound present before the reaction is ${\text{CaF}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$. The dissociation reaction of ${\text{CaF}}_{\text{2}}$ is,

${\text{CaF}}_{\text{2}}\left(s\right)\stackrel{}{\to }{\text{Ca}}^{\text{2}+}\left(aq\right)+{\text{2F}}^{-}\left(aq\right)$

Since, ${\text{CaF}}_{\text{2}}$ does not dissolved initially, hence,

${\left[{\text{Ca}}^{\text{2}+}\right]}_{\text{initial}}={\left[{\text{F}}^{-}\right]}_{\text{initial}}=\text{0}$

The solubility of ${\text{CaF}}_{\text{2}}$ can be calculated from the concentration of ions at equilibrium.

It is assumed that $s\text{mol/L}$ of solid is dissolved to reach the equilibrium. The meaning of $\text{1}:\text{2}$ stoichiometry of salt is,

$s\text{mol/L}{\text{CaF}}_{\text{2}}\stackrel{}{\to }s\text{mol/L}{\text{Ca}}^{\text{2}+}+\text{2}s\text{mol/L}\text{F}$

Make the ICE table for the dissociation reaction of ${\text{CaF}}_{\text{2}}$.

$\begin{array}{ccccc}& {\text{CaF}}_{\text{2}}\left(s\right)& \stackrel{}{\rightleftharpoons }& {\text{Ca}}^{\text{2}+}\left(aq\right)& {\text{2F}}^{-}\left(aq\right)\\ \text{Initial}\text{(M):}& & & 0& 0\\ \text{Chang}\left(\text{M}\right):& & & s& 2s\\ \text{Equilibrium}\left(\text{M}\right):& & & s& 2s\end{array}$

Formula

The solubility product of ${\text{CaF}}_{\text{2}}$ is calculated as,

$\begin{array}{c}{K}_{\text{sp}}=\left[{\text{Ca}}^{\text{2}+}\right]{\left[{\text{F}}^{-}\right]}^{\text{2}}\\ =\left(s\right){\left(\text{2}s\right)}^{\text{2}}\end{array}$

Where,

• $K_{\text{sp}}$ is solubility product.
• $\left[{\text{Ca}}^{\text{2+}}\right]$ is concentration of ${\text{Ca}}^{\text{2+}}$.
• $\left[{\text{F}}^{-}\right]$ is concentration of ${\text{F}}^{-}$.
• $s$ is the solubility.

Substitute the values of $K_{\text{sp}}$ in the above expression.

$\begin{array}{c}{K}_{\text{sp}}=\left(s\right){\left(\text{2}s\right)}^{\text{2}}\\ \text{4}\text{.0}\times {\text{10}}^{-\text{11}}=\text{4}{s}^{\text{3}}\\ s=\underset{\_}{2.1\times {10}^{-4}mol/L}\end{array}$

The solubility of ${\text{BaF}}_{\text{2}}$ is $\underset{\_}{1.8\times {10}^{-2}mol/L}$.

Given

Solubility product of ${\text{BaF}}_{\text{2}}$ is $\text{2}\text{.4}\times {\text{10}}^{-\text{5}}$.

Since, solid ${\text{BaF}}_{\text{2}}$ is placed in contact with water. Therefore, compound present before the reaction is ${\text{BaF}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$. The dissociation reaction of ${\text{BaF}}_{\text{2}}$ is,

${\text{BaF}}_{\text{2}}\left(s\right)\stackrel{}{\to }{\text{Ba}}^{\text{2}+}\left(aq\right)+{\text{2F}}^{-}\left(aq\right)$

Since, ${\text{BaF}}_{\text{2}}$ does not dissolved initially, hence,

${\left[{\text{Ba}}^{\text{2}+}\right]}_{\text{initial}}={\left[{\text{F}}^{-}\right]}_{\text{initial}}=\text{0}$

The solubility of ${\text{BaF}}_{\text{2}}$ can be calculated from the concentration of ions at equilibrium.

It is assumed that $s\text{mol/L}$ of solid is dissolved to reach the equilibrium. The meaning of $\text{1}:\text{2}$ stoichiometry of salt is,

$s\text{mol/L}{\text{BaF}}_{\text{2}}\stackrel{}{\to }s\text{mol/L}{\text{Ba}}^{\text{2}+}+\text{2}s\text{mol/L}\text{F}$

Make the ICE table for the dissociation reaction of ${\text{BaF}}_{\text{2}}$.

$\begin{array}{ccccc}& {\text{BaF}}_{\text{2}}\left(s\right)& \stackrel{}{\rightleftharpoons }& {\text{Ba}}^{\text{2}+}\left(aq\right)& {\text{2F}}^{-}\left(aq\right)\\ \text{Initial}\text{(M):}& & & 0& 0\\ \text{Chang}\left(\text{M}\right):& & & s& 2s\\ \text{Equilibrium}\left(\text{M}\right):& & & s& 2s\end{array}$

Formula

The solubility product of ${\text{BaF}}_{\text{2}}$ is calculated as,

$\begin{array}{c}{K}_{\text{sp}}=\left[{\text{Ba}}^{\text{2}+}\right]{\left[{\text{F}}^{-}\right]}^{\text{2}}\\ =\left(s\right){\left(\text{2}s\right)}^{\text{2}}\end{array}$

Where,

• $K_{\text{sp}}$ is solubility product.
• $\left[{\text{Ba}}^{\text{2+}}\right]$ is concentration of ${\text{Ba}}^{\text{2+}}$.
• $\left[{\text{F}}^{-}\right]$ is concentration of ${\text{F}}^{-}$.
• $s$ is the solubility.

Substitute the values of $K_{\text{sp}}$ in the above expression.

$\begin{array}{c}{K}_{\text{sp}}=\left(s\right){\left(\text{2}s\right)}^{\text{2}}\\ \text{2}\text{.4}\times {\text{10}}^{-\text{5}}=\text{4}{s}^{\text{3}}\\ s=\underset{\_}{1.8\times {10}^{-2}mol/L}\end{array}$

The ${\text{CaF}}_{\text{2}}$ has the smallest molar solubility in comparison of ${\text{BaF}}_{\text{2}}$.

The molar solubility of ${\text{CaF}}_{\text{2}}$ is $\text{2}\text{.1}\times {\text{10}}^{-\text{4}}\text{mol/L}$ whereas the molar solubility of ${\text{BaF}}_{\text{2}}$ is $\text{1}\text{.8}\times {\text{10}}^{-\text{2}}\text{mol/L}$. Hence, ${\text{CaF}}_{\text{2}}$ has the smallest molar solubility in comparison of ${\text{BaF}}_{\text{2}}$.

(b)

Interpretation Introduction

Interpretation: The solubility product of each pair of solids is given. The solid that has the smallest molar solubility is to be identified from each given pairs.

Concept introduction: Solubility is defined as the maximum amount of solute that can dissolve at a certain amount of solvent at certain temperature. The solubility product, $K_{\text{sp}}$ is the equilibrium constant that is applied when salt partially dissolve in a solvent. The solubility product of dissociation of ${\text{A}}_x{\text{B}}_y$ is calculated as,

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

Answer to Problem 35E

Answer

The compound ${\text{FePO}}_{\text{4}}$ has the smallest molar solubility as compared to ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$.

Explanation of Solution

Explanation

To identify: The solid that has the smallest molar solubility from ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ and ${\text{FePO}}_{\text{4}}$.

The solubility of ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is $\underset{\_}{1.6\times {10}^{-7}mol/L}$.

Given

Solubility product of ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is $\text{1}\text{.3}\times {\text{10}}^{-\text{32}}$.

Since, solid ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is placed in contact with water. Therefore, compound present before the reaction is ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$. The dissociation reaction of ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is,

${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)\stackrel{}{\to }{\text{3Ca}}^{\text{2}+}\left(aq\right)+{\text{2PO}}_{\text{4}}{}^{\text{3}-}\left(aq\right)$

Since, ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ does not dissolved initially, hence,

${\left[{\text{Ca}}^{\text{2}+}\right]}_{\text{initial}}={\left[{\text{PO}}_{\text{4}}{}^{\text{3}-}\right]}_{\text{initial}}=\text{0}$

The solubility of ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ can be calculated from the concentration of ions at equilibrium.

It is assumed that $s\text{mol/L}$ of solid is dissolved to reach the equilibrium. The meaning of $\text{3}:\text{2}$ stoichiometry of salt is,

$s\text{mol/L}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\stackrel{}{\to }\text{3}s\text{mol/L}{\text{Ca}}^{\text{2}+}+\text{2}s\text{mol/L}{\text{PO}}_{\text{4}}$

Make the ICE table for the dissociation reaction of ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$.

$\begin{array}{ccccc}& {\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)& \stackrel{}{\rightleftharpoons }& {\text{3Ca}}^{\text{2}+}\left(aq\right)& {\text{2PO}}_{\text{4}}\left(aq\right)\\ \text{Initial}\text{(M):}& & & 0& 0\\ \text{Chang}\left(\text{M}\right):& & & 3s& 2s\\ \text{Equilibrium}\left(\text{M}\right):& & & 3s& 2s\end{array}$

Formula

The solubility product of ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is calculated as,

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

Where,

• $K_{\text{sp}}$ is solubility product.
• $\left[{\text{Ca}}^{\text{2+}}\right]$ is concentration of ${\text{Ca}}^{\text{2+}}$.
• $\left[{\text{PO}}_{\text{4}}{}^{\text{3}-}\right]$ is concentration of ${\text{PO}}_{\text{4}}{}^{\text{3}-}$.
• $s$ is the solubility.

Substitute the values of $K_{\text{sp}}$ in the above expression.

$\begin{array}{c}{K}_{\text{sp}}={\left(\text{3}s\right)}^{\text{3}}{\left(\text{2}s\right)}^{\text{2}}\\ \text{1}\text{.3}\times {\text{10}}^{-\text{32}}=\text{108}{s}^{\text{5}}\\ s=\underset{\_}{1.6\times {10}^{-7}mol/L}\end{array}$

The solubility of ${\text{FePO}}_{\text{4}}$ is $\underset{\_}{1.0\times {10}^{-11}mol/L}$.

Given

Solubility product of ${\text{FePO}}_{\text{4}}$ is $\text{2}\text{.4}\times {\text{10}}^{-\text{5}}$.

Since, solid ${\text{FePO}}_{\text{4}}$ is placed in contact with water. Therefore, compound present before the reaction is ${\text{FePO}}_{\text{4}}$ and ${\text{H}}_{\text{2}}\text{O}$. The dissociation reaction of ${\text{FePO}}_{\text{4}}$ is,

${\text{FePO}}_{\text{4}}\left(s\right)\stackrel{}{\to }{\text{Fe}}^{\text{3}+}\left(aq\right)+{\text{PO}}_{\text{4}}{}^{\text{3}-}\left(aq\right)$

Since, ${\text{FePO}}_{\text{4}}$ does not dissolved initially, hence,

${\left[{\text{Fe}}^{\text{3}+}\right]}_{\text{initial}}={\left[{\text{PO}}_{\text{4}}{}^{\text{3}-}\right]}_{\text{initial}}=\text{0}$

The solubility of ${\text{FePO}}_{\text{4}}$ can be calculated from the concentration of ions at equilibrium.

It is assumed that $s\text{mol/L}$ of solid is dissolved to reach the equilibrium. The meaning of $\text{1}:\text{1}$ stoichiometry of salt is,

$s\text{mol/L}{\text{FePO}}_{\text{4}}\stackrel{}{\to }s\text{mol/L}{\text{Fe}}^{\text{3}+}+s\text{mol/L}{\text{PO}}_{\text{4}}$

Make the ICE table for the dissociation reaction of ${\text{FePO}}_{\text{4}}$.

$\begin{array}{ccccc}& {\text{FePO}}_{\text{4}}\left(s\right)& \stackrel{}{\rightleftharpoons }& {\text{Fe}}^{\text{3}+}\left(aq\right)& {\text{PO}}_{\text{4}}\left(aq\right)\\ \text{Initial}\text{(M):}& & & 0& 0\\ \text{Chang}\left(\text{M}\right):& & & s& s\\ \text{Equilibrium}\left(\text{M}\right):& & & s& s\end{array}$

Formula

The solubility product of ${\text{FePO}}_{\text{4}}$ is calculated as,

$\begin{array}{c}{K}_{\text{sp}}=\left[{\text{Fe}}^{\text{3}+}\right]\left[{\text{PO}}_{\text{4}}\right]\\ =\left(s\right)\left(s\right)\end{array}$

Where,

• $K_{\text{sp}}$ is solubility product.
• $\left[{\text{Fe}}^{\text{3+}}\right]$ is concentration of ${\text{Fe}}^{\text{3+}}$.
• $\left[{\text{PO}}_{\text{4}}{}^{\text{3}-}\right]$ is concentration of ${\text{PO}}_{\text{4}}{}^{\text{3}-}$.
• $s$ is the solubility.

Substitute the values of $K_{\text{sp}}$ in the above expression.

$\begin{array}{c}{K}_{\text{sp}}=\left(s\right)\left(s\right)\\ \text{1}\text{.0}\times {\text{10}}^{-\text{22}}=s^{\text{2}}\\ s=\underset{\_}{1.0\times {10}^{-11}mol/L}\end{array}$

The ${\text{FePO}}_{\text{4}}$ has the smallest molar solubility in comparison of ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$.

The molar solubility of ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is $\text{1}\text{.6}\times {\text{10}}^{-\text{7}}\text{mol/L}$ whereas the molar solubility of ${\text{FePO}}_{\text{4}}$ is $\text{1}\text{.0}\times {\text{10}}^{-\text{11}}\text{mol/L}$. Hence, ${\text{FePO}}_{\text{4}}$ has the smallest molar solubility in comparison of ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$.