Chapter 15, Problem 28E

(a)

Interpretation Introduction

Interpretation: The solubility product of ${\text{PbI}}_{\text{2}},{\text{CdCO}}_{\text{3}},$ and ${\text{Sr}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is given. By using these values, the solubility (in $\text{mol/L}$) of ${\text{PbI}}_{\text{2}},{\text{CdCO}}_{\text{3}},$ and ${\text{Sr}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is to be calculated.

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,

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

Where,

• $x$ is coefficient of concentration of $\text{A}$.
• $y$ is coefficient of concentration of $\text{B}$.

Answer to Problem 28E

The solubility of ${\text{PbI}}_{\text{2}}$ is $\underset{\_}{1.5\times {10}^{-3}mol/L}$.

Explanation of Solution

Explanation

To determine: The solubility of ${\text{PbI}}_{\text{2}}$ (in $\text{mol/L}$) from the given $K_{\text{sp}}$ value.

The solubility of ${\text{PbI}}_{\text{2}}$ is $\underset{\_}{1.5\times {10}^{-3}mol/L}$.

Given

Solubility product of ${\text{PbI}}_{\text{2}}$ is $\text{1}\text{.4}\times {\text{10}}^{-\text{8}}$.

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

${\text{PbI}}_{\text{2}}\left(s\right)\stackrel{}{\to }{\text{Pb}}^{\text{2}+}\left(aq\right)+{\text{2I}}^{-}\left(aq\right)$

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

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

The solubility of ${\text{PbI}}_{\text{2}}$ 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,

$s\text{mol/L}{\text{PbI}}_{\text{2}}\stackrel{}{\to }s\text{mol/L}{\text{Pb}}^{\text{2}+}+\text{2}s\text{mol/L}\text{I}$

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

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

Formula

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

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

Where,

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

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

$\begin{array}{c}{K}_{\text{sp}}=\left(s\right){\left(\text{2}s\right)}^{\text{2}}\\ \text{1}\text{.4}\times {\text{10}}^{-\text{8}}=\text{4}{s}^{\text{3}}\\ s=\underset{\_}{1.5\times {10}^{-3}mol/L}\end{array}$

(b)

Interpretation Introduction

Interpretation: The solubility product of ${\text{PbI}}_{\text{2}},{\text{CdCO}}_{\text{3}},$ and ${\text{Sr}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is given. By using these values, the solubility (in $\text{mol/L}$) of ${\text{PbI}}_{\text{2}},{\text{CdCO}}_{\text{3}},$ and ${\text{Sr}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is to be calculated.

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,

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

Where,

• $x$ is coefficient of concentration of $\text{A}$.
• $y$ is coefficient of concentration of $\text{B}$.

Answer to Problem 28E

The solubility of ${\text{CdCO}}_{\text{3}}$ is $\underset{\_}{2.3\times {10}^{-6}mol/L}$.

Explanation of Solution

Explanation

To determine: The solubility of ${\text{CdCO}}_{\text{3}}$ (in $\text{mol/L}$) from the given $K_{\text{sp}}$ value.

The solubility of ${\text{CdCO}}_{\text{3}}$ is $\underset{\_}{2.3\times {10}^{-6}mol/L}$.

Given

Solubility product of ${\text{CdCO}}_{\text{3}}$ is $\text{5}\text{.2}\times {\text{10}}^{-\text{12}}$.

Since, solid ${\text{CdCO}}_{\text{3}}$ is placed in contact with water. Therefore, compound present before the reaction is ${\text{CdCO}}_{\text{3}}$ and ${\text{H}}_{\text{2}}\text{O}$. The dissociation reaction of ${\text{CdCO}}_{\text{3}}$ is,

${\text{CdCO}}_{\text{3}}\left(s\right)\stackrel{}{\to }{\text{Cd}}^{\text{2}+}\left(aq\right)+{\text{CO}}_{\text{3}}{}^{\text{2}-}\left(aq\right)$

Since, ${\text{CdCO}}_{\text{3}}$ does not dissolved initially, hence,

${\left[{\text{Cd}}^{\text{2}+}\right]}_{\text{initial}}={\left[{\text{CO}}_{\text{3}}{}^{\text{2}-}\right]}_{\text{initial}}=\text{0}$

The solubility of ${\text{CdCO}}_{\text{3}}$ 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,

$s\text{mol/L}{\text{CdCO}}_{\text{3}}\stackrel{}{\to }s\text{mol/L}{\text{Cd}}^{\text{2}+}+s\text{mol/L}{\text{CO}}_{\text{3}}$

Make the ICE table for the dissociation reaction of ${\text{CdCO}}_{\text{3}}$.

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

Formula

The solubility product of ${\text{CdCO}}_{\text{3}}$ is calculated as,

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

Where,

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

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

$\begin{array}{c}{K}_{\text{sp}}=\left(s\right)\left(s\right)\\ \text{5}\text{.2}\times {\text{10}}^{-\text{12}}=s^{\text{2}}\\ s=\underset{\_}{2.3\times {10}^{-6}mol/L}\end{array}$

(c)

Interpretation Introduction

Interpretation: The solubility product of ${\text{PbI}}_{\text{2}},{\text{CdCO}}_{\text{3}},$ and ${\text{Sr}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is given. By using these values, the solubility (in $\text{mol/L}$) of ${\text{PbI}}_{\text{2}},{\text{CdCO}}_{\text{3}},$ and ${\text{Sr}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is to be calculated.

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,

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

Where,

• $x$ is coefficient of concentration of $\text{A}$.
• $y$ is coefficient of concentration of $\text{B}$.

Answer to Problem 28E

(c) The solubility of ${\text{Sr}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is $\underset{\_}{6.5\times {10}^{-7}mol/L}$.

Explanation of Solution

To determine: The solubility of ${\text{Sr}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ (in $\text{mol/L}$) from the given $K_{\text{sp}}$ value.

The solubility of ${\text{Sr}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is $\underset{\_}{2.5\times {10}^{-7}mol/L}$.

Given:

Solubility product of ${\text{Sr}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is $\text{1}\text{.0}\times {\text{10}}^{-\text{31}}$.

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

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

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

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

The solubility of ${\text{Sr}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ 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,

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

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

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

Formula:

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

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

Where,

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

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

$\begin{array}{c}{K}_{\text{sp}}={\left(\text{3}s\right)}^{\text{3}}{\left(\text{2}s\right)}^{\text{2}}\\ \text{1}\text{.0}\times {\text{10}}^{-\text{31}}=\text{108}{s}^{\text{5}}\\ s=\underset{\_}{2.5\times {10}^{-7}mol/L}\end{array}$