Chapter 16, Problem 41E

Calculate the solubility (in moles per liter) of Fe(OH)₃ (K_sp = 4 × 10⁻³⁸) in each of the following.

a. water

b. a solution buffered at pH = 5.0

c. a solution buffered at pH = 11.0

Interpretation Introduction

Interpretation: The solubility product of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ is given. The solubility (in $\text{mol/L}$) of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ is to be calculated for each given conditions.

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,

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

The ion product of water is calculated using the formula,

$K_{\text{w}}=\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]$

Answer to Problem 41E

Answer

The solubility of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ in water is $\underset{\_}{4\times {10}^{-17}mol/L}$.

Explanation of Solution

Explanation

To determine: The solubility of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ in water.

The solubility of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ in water is $\underset{\_}{4\times {10}^{-17}mol/L}$.

Given

Solubility product of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ is $\text{4}\times {\text{10}}^{-\text{38}}$.

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

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

Since, ${\text{Fe}}^{\text{3}+}$ does not dissolved initially, hence,

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

The solubility of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ 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 $\text{1}:\text{3}$ stoichiometry of salt is,

$s\text{mol/L}\text{Fe}{\left(\text{OH}\right)}_{\text{3}}\stackrel{}{\to }s\text{mol/L}{\text{Fe}}^{\text{3}+}+\text{3}s\text{mol/L}\text{OH}$

Since, in a neutral solution, the initial concentration of ${\text{OH}}^{-}$ is $\text{1}\times {\text{10}}^{-\text{7}}$. Hence, in a saturated solution, the concentration of ${\text{OH}}^{-}$ is $\text{1}\times {\text{10}}^{-\text{7}}+\text{3}s$. But the value of $s$ is expected to be very small, therefore, the concentration of ${\text{OH}}^{-}$ is approximated to,

$\text{1}\times {\text{10}}^{-\text{7}}+\text{3}s\approx \text{1}\times {\text{10}}^{-\text{7}}$

Make the ICE table for the dissociation reaction of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$.

$\begin{array}{ccccc}& \text{Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)& \stackrel{}{\rightleftharpoons }& {\text{Fe}}^{3+}\left(aq\right)& {\text{3OH}}^{-}\left(aq\right)\\ \text{Initial}\left(\text{M}\right):& & & 0& \text{1}\times {\text{10}}^{-\text{7}}\\ \text{Chang}\left(\text{M}\right):& & & s& 3s\\ \text{Equilibrium}\left(\text{M}\right):& & & s& \text{1}\times {\text{10}}^{-\text{7}}+3s\approx \text{1}\times {\text{10}}^{-\text{7}}\end{array}$

Formula

The solubility product of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ is calculated as,

$\begin{array}{c}{K}_{\text{sp}}=\left[{\text{Fe}}^{\text{3}+}\right]{\left[{\text{OH}}^{-}\right]}^{\text{3}}\\ =\left(s\right){\left(\text{1}\times {\text{10}}^{-\text{7}}\right)}^{\text{3}}\end{array}$

Where,

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

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

$\begin{array}{c}{K}_{\text{sp}}=\left(s\right){\left(\text{1}\times {\text{10}}^{-\text{7}}\right)}^{\text{3}}\\ \text{4}\times {\text{10}}^{-\text{38}}=s{\left(\text{1}\times {\text{10}}^{-\text{7}}\right)}^{\text{3}}\\ s=\underset{\_}{4\times {10}^{-17}mol/L}\end{array}$

Interpretation Introduction

Interpretation: The solubility product of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ is given. The solubility (in $\text{mol/L}$) of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ is to be calculated for each given conditions.

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,

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

The ion product of water is calculated using the formula,

$K_{\text{w}}=\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]$

Answer to Problem 41E

Answer

The solubility of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ in a buffered solution at $\text{pH}=\text{5}\text{.0}$ is $\underset{\_}{4\times {10}^{-11}mol/L}$.

Explanation of Solution

Explanation

To determine: The solubility of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ in a buffered solution at $\text{pH}=\text{5}\text{.0}$.

The concentration of ${\text{H}}^{+}$ is $\underset{\_}{1\times {10}^{-5}M}$.

Given

Solubility product of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ is $\text{4}\times {\text{10}}^{-\text{38}}$.

The $\text{pH}$ value of buffered solution is $\text{5}\text{.0}$.

Formula

The $\text{pH}$ is calculated using the formula,

$\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]$

Substitute the value of $\text{pH}$ in the above equation.

$\begin{array}{c}\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]\\ \text{5}\text{.0}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]\\ \left[{\text{H}}^{+}\right]=\underset{\_}{1\times {10}^{-5}M}\end{array}$

The concentration of ${\text{OH}}^{-}$ is $\underset{\_}{1\times 10^{-9}M}$.

The concentration of ${\text{H}}^{+}$ is $\text{1}\times {\text{10}}^{-\text{5}}\text{M}$.

Formula

The ion product of water is calculated using the formula,

$K_{\text{w}}=\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]$

Where,

• $K_{\text{w}}$ is ion product of water $\left(\text{1}\times {\text{10}}^{-\text{14}}\right)$.

Substitute the values of $\left[{\text{H}}^{+}\right]$ and $K_{\text{w}}$ in the above equation.

$\begin{array}{c}{K}_{\text{w}}=\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]\\ \left(\text{1}\times {\text{10}}^{-\text{14}}\right)=\left(\text{1}\times {\text{10}}^{-\text{5}}\text{M}\right)\left[{\text{OH}}^{-}\right]\\ \left[{\text{OH}}^{-}\right]=\underset{\_}{1\times 10^{-9}M}\end{array}$

The solubility of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ in a buffered solution at $\text{pH}=\text{5}\text{.0}$ is $\underset{\_}{4\times {10}^{-11}mol/L}$.

The solubility of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ 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 $\text{1}:\text{3}$ stoichiometry of salt is,

$s\text{mol/L}\text{Fe}{\left(\text{OH}\right)}_{\text{3}}\stackrel{}{\to }s\text{mol/L}{\text{Fe}}^{\text{3}+}+\text{3}s\text{mol/L}\text{OH}$

Make the ICE table for the dissociation reaction of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$.

$\begin{array}{ccccc}& \text{Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)& \stackrel{}{\rightleftharpoons }& {\text{Fe}}^{3+}\left(aq\right)& {\text{3OH}}^{-}\left(aq\right)\\ \text{Initial}\left(\text{M}\right):& & & 0& \text{1}\times {\text{10}}^{-\text{9}}\\ \text{Chang}\left(\text{M}\right):& & & s& 3s\\ \text{Equilibrium}\left(\text{M}\right):& & & s& \text{1}\times {\text{10}}^{-\text{9}}+3s\end{array}$

Formula

The solubility product of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ is calculated as,

$\begin{array}{c}{K}_{\text{sp}}=\left[{\text{Fe}}^{\text{3}+}\right]{\left[{\text{OH}}^{-}\right]}^{\text{3}}\\ =\left(s\right){\left(\text{1}\times {\text{10}}^{-\text{9}}+\text{3}s\right)}^{\text{3}}\end{array}$

Where,

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

Since, the value of $s$ is expected to be very small value, hence, it is assumed that,

$\text{1}\times {\text{10}}^{-\text{9}}+\text{3}s\approx \text{1}\times {\text{10}}^{-\text{9}}$

Substitute this value in the solubility product expression.

$\begin{array}{c}{K}_{\text{sp}}=\left(s\right){\left(\text{1}\times {\text{10}}^{-\text{9}}+\text{3}s\right)}^{\text{3}}\\ =\left(s\right){\left(\text{1}\times {\text{10}}^{-\text{9}}\right)}^{\text{3}}\end{array}$

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

$\begin{array}{c}{K}_{\text{sp}}=\left(s\right){\left(\text{1}\times {\text{10}}^{-\text{9}}\right)}^{\text{3}}\\ \text{4}\times {\text{10}}^{-\text{38}}=\left(s\right){\left(\text{1}\times {\text{10}}^{-\text{9}}\right)}^{\text{3}}\\ s=\underset{\_}{4\times {10}^{-11}mol/L}\end{array}$

Interpretation Introduction

Interpretation: The solubility product of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ is given. The solubility (in $\text{mol/L}$) of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ is to be calculated for each given conditions.

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,

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

The ion product of water is calculated using the formula,

$K_{\text{w}}=\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]$

Answer to Problem 41E

Answer

The solubility of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ in a buffered solution at $\text{pH}=\text{11}\text{.0}$ is $\underset{\_}{4\times {10}^{-29}mol/L}$.

Explanation of Solution

Explanation

To determine: The solubility of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ in a buffered solution at $\text{pH}=\text{11}\text{.0}$.

The concentration of ${\text{H}}^{+}$ is $\underset{\_}{1\times {10}^{-11}M}$.

Given

Solubility product of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ is $\text{4}\times {\text{10}}^{-\text{38}}$.

The $\text{pH}$ value of buffered solution is $\text{11}\text{.0}$.

Formula

The $\text{pH}$ is calculated using the formula,

$\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]$

Substitute the value of $\text{pH}$ in the above equation.

$\begin{array}{c}\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]\\ \text{11}\text{.0}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]\\ \left[{\text{H}}^{+}\right]=\underset{\_}{1\times {10}^{-11}M}\end{array}$

The concentration of ${\text{OH}}^{-}$ is $\underset{\_}{1\times 10^{-3}M}$.

The concentration of ${\text{H}}^{+}$ is $\text{1}\times {\text{10}}^{-\text{11}}\text{M}$.

Formula

The ion product of water is calculated using the formula,

$K_{\text{w}}=\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]$

Where,

• $K_{\text{w}}$ is ion product of water $\left(\text{1}\times {\text{10}}^{-\text{14}}\right)$.

Substitute the values of $\left[{\text{H}}^{+}\right]$ and $K_{\text{w}}$ in the above equation.

$\begin{array}{c}{K}_{\text{w}}=\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]\\ \left(\text{1}\times {\text{10}}^{-\text{14}}\right)=\left(\text{1}\times {\text{10}}^{-\text{11}}\text{M}\right)\left[{\text{OH}}^{-}\right]\\ \left[{\text{OH}}^{-}\right]=\underset{\_}{1\times 10^{-3}M}\end{array}$

The solubility of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ in a buffered solution at $\text{pH}=\text{11}\text{.0}$ is $\underset{\_}{4\times {10}^{-29}mol/L}$.

The solubility of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ 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 $\text{1}:\text{3}$ stoichiometry of salt is,

$s\text{mol/L}\text{Fe}{\left(\text{OH}\right)}_{\text{3}}\stackrel{}{\to }s\text{mol/L}{\text{Fe}}^{\text{3}+}+\text{3}s\text{mol/L}\text{OH}$

Make the ICE table for the dissociation reaction of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$.

$\begin{array}{ccccc}& \text{Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)& \stackrel{}{\rightleftharpoons }& {\text{Fe}}^{3+}\left(aq\right)& {\text{3OH}}^{-}\left(aq\right)\\ \text{Initial}\left(\text{M}\right):& & & 0& \text{1}\times {\text{10}}^{-\text{3}}\\ \text{Chang}\left(\text{M}\right):& & & s& 3s\\ \text{Equilibrium}\left(\text{M}\right):& & & s& \text{1}\times {\text{10}}^{-\text{3}}+3s\end{array}$

Formula

The solubility product of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ is calculated as,

$\begin{array}{c}{K}_{\text{sp}}=\left[{\text{Fe}}^{\text{3}+}\right]{\left[{\text{OH}}^{-}\right]}^{\text{3}}\\ =\left(s\right){\left(\text{1}\times {\text{10}}^{-\text{3}}+\text{3}s\right)}^{\text{3}}\end{array}$

Where,

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

Since, the value of $s$ is expected to be very small value, hence, it is assumed that,

$\text{1}\times {\text{10}}^{-\text{3}}+\text{3}s\approx \text{1}\times {\text{10}}^{-\text{3}}$

Substitute this value in the solubility product expression.

$\begin{array}{c}{K}_{\text{sp}}=\left(s\right){\left(\text{1}\times {\text{10}}^{-\text{3}}+\text{3}s\right)}^{\text{3}}\\ =\left(s\right){\left(\text{1}\times {\text{10}}^{-\text{3}}\right)}^{\text{3}}\end{array}$

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

$\begin{array}{c}{K}_{\text{sp}}=\left(s\right){\left(\text{1}\times {\text{10}}^{-\text{3}}\right)}^{\text{3}}\\ \text{4}\times {\text{10}}^{-\text{38}}=\left(s\right){\left(\text{1}\times {\text{10}}^{-\text{3}}\right)}^{\text{3}}\\ s=\underset{\_}{4\times {10}^{-29}mol/L}\end{array}$