Chapter 17, Problem 69QP

Calculate the molar solubility of $\text{Fe}{\left(\text{OH}\right)}_{\text{2}}$ in a solution buffered at (a) a pH of 8.00 and (b) a pH of 10.00.

Interpretation Introduction

Interpretation:

The molar solubility of $\text{Fe}{\left(\text{OH}\right)}_{\text{2}}$ in the solution buffered at $\text{pH}=8$ and $\text{pH}=10$ is to be calculated.

Concept introduction:

The amount of solute dissolved in a given volume of the solvent to form a saturated solution at a given temperature is termed as the solubility of the solute in the solvent at that temperature.

The solubility product of a sparinglysoluble salt is given as the product of the concentration of the ions raised to the power equal to the number of times the ion occurs in the equation after the dissociation of the electrolyte.

The number of moles of solute dissolved per litre of solution is called molar solubility.

Molar solubility of a substance is defined as the number of moles of solution present in one liter of saturated solution.

The unit of molar solubility is $\text{mol}/\text{L}$.

At a given temperature, the product of molar concentrations of the ions of salt present in the solution is known as the solubility product of the salt. It is represented by $K_{sp}$.

Higher is the value of solubility product of a salt, more is its solubility.

The presence of common ions in the solution decreases the solubility of a given compound.

For a general reaction: $\text{AB}\left(s\right)\rightleftharpoons {\text{A}}^{+}\left(aq\right)+{\text{B}}^{-}\left(aq\right)$

The solubility product can be calculated by the expression as: ${\text{K}}_{\text{sp}}{\text{= [A}}^{\text{+}}{\text{][B}}^{-}\text{]}$

Here, ${\text{K}}_{\text{sp}}$ is the solubility product constant and $sp$ stands for solubility product.

The expression of $\text{pOH}$ is represented as:

$\text{pOH}=-\text{log}\left[\text{OH}\right]$

$[{\text{OH}}^{-}]={10}^{-\text{pOH}}$

The relation between $\text{pH}$ and $\text{pOH}$ can be expressed as: $\text{pH}+\text{pOH}=14$

Answer to Problem 69QP

Solution:

a)

The molar solubility of $\text{Fe}{\left(\text{OH}\right)}_{\text{2}}$ is $\text{0}\text{.016}\text{M}$ at pH equal to $8$

b)

The molar solubility of $\text{Fe}{\left(\text{OH}\right)}_{\text{2}}$ is $1.6\times {10}^{-6}\text{M}$ at pH equal to $10$.

Explanation of Solution

a)ApH is $8.00$

The equation of the dissociation of $\text{Fe}{\left(\text{OH}\right)}_{\text{2}}$ is as follows:

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

The concentration of $\left[{\text{OH}}^{-}\right]$ is calculated as follows:

$\text{pOH}$ is calculated by using the expression as follows:

$\begin{array}{l}\text{pOH}=14-\text{pH}\\ =14-8\\ =6\end{array}$

The concentration of $\left[{\text{OH}}^{-}\right]$ is calculated by using the expression as follows:

$[{\text{OH}}^{-}]={10}^{-\text{pOH}}$.

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

$[{\text{OH}}^{-}]=1\times {10}^{-6}\text{M}$

The concentration of $\left[{\text{OH}}^{-}\right]$ is $1\times {10}^{-6}\text{M}$.

Summarise the concentrations at equilibrium as follows:

Consider s to be the molar solubility.

$\begin{array}{l}\text{Fe}{\left(\text{OH}\right)}_2\left(s\right)\rightleftharpoons {\text{Fe}}^{2+}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(M\right)& -& 0& 1\times {10}^{-6}\\ \text{Change}\left(M\right)& -s& +s& +2s\\ \text{Equilibrium}\left(M\right)& -& s& 1\times {10}^{-6}+2s\end{array}\end{array}$

The equilibrium expression for a reaction is written as follows:

${\text{K}}_{\text{sp}}={\text{ [Fe}}^{2+}{\text{][OH}}^{-}{\text{]}}^2$

Here, ${\text{K}}_{\text{sp}}$ is the solubility product constant, ${\text{[Fe}}^{2+}\text{]}$ is the concentration of the ferrous ion and ${\text{[OH}}^{-}\text{]}$ is the concentration of the hydroxide ion.

Substitute the values of ${\text{K}}_{\text{sp}}$, ${\text{[Fe}}^{2+}\text{]}$, and ${\text{[OH}}^{-}\text{]}$ in the above expression.

${\text{K}}_{\text{sp}}\text{= }\left(\text{s}\right){\left(\text{2s + 1}\times {\text{10}}^{-6}\right)}^2$

The value of s is very small as compared to $1\times {10}^{-6}$. It can be neglected.

$1\times {10}^{-6}+2s\approx 1\times {10}^{-6}$

On solving further,

$\begin{array}{c}\text{1}\text{.6}\times {\text{10}}^{-14}=s{\left(1\times {10}^{-6}\right)}^2\\ s=0.016\text{M}\end{array}$

Hence, the molar solubility of $\text{Fe}{\left(\text{OH}\right)}_{\text{2}}$ is $\text{0}\text{.016 M}$.

b) ApH is $10.00$

The equation of the dissociation of $\text{Fe}{\left(\text{OH}\right)}_{\text{2}}$ is as follows:

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

The value of $\text{pOH}$ is calculated by using the expression as follow:

$\begin{array}{l}\text{pOH}=14-\text{pH}\\ \text{=14-10}\\ \text{= 4 }\end{array}$

The concentration of $\left[{\text{OH}}^{-}\right]$ is calculated by using the expression as follows:

$[{\text{OH}}^{-}]={10}^{-\text{pOH}}$.

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

$[{\text{OH}}^{-}]=1\times {10}^{-4}\text{M}$

Summarise the concentrations at equilibrium as follows

Consider s to be the molar solubility.

$\begin{array}{l}\text{Fe}{\left(\text{OH}\right)}_2\left(s\right)\rightleftharpoons {\text{Fe}}^{2+}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(M\right)& -& 0& 1\times {10}^{-4}\\ \text{Change}\left(M\right)& -s& +s& +2s\\ \text{Equilibrium}\left(M\right)& -& s& 1\times {10}^{-4}+2s\end{array}\end{array}$

The equilibrium expression for a reaction is written as follows:

${\text{K}}_{\text{sp}}=[{\text{Fe}}^{2+}]{[{\text{OH}}^{-}]}^2$

Here, ${\text{K}}_{\text{sp}}$ is the solubility product constant, ${\text{[Fe}}^{2+}\text{]}$ is the concentration of the ferrous ion, and ${\text{[OH}}^{-}\text{]}$ is the concentration of the hydroxide ion.

Substitute the values of ${\text{K}}_{\text{sp}}$, ${\text{[Fe}}^{2+}\text{]}$, and ${\text{[OH}}^{-}\text{]}$ in the above expression.

${\text{K}}_{\text{sp}}\text{= }\left(\text{s}\right){\left(\text{2s + 1}\times {\text{10}}^{-4}\right)}^2$

The value of s is very small as compared to $1\times {10}^{-4}$. It can be neglected.

$1\times {10}^{-4}+2s\approx 1\times {10}^{-4}$

On solving further,

$\begin{array}{c}\text{1}\text{.6}\times {\text{10}}^{-14}=s{\left(1\times {10}^{-4}\right)}^2\\ s=1.6\times {10}^{-6}\text{M}\end{array}$ $$

Hence, the molar solubility of $\text{Fe}{\left(\text{OH}\right)}_{\text{2}}$ is $\text{1}\text{.6}\times {\text{10}}^{-6}\text{M}$.