Chapter 16, Problem 16.81QP

Interpretation Introduction

Interpretation: The $\text{pH}$ of the given concentration of $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ is to be calculated.

Concept introduction: The acid dissociation constant is calculated by the formula,

$K_{\text{a}}=\frac{\left[\text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}{\left({\text{OH}}^{-}\right)}^{2+}\right]\left[{\text{H}}^{+}\right]}{\left[\text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}{}^{3+}\right]}$

The $\text{pH}$ of the solution is calculated by the formula,

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

To determine: The $\text{pH}$ of the given concentration of $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$.

Answer to Problem 16.81QP

Solution:

The $\text{pH}$ of the given concentration of $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ is $\underset{\_}{2.73}$.

Explanation of Solution

Given

The concentration of $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ is $0.25\text{M}$.

Aluminium nitrate on dissolving in the aqueous solution causes the formation of aluminium and nitrate ions.

Aluminium ion in the aqueous solution is surrounded by the six water molecules. Therefore concentration of $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ and $\text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}{}^{3+}$ is same.

Aluminium ion due to its positive charge extracts the hydroxide ion of one of the water molecules to form a proton. Therefore, aluminium ion forms an acidic solution in water.

$\text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}{}^{3+}\underset{}{\rightleftharpoons }\text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}{\left({\text{OH}}^{-}\right)}^{2+}+{\text{H}}^{+}$

The acid dissociation constant, $K_{\text{a}}$ of $\text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}{}^{3+}$ is $1.4\times {10}^{-5}$.

The acid dissociation constant is calculated by the formula,

$K_{\text{a}}=\frac{\left[\text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}{\left({\text{OH}}^{-}\right)}^{2+}\right]\left[{\text{H}}^{+}\right]}{\left[\text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}{}^{3+}\right]}$

The concentration of $\text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}{\left({\text{OH}}^{-}\right)}^{2+}$ ion and ${\text{H}}^{+}$ is same.

Therefore, the above equation becomes,

$K_{\text{a}}=\frac{{\left[{\text{H}}^{+}\right]}^2}{\left[\text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}{}^{3+}\right]}$

Substitute the value of $K_{\text{a}}$ and $\left[\text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}{}^{3+}\right]$ in the above equation.

$\begin{array}{c}1.4\times {10}^{-5}=\frac{{\left[{\text{H}}^{+}\right]}^2}{0.25\text{M}}\\ {\left[{\text{H}}^{+}\right]}^2=1.4\times {10}^{-5}\times 0.25\\ =3.5\times {10}^{-6}\\ \left[{\text{H}}^{+}\right]=1.871\times {10}^{-3}\end{array}$

The $\text{pH}$ of the solution is calculated by the formula,

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

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above equation.

$\begin{array}{c}\text{pH}=-\log \left(1.871\times {10}^{-3}\right)\\ =\underset{\_}{2.73}\end{array}$

Therefore, the $\text{pH}$ of the given concentration of $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ is $\underset{\_}{2.73}$.

Conclusion:

The $\text{pH}$ of the given concentration of $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ is $\underset{\_}{2.73}$