Question

Calculate the pHof a$\mathbf{0}\mathbf{.}\mathbf{050}\mathbf{}\mathbf{M}$ $\mathbf{Al}\mathbf{(}\mathbf{N}{\mathbf{O}}_{\mathbf{3}}{\mathbf{)}}_{\mathbf{3}}$solution. The ${\mathbf{K}}_{\mathbf{a}}$value for $\mathbf{Al}\mathbf{(}{\mathbf{H}}_{\mathbf{2}}{\mathbf{O}\mathbf{)}}_{\mathbf{6}}^{\mathbf{3}\mathbf{+}}$is$\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}$.

Short answer

The pH value of $0.050\mathrm{M}\mathrm{Al}(\mathrm{N}{\mathrm{O}}_3{)}_3$the solution is $\mathrm{pH}=3.08$.

Step-by-step solution

Concept Introduction

The negative logarithm of ${\mathbf{H}}^{\mathbf{+}}$ion concentration is used to calculate pH. As a result, the meaning of pHis justified as hydrogen power.

Reaction of Aluminium nitrate

The aluminum nitrate$\mathrm{Al}{\left({\mathrm{NO}}_3\right)}_3$dissociates in water.

$\mathrm{Al}{\left({\mathrm{NO}}_3\right)}_3\left(\mathrm{aq}\right)\to {\mathrm{Al}}^{3+}{\left(\mathrm{aq}\right)+3\mathrm{NO}}_3^{-}\left(\mathrm{aq}\right)$

An aqueous solution${\mathrm{Al}}^{3+}$exists as hexahydrate, ${\left[\mathrm{Al}({\mathrm{H}}_2\mathrm{O}{)}_6\right]}^{3+}$.

${\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}\rightleftharpoons {\left[\mathrm{Al}\left(\mathrm{OH}\right){\left({\mathrm{H}}_2\mathrm{O}\right)}_5\right]}^{2+}+{\mathrm{H}}^{+}$

The initial concentration of ${\left[\mathrm{Al}({\mathrm{H}}_2\mathrm{O}{)}_6\right]}^{3+}$is equal to the given concentration of $\mathrm{Al}(\mathrm{N}{\mathrm{O}}_3{)}_3$solution.

${\left[{\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}\right]}_0=0.050\mathrm{M}$

Equilibrium concentration of ${\left[\mathrm{Al}({\mathrm{H}}_2\mathrm{O}{)}_6\right]}^{3+}$equal to the initial concentration of ${\left[\mathrm{Al}({\mathrm{H}}_2\mathrm{O}{)}_6\right]}^{3+}$equilibrium concentration of role="math" localid="1663866262015" ${\left[\mathrm{Al}\left(\mathrm{OH}\right){\left({\mathrm{H}}_2\mathrm{O}\right)}_5\right]}^{2+}$.

$$\begin{gathered} \left[{\left[\mathrm{Al}\left(\mathrm{OH}\right){\left({\mathrm{H}}_2\mathrm{O}\right)}_5\right]}^{2+}\right]=\left[{\mathrm{H}}^{+}\right]=\mathrm{x} \\ \left[{\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}\right]={\left[{\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}\right]}_0-\mathrm{x} \\ \left[{\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}\right]=0.050\mathrm{M}-\mathrm{x} \end{gathered}$$

Calculation for pH

The value of the unknown x can be calculated from the expression of the acid dissociation constant.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\left[\mathrm{Al}\left(\mathrm{OH}\right){\left({\mathrm{H}}_2\mathrm{O}\right)}_5\right]}^{2+}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}\right]} \\ 1.4\times {10}^{-5}=\frac{{\mathrm{x}}^2}{(0.050-\mathrm{x})} \\ \mathrm{x}=8.37\times {10}^{-4}\mathrm{M} \\ \left[{\mathrm{H}}^{+}\right]=\mathrm{x}=\left[8.37\times {10}^{-4}\mathrm{M}\right] \end{gathered}$$

The pH of a solution measures the hydronium ion concentration, which in turn reflects the solution's acidity.

The pH is calculated according to the equation shown below.

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ \mathrm{pH}=-\log \left[8.37\times {10}^{-4}\mathrm{M}\right] \\ \mathrm{pH}=3.08 \end{gathered}$$

Therefore, the pH value is obtained as 3.08.