Question

When ${\mathbf{\text{Fe}}}^{\mathbf{\text{3 +}}}$salts are dissolved in water, the solution becomes acidic due to formation of$\mathbf{\text{Fe}}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\mathbf{\text{5}}}{\mathbf{\text{OH}}}^{\mathbf{\text{2 +}}}$and${\mathbf{\text{H}}}_{\mathbf{\text{3}}}{\mathbf{\text{O}}}^{\mathbf{\text{+}}}$.The overall process involves both Lewis and Brønsted-Lowry acid-base reactions. Write the equations for the process.

Short answer

The equation of the process is as shown below:

${\text{Fe}}^{3+}+6{\text{H}}_2\text{O}\rightleftharpoons \text{Fe}{\left({\text{H}}_2\text{O}\right)}_6^{3+}.$

$\text{Fe}{\left({\text{H}}_2\text{O}\right)}_6^{3+}+{\text{H}}_2\text{O}\rightleftharpoons \text{Fe}{\left({\text{H}}_2\text{O}\right)}_5{\text{OH}}^{2+}+{\text{H}}_3{\text{O}}^{+}.$

Step-by-step solution

Write the equations for the process

First, the ${\text{Fe}}^{\text{3 +}}$from its salt will form a metal hydrate with water.

${\text{Fe}}^{3+}+6{\text{H}}_2\text{O}\rightleftharpoons \text{Fe}{\left({\text{H}}_2\text{O}\right)}_6^{3+}$

In this reaction, the lone pairs in the${\text{H}}_2\text{O}$are being donated to the${\text{Fe}}^{\text{3 +}}$to form the hydrate. Since there is an electron transfer that happened between the reactants, this is a Lewis acid base reaction.${\text{Fe}}^{\text{3 +}}$is a Lewis acid and ${\text{H}}_2\text{O}$is a Lewis base. They form an adduct, which is the metal hydrate, $\text{Fe}{\left({\text{H}}_2\text{O}\right)}_{\text{6}}^{\text{3 +}}$.

Next, the metal hydrate will react with water again.

$\text{Fe}{\left({\text{H}}_2\text{O}\right)}_6^{3+}+{\text{H}}_2\text{O}\rightleftharpoons \text{Fe}{\left({\text{H}}_2\text{O}\right)}_5{\text{OH}}^{2+}+{\text{H}}_3{\text{O}}^{+}$

The metal hydrate donated a${\text{H}}^{+}$ to ${\text{H}}_2\text{O}$to form $\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5{\mathrm{OH}}^{2+}.$Since there is a proton transfer that happened between the reactants, this is a Bronsted-Lowry acid-base reaction.$\text{Fe}{\left({\text{H}}_2\text{O}\right)}_{\text{6}}^{\text{3 +}}$ is a Bronsted-Lowry acid and ${\text{H}}_2\text{O}$is a Bronsted-Lowry base.

Hence, the equation of the process is:

${\text{Fe}}^{3+}+6{\text{H}}_2\text{O}\rightleftharpoons \text{Fe}{\left({\text{H}}_2\text{O}\right)}_6^{3+}.$

$\text{Fe}{\left({\text{H}}_2\text{O}\right)}_6^{3+}+{\text{H}}_2\text{O}\rightleftharpoons \text{Fe}{\left({\text{H}}_2\text{O}\right)}_5{\text{OH}}^{2+}+{\text{H}}_3{\text{O}}^{+}.$