Chemistry » Acid-Base Equilibria » Hydrolysis of Salt Solutions

# The Ionization of Hydrated Metal Ions

## The Ionization of Hydrated Metal Ions

If we measure the pH of the solutions of a variety of metal ions we will find that these ions act as weak acids when in solution. The aluminum ion is an example. When aluminum nitrate dissolves in water, the aluminum ion reacts with water to give a hydrated aluminum ion, $$\text{Al}\left({\text{H}}_{2}\text{O}\right)_{6}^{3+},$$ dissolved in bulk water. What this means is that the aluminum ion has the strongest interactions with the six closest water molecules (the so-called first solvation shell), even though it does interact with the other water molecules surrounding this $$\text{Al}\left({\text{H}}_{2}\text{O}\right)_{6}^{3+}$$ cluster as well:

$$\text{Al}\left({\text{NO}}_{3}\right)_{3}\left(s\right)+6{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Al}\left({\text{H}}_{2}\text{O}\right)_{6}^{3+}\left(aq\right)+3{\text{NO}}_{3}^{\text{−}}\left(aq\right)$$

We frequently see the formula of this ion written simply as “Al3+(aq)”, without explicitly noting that six water molecules are covalently bonded to the aluminum ion. This is similar to the simplification of the formula of the hydronium ion, H3O+ to H+. In this case, a bonded water molecule acts as a weak acid (see the figure below) and donates a proton to a water molecule.

$$\text{Al(H}_2\text{O})_6^{\;\;3+}(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{Al(H}_2\text{O})_5(\text{OH})^{2+}(aq)\;\;\;\;\;\;\;K_{\text{a}} = 1.4\;\times\;10^{-5}$$

The conjugate base produced by this process contains five other bonded water molecules capable of acting as acids, and so the sequential or step-wise transfer of protons is possible as depicted in the equations below:

$$\begin{array}{rl} \text{Al(H}_2\text{O})_6^{\;\;3+}(aq)\;+\;\text{H}_2\text{O}(l) \rightleftharpoons & \text{H}_3\text{O}^{+}(aq)\;+\;\text{Al(H}_2\text{O})_5(\text{OH})^{2+}(aq) \\[0.5em] \text{Al(H}_2\text{O})_5(\text{OH})^{2+}(aq)\;+\;\text{H}_2\text{O}(l) \rightleftharpoons & \text{H}_3\text{O}^{+}(aq)\;+\;\text{Al(H}_2\text{O})_4(\text{OH})_2^{\;\;+}(aq) \\[0.5em] \text{Al(H}_2\text{O})_4(\text{OH})_2^{\;\;+}(aq)\;+\;\text{H}_2\text{O}(l) \rightleftharpoons & \text{H}_3\text{O}^{+}(aq)\;+\;\text{Al(H}_2\text{O})_3(\text{OH})_3(aq) \end{array}$$

This is an example of a polyprotic acid, the topic of discussion in a later section of this tutorial.

When an aluminum ion reacts with water, the hydrated aluminum ion becomes a weak acid.

Additional examples of the first stage in the ionization of hydrated metal ions are:

$$\begin{array}{r @{{}\rightleftharpoons{}} ll} \text{Fe(H}_2\text{O})_6^{\;\;3+}(aq)\;+\;\text{H}_2\text{O}(l) & \text{H}_3\text{O}^{+}(aq)\;+\;\text{Fe(H}_2\text{O})_5(\text{OH})^{2+}(aq) & K_{\text{a}} = 2.74 \\[0.5em] \text{Cu(H}_2\text{O})_6^{\;\;2+}(aq)\;+\;\text{H}_2\text{O}(l) & \text{H}_3\text{O}^{+}(aq)\;+\;\text{Cu(H}_2\text{O})_5(\text{OH})^{+}(aq) & K_{\text{a}} = {\sim}6.3 \\[0.5em] \text{Zn(H}_2\text{O})_4^{\;\;2+}(aq)\;+\;\text{H}_2\text{O}(l) & \text{H}_3\text{O}^{+}(aq)\;+\;\text{Zn(H}_2\text{O})_3(\text{OH})^{+}(aq) & K_{\text{a}} = 9.6 \end{array}$$

## Example

### Hydrolysis of [Al(H2O)6]3+Calculate the p

H of a 0.10-M solution of aluminum chloride, which dissolves completely to give the hydrated aluminum ion $${\left[\text{Al}\left({\text{H}}_{2}\text{O}\right)_{6}\right]}^{3+}$$ in solution.

### Solution

In spite of the unusual appearance of the acid, this is a typical acid ionization problem.

1. Determine the direction of change. The equation for the reaction and Ka are:

$$\text{Al(H}_2\text{O})_6^{\;\;3+}(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{Al(H}_2\text{O})_5(\text{OH})^{2+}(aq)\;\;\;\;\;\;\;K_{\text{a}} = 1.4\;\times\;10^{-5}$$

The reaction shifts to the right to reach equilibrium.

2. Determine x and equilibrium concentrations. Use the table:

3. Solve for x and the equilibrium concentrations. Substituting the expressions for the equilibrium concentrations into the equation for the ionization constant yields:

$$K_{\text{a}} = \cfrac{[\text{H}_3\text{O}^{+}][\text{Al(H}_2\text{O})_5(\text{OH})^{2+}]}{[\text{Al(H}_2\text{O})_6^{\;\;3+}]}$$

$$= \cfrac{(x)(x)}{0.10\;-\;x} = 1.4\;\times\;10^{-5}$$

Solving this equation gives:

$$x=1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}M$$

From this we find:

$$\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=0+x=1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}M$$

$$\text{pH}=\text{−log}\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\phantom{\rule{0.2em}{0ex}}=2.92\phantom{\rule{0.2em}{0ex}}\left(\text{an acidic solution}\right)$$

4. Check the work. The arithmetic checks; when 1.2 $$×$$ 10−3M is substituted for x, the result = Ka.

The constants for the different stages of ionization are not known for many metal ions, so we cannot calculate the extent of their ionization. However, practically all hydrated metal ions other than those of the alkali metals ionize to give acidic solutions. Ionization increases as the charge of the metal ion increases or as the size of the metal ion decreases.