As we have seen in the section on chemical reactions, when an acid and base are mixed, they undergo a neutralization reaction. The word “neutralization” seems to imply that a stoichiometrically equivalent solution of an acid and a base would be neutral. This is sometimes true, but the salts that are formed in these reactions may have acidic or basic properties of their own, as we shall now see.

Acid-Base Neutralization

A solution is neutral when it contains equal concentrations of hydronium and hydroxide ions. When we mix solutions of an acid and a base, an acid-base neutralization reaction occurs. However, even if we mix stoichiometrically equivalent quantities, we may find that the resulting solution is not neutral. It could contain either an excess of hydronium ions or an excess of hydroxide ions because the nature of the salt formed determines whether the solution is acidic, neutral, or basic. The following four situations illustrate how solutions with various pH values can arise following a neutralization reaction using stoichiometrically equivalent quantities:

1. A strong acid and a strong base, such as HCl(aq) and NaOH(aq) will react to form a neutral solution since the conjugate partners produced are of negligible strength: [latex]\text{HCl}\left(aq\right)+\text{NaOH}\left(aq\right)\rightleftharpoons \text{NaCl}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)[/latex]
2. A strong acid and a weak base yield a weakly acidic solution, not because of the strong acid involved, but because of the conjugate acid of the weak base.
3. A weak acid and a strong base yield a weakly basic solution. A solution of a weak acid reacts with a solution of a strong base to form the conjugate base of the weak acid and the conjugate acid of the strong base. The conjugate acid of the strong base is a weaker acid than water and has no effect on the acidity of the resulting solution. However, the conjugate base of the weak acid is a weak base and ionizes slightly in water. This increases the amount of hydroxide ion in the solution produced in the reaction and renders it slightly basic.
4. A weak acid plus a weak base can yield either an acidic, basic, or neutral solution. This is the most complex of the four types of reactions. When the conjugate acid and the conjugate base are of unequal strengths, the solution can be either acidic or basic, depending on the relative strengths of the two conjugates. Occasionally the weak acid and the weak base will have the same strength, so their respective conjugate base and acid will have the same strength, and the solution will be neutral. To predict whether a particular combination will be acidic, basic or neutral, tabulated K values of the conjugates must be compared.

Culinary Aspects of Chemistry

Cooking is essentially synthetic chemistry that happens to be safe to eat. There are a number of examples of acid-base chemistry in the culinary world. One example is the use of baking soda, or sodium bicarbonate in baking. NaHCO₃ is a base. When it reacts with an acid such as lemon juice, buttermilk, or sour cream in a batter, bubbles of carbon dioxide gas are formed from decomposition of the resulting carbonic acid, and the batter “rises.” Baking powder is a combination of sodium bicarbonate, and one or more acid salts that react when the two chemicals come in contact with water in the batter.

Many people like to put lemon juice or vinegar, both of which are acids, on cooked fish (Figure 1). It turns out that fish have volatile amines (bases) in their systems, which are neutralized by the acids to yield involatile ammonium salts. This reduces the odor of the fish, and also adds a “sour” taste that we seem to enjoy.

Figure 1. A neutralization reaction takes place between citric acid in lemons or acetic acid in vinegar, and the bases in the flesh of fish.

 

Pickling is a method used to preserve vegetables using a naturally produced acidic environment. The vegetable, such as a cucumber, is placed in a sealed jar submerged in a brine solution. The brine solution favors the growth of beneficial bacteria and suppresses the growth of harmful bacteria. The beneficial bacteria feed on starches in the cucumber and produce lactic acid as a waste product in a process called fermentation. The lactic acid eventually increases the acidity of the brine to a level that kills any harmful bacteria, which require a basic environment. Without the harmful bacteria consuming the cucumbers they are able to last much longer than if they were unprotected. A byproduct of the pickling process changes the flavor of the vegetables with the acid making them taste sour.

Salts of Weak Bases and Strong Acids

When we neutralize a weak base with a strong acid, the product is a salt containing the conjugate acid of the weak base. This conjugate acid is a weak acid. For example, ammonium chloride, NH₄Cl, is a salt formed by the reaction of the weak base ammonia with the strong acid HCl:

[latex]{\text{NH}}_{3}\left(aq\right)+\text{HCl}\left(aq\right)\longrightarrow {\text{NH}}_{4}\text{Cl}\left(aq\right)[/latex]

A solution of this salt contains ammonium ions and chloride ions. The chloride ion has no effect on the acidity of the solution since HCl is a strong acid. Chloride is a very weak base and will not accept a proton to a measurable extent. However, the ammonium ion, the conjugate acid of ammonia, reacts with water and increases the hydronium ion concentration:

[latex]{\text{NH}}_{4}{}^{\text{+}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{NH}}_{3}\left(aq\right)[/latex]

The equilibrium equation for this reaction is simply the ionization constant. K_a, for the acid [latex]{\text{NH}}_{4}{}^{\text{+}}:[/latex]

[latex]\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{NH}}_{3}\right]}{\left[{\text{NH}}_{4}{}^{\text{+}}\right]}={K}_{\text{a}}[/latex]

We will not find a value of K_a for the ammonium ion in Ionization Constants of Weak Acids. However, it is not difficult to determine K_a for [latex]{\text{NH}}_{4}{}^{\text{+}}[/latex] from the value of the ionization constant of water, K_w, and K_b, the ionization constant of its conjugate base, NH₃, using the following relationship:

[latex]{K}_{\text{w}}={K}_{\text{a}}\times {K}_{\text{b}}[/latex]

This relation holds for any base and its conjugate acid or for any acid and its conjugate base.

Salts of Weak Acids and Strong Bases

When we neutralize a weak acid with a strong base, we get a salt that contains the conjugate base of the weak acid. This conjugate base is usually a weak base. For example, sodium acetate, NaCH₃CO₂, is a salt formed by the reaction of the weak acid acetic acid with the strong base sodium hydroxide:

[latex]{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+\text{NaOH}\left(aq\right)\longrightarrow {\text{NaCH}}_{3}{\text{CO}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(aq\right)[/latex]

A solution of this salt contains sodium ions and acetate ions. The sodium ion, as the conjugate acid of a strong base, has no effect on the acidity of the solution. However, the acetate ion, the conjugate base of acetic acid, reacts with water and increases the concentration of hydroxide ion:

[latex]{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{-}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{OH}}^{\text{-}}\left(aq\right)[/latex]

The equilibrium equation for this reaction is the ionization constant, K_b, for the base [latex]{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{-}}[/latex]. The value of K_b can be calculated from the value of the ionization constant of water, K_w, and K_a, the ionization constant of the conjugate acid of the anion using the equation:

[latex]{K}_{\text{w}}={K}_{\text{a}}\times {K}_{\text{b}}[/latex]

For the acetate ion and its conjugate acid we have:

[latex]{K}_{\text{b}}\left(\text{for}{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{-}}\right)=\frac{{K}_{\text{w}}}{{K}_{\text{a}}\left(\text{for}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right)}=\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}=5.6\times {10}^{-10}[/latex]

Some handbooks do not report values of K_b. They only report ionization constants for acids. If we want to determine a K_b value using one of these handbooks, we must look up the value of K_a for the conjugate acid and convert it to a K_b value.

Equilibrium in a Solution of a Salt of a Weak Acid and a Weak Base

In a solution of a salt formed by the reaction of a weak acid and a weak base, to predict the pH, we must know both the K_a of the weak acid and the K_b of the weak base. If K_a > K_b, the solution is acidic, and if K_b > K_a, the solution is basic.

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, [latex]\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}[/latex], 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 [latex]\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}[/latex] cluster as well:

[latex]\text{Al}{\left({\text{NO}}_{3}\right)}_{3}\left(s\right)+6{\text{H}}_{2}\text{O}\left(l\right)\longrightarrow \text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}\left(aq\right)+3{\text{NO}}_{3}{}^{\text{-}}\left(aq\right)[/latex]

We frequently see the formula of this ion simply as “Al³⁺(aq)”, without explicitly noting the six water molecules that are the closest ones to the aluminum ion and just describing the ion as being solvated in water (hydrated). This is similar to the simplification of the formula of the hydronium ion, [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex] to H⁺. However, in this case, the hydrated aluminum ion is a weak acid (Figure 2) and donates a proton to a water molecule. Thus, the hydration becomes important and we may use formulas that show the extent of hydration:

[latex]\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}{\left(\text{OH}\right)}^{2+}\left(aq\right){K}_{\text{a}}=1.4\times {10}^{-5}[/latex]

As with other polyprotic acids, the hydrated aluminum ion ionizes in stages, as shown by:

[latex]\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}{\left(\text{OH}\right)}^{2+}\left(aq\right)[/latex]
[latex]\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}{\left(\text{OH}\right)}^{2+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{4}{\left(\text{OH}\right)}_{2}{}^{\text{+}}\left(aq\right)[/latex]
[latex]\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{4}{\left(\text{OH}\right)}_{2}{}^{\text{+}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{3}{\left(\text{OH}\right)}_{3}\left(aq\right)[/latex]

Note that some of these aluminum species are exhibiting amphiprotic behavior, since they are acting as acids when they appear on the right side of the equilibrium expressions and as bases when they appear on the left side.

Figure 3. When an aluminum ion reacts with water, the hydrated aluminum ion becomes a weak acid.

However, the ionization of a cation carrying more than one charge is usually not extensive beyond the first stage. Additional examples of the first stage in the ionization of hydrated metal ions are:

[latex]\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{5}{\left(\text{OH}\right)}^{2+}\left(aq\right){K}_{\text{a}}=2.74[/latex]
[latex]\text{Cu}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{2+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Cu}{\left({\text{H}}_{2}\text{O}\right)}_{5}{\left(\text{OH}\right)}^{\text{+}}\left(aq\right){K}_{\text{a}}=~6.3[/latex]
[latex]\text{Zn}{\left({\text{H}}_{2}\text{O}\right)}_{4}{}^{2+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Zn}{\left({\text{H}}_{2}\text{O}\right)}_{3}{\left(\text{OH}\right)}^{\text{+}}\left(aq\right){K}_{\text{a}}=9.6[/latex]

Example 4: Hydrolysis of [Al(H₂O)₆]³⁺

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

Show Answer

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 K_a are:[latex]\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}{\left(\text{OH}\right)}^{2+}\left(aq\right){K}_{\text{a}}=1.4\times {10}^{-5}[/latex]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:[latex]{K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}{\left(\text{OH}\right)}^{2+}\right]}{\left[\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}\right]}[/latex][latex]=\frac{\left(x\right)\left(x\right)}{0.10-x}=1.4\times {10}^{-5}[/latex]Solving this equation gives:[latex]x=1.2\times {10}^{-3}M[/latex]From this we find:[latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=0+x=1.2\times {10}^{-3}M[/latex][latex]\text{pH}=\text{-log}\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=2.92\left(\text{an acidic solution}\right)[/latex]
4. Check the work. The arithmetic checks; when 1.2 [latex]\times [/latex] 10⁻³M is substituted for x, the result = K_a.

Check Your Learning

What is [latex]\left[\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}{\left(\text{OH}\right)}^{2+}\right][/latex] in a 0.15-M solution of Al(NO₃)₃ that contains enough of the strong acid HNO₃ to bring [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] to 0.10 M?

Show Answer

2.1 [latex]\times [/latex] 10⁻⁵M

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.