Chapter 15: Acid-Based Equilibria

15.4 Hydrolysis of Salts

Learning Outcomes

  • Predict whether a salt solution will be acidic, basic, or neutral
  • Calculate the concentrations of the various species in a salt solution
  • Describe the acid ionization of hydrated metal ions

Salts with Acidic Ions

Salts are ionic compounds composed of cations and anions, either of which may be capable of undergoing an acid or base ionization reaction with water. Aqueous salt solutions, therefore, may be acidic, basic, or neutral, depending on the relative acid-base strengths of the salt’s constituent ions. For example, dissolving ammonium chloride in water results in its dissociation, as described by the equation

[latex]\text{NH}_4\text{Cl}(s) \rightleftharpoons \text{NH}_4^+(aq)+ \text{Cl}^-(aq)[/latex]

The ammonium ion is the conjugate acid of the base ammonia, NH3; its acid ionization (or acid hydrolysis) reaction is represented by

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

Since ammonia is a weak base, Kb is measurable and Ka > 0 (ammonium ion is a weak acid).

The chloride ion is the conjugate base of hydrochloric acid, and so its base ionization (or base hydrolysis) reaction is represented by

[latex]{\text{Cl}}^{\text{-}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{HCl}}\left(aq\right)+{\text{OH}}^{\text{-}}\left(aq\right)[/latex]       [latex]{K}_{\text{b}}={K}_{\text{w}}/ {K}_{\text{a}}[/latex]

Since HCl is a strong acid, Ka is immeasurably large and Kb ≈ 0 (chloride ions don’t undergo appreciable hydrolysis).

Thus, dissolving ammonium chloride in water yields a solution of weak acid cations (NH4+) and inert anions (Cl−), resulting in an acidic solution.

Example 15.4.1: The pH of a Solution of a Salt of a Weak Base and a Strong Acid

Aniline is an amine that is used to manufacture dyes. It is isolated as aniline hydrochloride, [latex]\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right]\text{Cl}[/latex], a salt prepared by the reaction of the weak base aniline and hydrochloric acid. What is the pH of a 0.233 M solution of aniline hydrochloride?

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

Show Solution

The new step in this example is to determine Ka for the [latex]{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}[/latex] ion. The [latex]{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}[/latex] ion is the conjugate acid of a weak base. The value of Ka for this acid is not listed in Ionization Constants of Weak Acids, but we can determine it from the value of Kb for aniline, C6H5NH2, which is given as 4.6 [latex]\times[/latex] 10−10 (Relative Strengths of Acids and Bases and Ionization Constants of Weak Bases):

[latex]{K}_{\text{a}}\left(\text{ for }{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right)\times {K}_{\text{b}}\left(\text{ for }{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\right)={K}_{\text{w}}=1.0\times {10}^{-14}[/latex]
[latex]{K}_{\text{a}}\left(\text{for}{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right)=\dfrac{{K}_{\text{w}}}{{K}_{\text{b}}\left(\text{ for }{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\right)}=\dfrac{1.0\times {10}^{-14}}{4.6\times {10}^{-10}}=2.2\times {10}^{-5}[/latex]

Now we have the ionization constant and the initial concentration of the weak acid, the information necessary to determine the equilibrium concentration of [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex], and the pH:

Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled “Determine the direction of change.” The second is labeled “Determine x and the equilibrium concentrations.” The third is labeled “Solve for x and the equilibrium concentrations.” The fourth is labeled “Check the math.”

Using the provided information, an ICE table for this system is prepared:

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium ( M ). The second column has the header of “C subscript 6 H subscript 5 N H subscript 3 superscript positive sign plus sign H subscript 2 O equilibrium sign C subscript 6 H subscript 5 N H subscript 2 plus sign H subscript 3 O superscript positive sign.” Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.233, negative x, 0.233 minus x. The second column is blank for all three rows. The third column has the following: 0, positive x, x. The fourth column has the following: approximately 0, positive x, x.

Substituting these equilibrium concentration terms into the Ka expression gives

[latex]\text{K}_a = [\text{C}_6\text{H}_5\text{NH}_2][\text{H}_3\text{O}^+]/[\text{C}_6\text{H}_5\text{NH}_3^+][/latex]

[latex]2.3 \text{x} 10^{-5} = (x)(x)/(0.233-x)[/latex]

Assuming x << 0.233, the equation is simplified and solved for x:

[latex]2.3 \text{x} 10^{-5} = x^2/0.233[/latex]

[latex]x = 0.0023 M[/latex]

The ICE table defines x as the hydronium ion molarity, and so the pH is computed as

[latex]\text{pH} = -\text{log}[\text{H}_3\text{O}^+] = -\text{log}(0.0023) = 2.64[/latex]

Check Your Learning

Salts with Basic Ions

As another example, consider dissolving sodium acetate in water:

[latex]\text{NaCH}_3\text{CO}_2 (s) \rightleftharpoons \text{Na}^+ (aq) +\text{CH}_3\text{CO}_2^-(aq)[/latex]

The sodium ion does not undergo appreciable acid or base ionization and has no effect on the solution pH. This may seem obvious from the ion’s formula, which indicates no hydrogen or oxygen atoms, but some dissolved metal ions function as weak acids, as addressed later in this section.

The acetate ion, [latex]\text{CH}_3\text{CO}_2^-[/latex] is the conjugate base of acetic acid, CH3CO2H, and so its base ionization (or base hydrolysis) reaction is represented by

[latex]\text{CH}_3\text{CO}_2^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{CH}_3\text{CO}_2\text{H}(aq) + \text{OH}^-(aq)[/latex]        [latex]K_b=K_w/K_a[/latex]

Because acetic acid is a weak acid, its Ka is measurable and Kb > 0 (acetate ion is a weak base).

Dissolving sodium acetate in water yields a solution of inert cations (Na+) and weak base anions ([latex]\text{CH}_3\text{CO}_2^-[/latex]), resulting in a basic solution.

Example 15.4.2: Equilibrium in a Solution of a Salt of a Weak Acid and a Strong Base

Determine the acetic acid concentration in a solution with [latex]\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{-}}\right]=0.050M[/latex] and [OH−] = 2.5 [latex]\times[/latex] 10−6M at equilibrium. The reaction is:

[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]

Show Solution

We are given two of three equilibrium concentrations and asked to find the missing concentration. If we can find the equilibrium constant for the reaction, the process is straightforward.

The acetate ion behaves as a base in this reaction; hydroxide ions are a product. We determine Kb as follows:

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

Now find the missing concentration:

[latex]{K}_{\text{b}}=\dfrac{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{-}}\right]}=5.6\times {10}^{-10}[/latex]
[latex]=\dfrac{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]\left(2.5\times {10}^{-6}\right)}{\left(0.050\right)}=5.6\times {10}^{-10}[/latex]

Solving this equation we get [CH3CO2H] = [latex]1.1\times{10}^{-5}M[/latex].

Check Your Learning

Salts with Acidic and Basic Ions

Some salts are composed of both acidic and basic ions, and so the pH of their solutions will depend on the relative strengths of these two species. Likewise, some salts contain a single ion that is amphiprotic, and so the relative strengths of this ion’s acid and base character will determine its effect on solution pH. For both types of salts, a comparison of the Ka and Kb values allows prediction of the solution’s acid-base status, as illustrated in the following example exercise.

Example 15.4.3: Determining the Acidic or Basic Nature of Salts

Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:

  1. KBr
  2. NaHCO3
  3. NH4Cl
  4. Na2HPO4
  5. NH4F
Show Solution

Consider each of the ions separately in terms of its effect on the pH of the solution, as shown here:

  1. The K+ cation and the Br− anion are both spectators, since they are the cation of a strong base (KOH) and the anion of a strong acid (HBr), respectively. The solution is neutral.
  2. The Na+ cation is a spectator, and will not affect the pH of the solution; while the [latex]{\text{HCO}}_{3}{}^{\text{-}}[/latex] anion is amphiprotic, it could either behave as an acid or a base. The Ka of [latex]{\text{HCO}}_{3}{}^{\text{-}}[/latex] is 4.7 [latex]\times[/latex] 10−11, so the Kb of its conjugate base is [latex]\frac{1.0\times {10}^{-14}}{4.7\times {10}^{-11}}=2.1\times {10}^{-4}[/latex].
    Since Kb >> Ka, the solution is basic.
  3. The [latex]{\text{NH}}_{4}{}^{\text{+}}[/latex] ion is acidic and the Cl− ion is a spectator. The solution will be acidic.
  4. The Na+ ion is a spectator, while the [latex]{\text{HPO}}_{4}{}^{-}[/latex] ion is amphiprotic, with a Ka of 3.6 [latex]\times[/latex] 10−13
    so that the Kb of its conjugate base is [latex]\frac{1.0\times {10}^{-14}}{3.6\times {10}^{-13}}=2.8\times {10}^{-2}[/latex]. Because Kb >> Ka, the solution is basic.
  5. The [latex]{\text{NH}}_{4}{}^{\text{+}}[/latex] ion is listed as being acidic, and the F− ion is listed as a base, so we must directly compare the Ka and the Kb of the two ions. Ka of [latex]{\text{NH}}_{4}{}^{\text{+}}[/latex] is 5.6 [latex]\times[/latex] 10−10, which seems very small, yet the Kb of F− is 1.4 [latex]\times[/latex] 10−11, so the solution is acidic, since Ka > Kb.

Check Your Learning

The Ionization of Hydrated Metal Ions

Unlike the group 1 and 2 metal ions of the preceding examples (Na+, Ca2+, etc.), some metal ions function as acids in aqueous solutions. These ions are not just loosely solvated by water molecules when dissolved, instead they are covalently bonded to a fixed number of water molecules to yield a complex ion (see chapter on coordination chemistry). As an example, the dissolution of aluminum nitrate in water is typically represented as

[latex]\text{Al}{\left({\text{NO}}_{3}\right)}\left(s\right)\leftrightharpoons \text{Al}^{3}+\left(aq\right)+3{\text{NO}_{3}}^{\text{-}}\left(aq\right)[/latex]

However, the aluminum(III) ion actually reacts with six water molecules to form a stable complex ion, and so the more explicit representation of the dissolution process is

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

As shown in Figure 15.4.1, the [latex]{\text{Al(H}_{2}\text{O)}_{6}}^{3+}[/latex] ions involve bonds between a central Al atom and the O atoms of the six water molecules. Consequently, the bonded water molecules’ O–H bonds are more polar than in nonbonded water molecules, making the bonded molecules more prone to donation of a hydrogen ion:

[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)\qquad{K}_{\text{a}}=1.4\times {10}^{-5}[/latex]

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 few equations below:

[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]

This is an example of a polyprotic acid,

A reaction is shown using ball and stick models. On the left, inside brackets with a superscript of 3 plus outside to the right is structure labeled “[ A l ( H subscript 2 O ) subscript 6 ] superscript 3 plus.” Inside the brackets is s central grey atom to which 6 red atoms are bonded in an arrangement that distributes them evenly about the central grey atom. Each red atom has two smaller white atoms attached in a forked or bent arrangement. Outside the brackets to the right is a space-filling model that includes a red central sphere with two smaller white spheres attached in a bent arrangement. Beneath this structure is the label “H subscript 2 O.” A double sided arrow follows. Another set of brackets follows to the right of the arrows which have a superscript of two plus outside to the right. The structure inside the brackets is similar to that on the left, except a white atom is removed from the structure. The label below is also changed to “[ A l ( H subscript 2 O ) subscript 5 O H ] superscript 2 plus.” To the right of this structure and outside the brackets is a space filling model with a central red sphere to which 3 smaller white spheres are attached. This structure is labeled “H subscript 3 O superscript plus.”
Figure 15.4.1. When an aluminum ion reacts with water, the hydrated aluminum ion becomes a weak acid.

Aside from the alkali metals (group 1) and some alkaline earth metals (group 2), most other metal ions will undergo acid ionization to some extent when dissolved in water. The acid strength of these complex ions typically increases with increasing charge and decreasing size of the metal ions. The first-step acid ionization equations for a few other acidic metal ions are shown below:

[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)\qquad{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)\qquad{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)\qquad{K}_{\text{a}}=9.6[/latex]

Example 15.4.4: 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 Solution

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

Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled “Determine the direction of change.” The second is labeled “Determine x and the equilibrium concentrations.” The third is labeled “Solve for x and the equilibrium concentrations.” The fourth is labeled “Check the math.”

Step 1. Determine the direction of change. The equation for the reaction and Ka 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)\qquad{K}_{\text{a}}=1.4\times {10}^{-5}[/latex]

the reaction shifts to the right to reach equilibrium.

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

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of “A l ( H subscript 2 O ) subscript 6 superscript 3 positive sign plus H subscript 2 O equilibrium arrow H subscript 3 O superscript positive sign plus A l ( H subscript 2 O ) subscript 5 ( O H ) superscript 2 positive sign.” Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.10 (which appears in red), negative x, 0.10 minus x. The second column is blank. The third column has the following: approximately 0, x, x. The fourth column has the following: 0, x, x.

Step 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}}=\dfrac{\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]=\dfrac{\left(x\right)\left(x\right)}{0.10-x}=1.4\times {10}^{-5}[/latex]

Assuming x << 0.10 and solving the simplified equation gives:

[latex]x=1.2\times {10}^{-3}M[/latex]

The ICE table defined x as equal to the hydronium ion concentration, and so the pH is calculated to be

[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]

Step 4. Check the work. The arithmetic checks; when [latex][/latex] is substituted for x, the result = Ka.

Check Your Learning

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.

Key Concepts and Summary

The ions composing salts may possess acidic or basic character, ionizing when dissolved in water to yield acidic or basic solutions. Acidic cations are typically the conjugate partners of weak bases, and basic anions are the conjugate partners of weak acids. Many metal ions bond to water molecules when dissolved to yield complex ions that may function as acids.

Try It

  1. Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
    1. FeCl3
    2. K2CO3
    3. NH4Br
    4. KClO4
  2. Novocaine, C13H21O2N2Cl, is the salt of the base procaine and hydrochloric acid. The ionization constant for procaine is 7 [latex]\times[/latex] 10−6. Is a solution of novocaine acidic or basic? What are [H3O+], [OH−], and pH of a 2.0% solution by mass of novocaine, assuming that the density of the solution is 1.0 g/mL.
Show Solutions

1. The solutions can be categorized as follows:

  1. FeCl3 dissociates into Fe3+ ions (acidic metal cation) and Cl− ions (the conjugate base of a strong acid and therefore essentially neutral). The aqueous solution is therefore acidic.
  2. K2CO3 dissociates into K+ ions (neutral metal cation) and [latex]{\text{CO}}_{3}{}^{2-}[/latex] ions (the conjugate base of a weak acid and therefore basic). The aqueous solution is therefore basic.
  3. NH4Br dissociates into NH4+ ions (a weak acid) and Br− ions (the conjugate base of a strong acid and therefore essentially neutral). The aqueous solution is therefore acidic.
  4. KClO4 dissociates into K+ ions (neutral metal cation) and [latex]{\text{ClO}}_{4}{}^{-}[/latex] ions (the conjugate base of a strong acid and therefore neutral). The aqueous solution is therefore neutral.
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