14.4 Hydrolysis of Salts

 

Learning Objectives

By the end of this section, you will be able to:

  • 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

NH4Cl(s) → NH4+(aq) + Cl(aq)

The ammonium ion is the conjugate acid of the weak base ammonia, NH3, and so it will undergo acid ionization (or acid hydrolysis):

NH4+(aq) + H2O(l) ⇌ H3O+(aq) + NH3(aq)          Ka = Kw/Kb

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

Cl(aq) + H2O(l) ⇌ HCl(aq) + OH(aq)           Kb = Kw/Ka

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.

Calculating the pH of an Acidic Salt Solution

Aniline is an amine that is used to manufacture dyes. It is isolated as anilinium chloride, [C6H5NH3]Cl, 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 anilinium chloride?

C6H5NH3+(aq) + H2O(l) ⇌ H3O+(aq) + C6H5NH2(aq)

Solution:

The Ka for anilinium ion is derived from the Kb for its conjugate base, aniline (see Appendix H):

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

Ka = [C6H5NH2][H3O+]/[C6H5NH3+]
2.3 × 10−5 = (x)(x)/(0.233-x)

Assuming x < 0.05 × 0.233M, i.e. x < 0.0116 M, the equation is simplified and solved for x:

2.3 × 10−5 = x2/0.233
x = 0.0023 M          ASSUMPTION VALID

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

pH = -log[H3O+] = −log(0.0023 M) = 2.64

Check Your Learning:

What is the hydronium ion concentration in a 0.100-M solution of ammonium nitrate, NH4NO3, a salt composed of the ions NH4+ and NO3. Which is the stronger acid, C6H5NH3+ or NH4+?

 

Answer:

[H3O+] = 7.5 × 10−6M; C6H5NH3+ is the stronger acid.

Salts with Basic Ions

As another example, consider dissolving sodium acetate in water:

NaCH3CO2(s) → Na+(aq) + CH3CO2(aq)

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, CH3CO2, is the conjugate base of acetic acid, CH3CO2H, and so its base ionization (or base hydrolysis) reaction is represented by

CH3CO2(aq) + H2O(l) ⇌ CH3CO2H(aq) + OH(aq)          Kb = Kw/Ka

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 (CH3CO2), resulting in a basic solution.

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

Determine the acetic acid concentration in a solution with [CH3CO2] = 0.050 M and [OH] = 2.5 × 10−6M at equilibrium.

The reaction is:

CH3CO2(aq) + H2O(l) ⇌ CH3CO2H(aq) + OH(aq)

Solution:

The provided equilibrium concentrations and a value for the equilibrium constant will permit calculation of the missing equilibrium concentration. The process in question is the base ionization of acetate ion, for which

Substituting the available values into the Kb expression gives

Solving the above equation for the acetic acid molarity yields [CH3CO2H] = 1.1 × 10−5M.

Check Your Learning:

What is the pH of a 0.083-M solution of NaCN?

 

Answer:

11.11

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.

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