2.11: Hydrolysis of Salts

Learning Objectives

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

• 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

\[\ce{NH4Cl(s) \rightleftharpoons NH4^{+}(aq) + Cl^{-}(aq)} \nonumber \]

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

\[\ce{NH4^{+}(aq) + H2O(l) \rightleftharpoons H3O^{+}(aq) + NH3(aq)} \quad K_{ a }=K_{ w } / K_{ b } \nonumber \]

Since ammonia is a weak base, K_b is measurable and K_a > 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

\[\ce{Cl^{-}(aq) + H2O(l) \rightleftharpoons HCl(aq) + OH^{-}(aq)} \quad K_{ b }=K_{ w } / K_{ a } \nonumber \]

Since \(\ce{HCl}\) is a strong acid, K_a is immeasurably large and K_b ≈ 0 (chloride ions don’t undergo appreciable hydrolysis).

Thus, dissolving ammonium chloride in water yields a solution of weak acid cations (\(NH_4^{+}\)) and inert anions (\(Cl^{-})\), resulting in an acidic solution.

Example \(\PageIndex{1}\): Calculating the pH of an Acidic Salt Solution

Aniline is an amine that is used to manufacture dyes. It is isolated as anilinium chloride, 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

\[\ce{C6H5NH3^{+}(aq) + H2O(l) \rightleftharpoons H3O^{+}(aq) + C6H5NH2(aq)} \nonumber \]

Solution

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

\[K_{ a }=\frac{K_{ w }}{K_{ b }}=\frac{1.0 \times 10^{-14}}{4.3 \times 10^{-10}}=2.3 \times 10^{-5} \nonumber \]

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

Substituting these equilibrium concentration terms into the K_a expression gives

\[\begin{align*}
K_a & =\dfrac{\left[ \ce{C6H5NH2} \right]\left[ \ce{H3O^{+}} \right] }{\left[ \ce{C6H5NH3^{+}}\right]} \\[4pt]
2.3 \times 10^{-5} & = \dfrac{(x)(x)}{0.233-x}
\end{align*} \nonumber \]

Assuming \(x \ll 0.233\), the equation is simplified and solved for \(x\):

\[\begin{align*}
& 2.3 \times 10^{-5}= \dfrac{x^2}{0.233} \\[4pt]
& x=0.0023 ~\text{M}
\end{align*} \nonumber \]

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

\[\text{pH} =-\log \left[ \ce{H3O^{+}} \right]=-\log (0.0023)=2.64 \nonumber \]

Exercise \(\PageIndex{1}\)

What is the hydronium ion concentration in a 0.100-M solution of ammonium nitrate, \(\ce{NH4NO3}\), a salt composed of the ions \(NH_4^{+}\) and \(NO_3^{-}\) and.

Which is the stronger acid \(\ce{C6H5NH3+}\) or \(\ce{NH4+}\)?

Answer

\([\ce{H3O^{+}}] = 7.5 \times 10^{−6} ~\text{M}\); is the stronger acid.

Salts with Basic Ions

As another example, consider dissolving sodium acetate in water:

\[\ce{NaCH_3CO_2(s) \rightleftharpoons Na^{+}(aq) + CH_3CO_2^{-}(aq)} \nonumber \]

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, \(\ce{CH3CO2^{-}(aq)}\), is the conjugate base of acetic acid, \(\ce{CH3CO2H}\), and so its base ionization (or base hydrolysis) reaction is represented by

\[\ce{CH3CO2^{-}(aq) + H2O(l) \rightleftharpoons CH_3 CO_2 H (aq) + OH -(aq)} \quad K_{ b }=K_{ w } / K_{ a } \nonumber \]

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

Dissolving sodium acetate in water yields a solution of inert cations (Na⁺) and weak base anions \(( \ce{CH3CO2^{-}})\), resulting in a basic solution.

Example \(\PageIndex{2}\): Equilibrium in a Solution of a Salt of a Weak Acid and a Strong Base

Determine the acetic acid concentration in a solution with and \([\ce{OH^{-}}] = 0.050~\text{M}\) and \([\ce{OH^{-}}] = 2.5 \times 10^{−6} M\) at equilibrium. The reaction is:

\[\ce{CH_3CO_2^{-}(aq) + H2O(l) \rightleftharpoons CH_3CO_2H(aq) + OH^{-}(aq)} \nonumber \]

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

\[\ce{CH_3CO_2^{-}(aq) + H2O(l) \rightleftharpoons CH_3CO_2H(aq) + OH^{-}(aq)} \nonumber \]

Substituting the available values into the K_b expression gives

\[\begin{align*}
K_{ b } &=\frac{\left[ CH_3 CO_2 H \right]\left[ \ce{OH^{-}} \right]}{\left[ \ce{CH3CO2^{-}} \right]}=5.6 \times 10^{-10} \\[4pt]
&=\frac{\left[ \ce{CH3CO2H} \right]\left(2.5 \times 10^{-6}\right)}{(0.050)}=5.6 \times 10^{-10}
\end{align*} \nonumber \]

Solving the above equation for the acetic acid molarity yields \([\ce{CH3CO2H}] = 1.1 \times 10^{−5}~ \text{M}\).

Exercise \(\PageIndex{2}\)

What is the pH of a 0.083-M solution of \(\ce{NaCN}\)?

Answer

11.11

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 K_a and K_b values allows prediction of the solution’s acid-base status, as illustrated in the following example exercise.

Example \(\PageIndex{3}\): 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. Na₂HPO₄
4. NH₄F

Solution

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

(a) The \(\ce{K^{+}}\) cation is inert and will not affect pH. The bromide ion is the conjugate base of a strong acid, and so it is of negligible base strength (no appreciable base ionization). The solution is neutral.

(b) The \(\ce{Na^{+}}\) cation is inert and will not affect the pH of the solution; while the \(\ce{HCO3^{−}}\) anion is amphiprotic. The \(K_a\) of \(\ce{HCO3^{-}}\) is \(4.7 \times 10^{−11}\), and its \(K_b\) is:

\[\frac{1.0 \times 10^{-14}}{4.3 \times 10^{-7}}=2.3 \times 10^{-8} \nonumber \]

Since \(K_b \gg K_a\), the solution is basic.

(c) The \(\ce{Na^{+}}\) cation is inert and will not affect the pH of the solution, while the \(\ce{HPO4^{2−}}\) anion is amphiprotic. The \(K_a\) of \(\ce{HPO4^{2−}}\) is \(4.2 \times 10^{−13}\), and its \(K_b\) is:

\[\frac{1.0 \times 10^{-14}}{6.2 \times 10^{-8}}=1.6 \times 10^{-7} \nonumber \]

Because \(K_b \gg K_a\), the solution is basic.

(d) The \(\ce{NH4^{+}}\) ion is acidic (see above discussion) and the \(\ce{F^{−}}\) ion is basic (conjugate base of the weak acid \(\ce{HF}\)). Comparing the two ionization constants: \(K_a\) of \(\ce{NH4^{+}}\) is 5.6 × 10−10 and the \(K_b\) of \(\ce{F^{−}}\) is \(1.6 \times 10^{−11}\), so the solution is acidic, since \(K_a > K_b\).

Exercise \(\PageIndex{3}\)

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

1. K₂CO₃
2. CaCl₂
3. KH₂PO₄
4. (NH₄)₂CO₃

Answer

(a) basic; (b) neutral; (c) acidic; (d) basic