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14.5: Hydrolysis of Salt Solutions

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    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, 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

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

    Since \(\ce{HCl}\) is a strong acid, 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 (\(NH_4^{+}\)) 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 Ka for anilinium ion is derived from the Kb 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:

    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

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

    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 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 \(( \ce{CH3CO2^{-}})\), 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 Kb 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 Ka and Kb 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

    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

    \[\ce{Al(NO3)3(s) \rightleftharpoons Al^{3+}(aq) +3 NO3^{-}(aq)} \nonumber \]

    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

    \[\ce{Al(NO3)3(s) +6 H2O(l) \rightleftharpoons Al(H2O)6^{3+}(aq) +3 NO_3^{-}(aq)} \nonumber \]

    As shown in Figure \(\PageIndex{1}\), the 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:

    \[\ce{Al(H2O)6^{3+}(aq) + H2O(l) \rightleftharpoons H3O^{+}(aq) + Al(H2O)5(OH)^{2+}(aq)} \quad \quad K_{ a }=1.4 \times 10^{-5} \nonumber \]

    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:

    \[\begin{align*}
    \ce{Al(H2O)_6^{3+}(aq) + H2O(l) &\rightleftharpoons H3O^{+}(aq) + Al(H2O)_5(OH)^{2+}(aq)} \\[4pt]
    \ce{Al(H2O)_5(OH)^{2+}(aq) + H2O(l) &\rightleftharpoons H3O^{+}(aq) + Al(H2O)_4(OH)_2^{+}(aq)} \\[4pt]
    \ce{Al(H2O)_4(OH)_2^{+}(aq) + H2O(l) &\rightleftharpoons H3O^{+}(aq) + Al(H2O)_3(OH)_3(aq)}
    \end{align*} \nonumber \]

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

    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 \(\PageIndex{1}\): 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:

    \[\begin{align*}
    \ce{Fe(H2O)_6^{3+}(aq) + H2O(l) &\rightleftharpoons H3O^{+}(aq) + Fe(H2O)_5(OH)^{2+}(aq)} && p K_{ a }=2.74 \\[4pt]
    \ce{Cu(H2O)_6^{2+}(aq) + H2O(l) &\rightleftharpoons H3O^{+}(aq) + Cu(H2O)_5(OH)^{+}(aq)} && p K_{ a }= \sim 6.3 \\[4pt]
    \ce{Zn(H2O)_4^{2+}(aq) + H2O(l) &\rightleftharpoons H3O^{+}(aq) + Zn(H2O)_3(OH)^{+}(aq)} && p K_{ a }=9.6
    \end{align*} \nonumber \]

    Example \(\PageIndex{4}\): 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 (\(\ce{[Al(H2O)6]^{3+}}\)) in solution.

    Solution

    The equation for the reaction and Ka are:

    \[\ce{Al(H2O)6^{3+}(aq) + H2O(l) \rightleftharpoons H3O^{+}(aq) + Al(H2O)5(OH)^{2+}(aq)} \quad K_{ a }=1.4 \times 10^{-5} \nonumber \]

    An ICE table with the provided information is

    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 three columns and three rows. The first column has the following: 0.10, negative x, 0.10 minus x. The second column has the following: approximately 0, positive x, x. The third column has the following: 0, positive x, x.

    Substituting the expressions for the equilibrium concentrations into the equation for the ionization constant yields:

    \[\begin{align*}
    K_{ a } &=\frac{\left[ \ce{H3O^{+}} \right]\left[ \ce{Al(H2O)5(OH)^{2+}} \right]}{\left[ \ce{Al(H2O)6^{3+}} \right]}\\[4pt]
    &=\frac{(x)(x)}{0.10-x}=1.4 \times 10^{-5}
    \end{align*} \nonumber \]

    Assuming \(x \ll 0.10\) and solving the simplified equation gives:

    \[x=1.2 \times 10^{-3} ~\text{M} \nonumber \]

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

    \[[\ce{H3O^{+}}]=0+x=1.2 \times 10^{-3}~\text{M} \nonumber \]

    \[\text{pH} =-\log \left[ \ce{H3O^{+}} \right]=2.92 \nonumber \]

    Therefore this is an acidic solution.

    Exercise \(\PageIndex{4}\)

    What is \([\ce{Al(H2O)5(OH)^{2+}}]\) in a 0.15-M solution of \(\ce{Al(NO3)3}\) that contains enough of the strong acid \(\ce{HNO3}\) to bring [\(\ce{H3O^{+}}\)] to 0.10 M?

    Answer

    \(2.1 \times 10^{−5} M\)

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