Hydrolysis

Acids-Base Properties of Aqueous Solutions of Salts  

We can assume that when salts dissolve in water, they are completely ionized; Most salts are strong electrolytes.  Consequently, any acid-base properties shown by a salt solution are due to the behavior of cations and anions.   Many ions are able to react with water to produce H₃O⁺(aq) or OH^-(aq).  This type of reaction is called hydroysis.

  Ions as Acids 

All cations that are the conjugate acids of weak bases act as acids in aqueous solutions.  For example, the ammonium ion, NH₄⁺, the conjugate acid of the weak base NH₃, is an acid:

NH₄⁺(aq)  +  H₂O(l)  <==>  H₃O⁺(aq)  +  NH₃(aq)

The following table summarizes  the behavior of some common cations in water.

  Ions as Bases

All anions that are the conjugate bases of weak acids act as proton acceptors, so we expect them to give basic solutions.  For example, formate ion, the conjugate base of formic acid will act as a base in water.

H₂O(l)  +  HCO₂^-(aq)  <==>    HCH₂O₂(aq)  +   OH^-(aq)

The following table summarizes  the behavior of some common anions in water.

We can summarize the facts presented above in a set of rules for predicting whether a salt solution will be neutral, acidic, or basic. These rules apply to salts that do not have an acidic hydrogen atom.

1.   A salt of a strong base and a strong acid will be neutral.  An example is NaBr.

2.   A salt of a strong base and a weak acid will be basic.  An example is NaF.

3.   A salt of a weak base and a strong acid will be acidic.  An example is NH₄Br.

4.   A salt of a weak base and a weak acid can be neutral, acidic or basic depending on the relative strength of the acid and base.

     a.   If K_a(cation) > K_b(anion) the solution of the salt is acidic.

     b.   If K_a(cation) = K_b(anion) the solution of the salt is neutral.  

     c.   If K_a(cation) < K_b(anion) the solution of the salt is basic.

Calculating the pH of a Salt Solution

First let us examine the relationship between the K_a for a weak acid and the K_b for its conjugate base.  To see this relationship between K_a and K_b for conjugate acid-base pairs, consider the acid ionization of a generic weak acid, HA, and the hydrolysis of its conjugate base, A^-.  When these two reactions are added you get the ionization of water.

HA(aq)  +  H₂O(l)  <==>    H₃O⁺(aq)  +  A^-(aq)   K_a 

A-(aq)  +  H₂O(l)  <==>    HA(aq)  +  OH^-(aq)     K_b 

---------------------------------------------------------------

           2 H₂O(l) <==>  H₃O⁺(aq)  +  OH^-(aq)       K_w 

In our discussion of equilibrium we demonstrated that when two reactions are added, their equilibrium constants are multiplied.  Therefore

KaKb = Kw 

For a solution of a salt in which only one of the ions hydrolyzes, the calculation of the equilibrium composition is very similar to that for a weak acid or weak base.  The main difference being that you must first calculate K_a or K_b for the ion that hydrolyzes from the relationship above, since values for K_a or K_b for salts are generally not tabulated.

Example 1 

Sodium fluoride is a salt from the reaction of sodium hydroxide, a strong base, and hydrofluoric acid a weak acid, therefore we expect the solution to be basic.  The fluoride ion will hydrolyze so we need to determine its K_b.  From a table of acid-ionization constants we see that the K_a for HF is 6.8 x 10^-4.  We use this and the K_w value to determine the value of K_b for F^-.

     1.0 x 10-14 
Kb = -------------- = 1.5 x 10-11
       6.8 x 10-4  

Next we construct our usual table.

The next step is to substitute the equilibrium concentrations into the equilibrium constant expression.

     [HF][OH-]                 (x)(x)
Kb = --------- = 1.5 x 10-11 ----------
       [F-]                   0.10 - x

x² = 1.5 x 10^-12

x = [OH^-] = 1.2 x 10^-6

pH = 14.00 - 5.92 = 8.08

Example 2  

What is the pH of a 0.10 M solution of ammonium chloride? 

Ammonium chloride is a salt from the reaction of ammonia, a weak base, and hydrochloric acid a strong acid, therefore we expect the solution to be acidic.  The ammonium  ion will hydrolyze so we need to determine its K_a.  From a table of base-ionization constants we see that the K_b for NH₃ is 1.8 x 10^-5.  We use this and the K_w value to determine the value of K_a for NH₄⁺.

     1.0 x 10-14 
Ka = -------------- = 5.6 x 10-10
       1.8 x 10-5  

Next we construct our usual table.

The next step is to substitute the equilibrium concentrations into the equilibrium constant expression.

     [H3O+][NH3]                 (x)(x)
Ka = ----------- = 5.6 x 10-10 ----------
       [NH4+]                   0.10 - x

x² = 5.6 x 10^-11

x = [H₃O⁺] = 7.5 x 10^-6