14.4 Hydrolysis of Salts

Salts are compounds made of the reactions of acids and base in aqueous or none aqueous solution.

Examples:

Salts formed in aqueous solutions:

HCl (aq)   +   AgOH {aq)      AgCl (s)    +  H₂O  (l)

AgCl (s) is solid salt.

H₂SO₄ (aq)    +   Ba(OH)₂          BaSO₄ (s)    +     2 H₂O (l)

Salts formed in non-aqueous solutions:

HCl (aq)                 +        C₆H₅-NH₂       C₆H₅-NH₃ ⁺ Cl ^– (s)

Hydrochloric acid  +        Anillin             Anilinium chloride (not soluble in water).

The salts can be acidic, basic or neutral depending on their hydrolysis in water.

Hydrolysis is the dissociation of a salt in water to produce weak acid or weak base.

Strong acids or strong base should not yield a hydrolysis and the reaction of the cation or the anion of the salt are considered Not Reacting with water.

Example of an acidic salt: NH₄Cl (ammonium chloride)

NH₄Cl (s)    +    H₂O (l)        NH₄ ⁺ (aq)   +    Cl  ^– (aq)   +   H₂O (l)

Now let both cation and anion hydrolyzed in water and see what both are producing weak or strong acid or base:

NH₄ ⁺ (aq)  + H₂O (l)       NH₄(OH) (aq)    +  H ⁺ (aq) -à  Hydrolysis occurs (weak base is produced)

Cl  ^– (aq)  + H₂O (l)     HCl (aq)    +  OH ⁺ (aq)  -à No Hydrolysis (strong acid is produced)

NH₄(OH) (aq)   is a weak base. Then hydrolysis does occur and it produces H ⁺ (aq). Thus NH₄Cl (s)   is a acidic compound.

Example of a basic salt: NaCH₃COO (sodium acetate)

NaCH₃COO (s)  +    H₂O (l)        Na⁺ (aq)   +    CH₃COO  ^– (aq)   +   H₂O (l)

Now let both cation and anion hydrolyzed in water and see what both are producing weak or strong acid or base:

Na⁺ (aq)  + H₂O (l)       NaOH (aq)    +  H ⁺ (aq)   -à  No Hydrolysis (strong base is produced)

CH₃COO  ^– (aq) + H₂O (l)      CH₃COOH (aq)    +  OH ^– (aq)   -à  Hydrolysis (weak acid is produced)

CH₃COOH (aq)   is a weak acid. Then hydrolysis does occur and it produces OH ^– (aq). Thus NaCH₃COO (s)   is a basic compound.

Example of a neutral salt: NH₄CH₃COO (ammonium acetate)

Hydrolysis of NH₄( aq):

NH₄ ⁺ (aq)  + H₂O (l)     NH₄(OH) (aq)    +  H ⁺ (aq) -à  Hydrolysis occurs (weak base is produced)

NH₄ ⁺ (aq) is hydrolyzed in H₂O and it is acidic.

Hydrolysis of CH₃COO  ^– (aq):

CH₃COO  ^– (aq) + H₂O (l)      CH₃COOH (aq)    +  OH ^– (aq)   -à  Hydrolysis (weak acid is produced)

CH₃COO  ^– (aq) is hydrolyzed in H₂O and it is basic.

Both the cation NH₄ ⁺ and the anion CH₃COO ^–  are hydrolyzed in H₂O.

How can we determine if the NH₄CH₃COO (ammonium acetate) is acidic or basic?

Answer:

In order to determine if the ammonium acetate is acidic or basic, one should use the Ka and Kb values:

NH₄ ⁺ (aq)  + H₂O (l)       NH₄(OH) (aq)    +  H ⁺ (aq)                  Ka = 5.60 x 10 ^{– 10}

CH₃COO  ^– (aq) + H₂O (l)      CH₃COOH (aq)    +  OH ^– (aq)     Kb = 5.60 x 10 ^{– 10}

Since Ka = Kb, then NH₄CH₃COO (ammonium acetate) is neutral.

Now, we can cover the calculations including salts in solutions.

Calculating the pH of acidic salt solution:

Example:

What is the pH of a 0.0235 M NH₄Cl solution?

NH₄ ⁺ (aq) is hydrolyzed and Cl  ^– (aq) is not.

NH₄ ⁺ (aq)  + H₂O (l)       NH₄(OH) (aq)    +  H ⁺ (aq)                  Ka = 5.6 x 10 ^{– 10}

Using the ICE Table

NH4 + (aq)  + H2O (l)       NH4(OH) (aq)    +  H + (aq)

Initial:                  0.0235 M            –                           0                          0

Change:                – X                                                 +X                      +X

Equilibrium:           0.0235 – X                                   +X                      +X

Ka  =  { [NH₄(OH) (aq)] * [H ⁺ (aq)] }  /  { [NH₄ ⁺ (aq)] }     =  5.60 x 10 ^{– 10}

Ka  =  { [X] * [X] }  /  { [0.0235 – X)] }     =  5.6 x 10 ^{– 10}

Using the simplified assumption:       0.0235  >>>  X

Ka  =  { [X] * [X] }  /  { [0.0235] }     =  5.60 x 10 ^{– 10}

[X] ²   =  [5.6 x 10 ^{– 10}] * [0.0235]

[X] = 3.63 x 10  ^{– 6} = [H ⁺]

pH = – log [H ⁺] = 5.44

Example:

What is the pH of a 0.0235 M NaCH₃COO solution?

CH₃COO  ^–  is hydrolyzed but Na  ⁺ is not.

CH₃COO  ^– (aq) + H₂O (l)    CH₃COOH (aq)    +  OH ^– (aq)     Kb = 5.60 x 10 ^{– 10}

Using the ICE Table

CH3COO  – (aq) + H2O (l)    CH3COOH (aq)    +  OH – (aq)

Initial:                  0.0235 M            –                               0                                0

Change:                – X                                                     +X                              +X

Equilibrium:           0.0235 – X                                       +X                              +X

Ka  =  { [NH₄(OH) (aq)] * [H ⁺ (aq)] }  /  { [NH₄ ⁺ (aq)] }     =  5.60 x 10 ^{– 10}

Ka  =  { [X] * [X] }  /  { [0.0235 – X)] }     =  5.6 x 10 ^{– 10}

Using the simplified assumption:       0.0235   >>>   X

Ka  =  { [X] * [X] }  /  { [0.0235] }     =  5.60 x 10 ^{– 10}

[X] ²   =  [5.6 x 10 ^{– 10}] * [0.0235]

[X] = 3.63 x 10  ^{– 6} = [OH ^–]

pOH = – log [OH ^–] = 5.44

pH = 14.0 – 5.44 = 8.56

The Ionization of Hydrated Metal Ions:

Hydrated metals ions (not group 1 A and 2 A metal ions) are hydrolyzed in water

[Co(H₂O)₆] ^{2 +} cobalt hexahydrate ion complex exhibits the following ionization:

[Co(H₂O)₆] ^{2 +} (aq)    +   H₂O (l)        [Co(H₂O)₅(OH)] ^{2 +} (aq)   +  H₃O  ⁺ (aq)    Ka = 3.0 X 10  ^{– 10}

The ionization continues till a very stable coordination complex is established. The complexes transfer the protons:

[Co(H₂O)₅(OH)] ^{2 +} (aq)   +  H₃O  ⁺ (aq)    [Co(H₂O)₄(OH)₂] ^{2 +} (aq)   +  H₃O  ⁺ (aq) 

[Co(H₂O)₄(OH)₂] ^{2 +} (aq)   +  H₃O  ⁺ (aq)    [Co(H₂O)₃(OH)₃] ^{2 +} (aq)   +  H₃O  ⁺ (aq) 

[Co(H₂O)₃(OH)₃] ^{2 +} (aq)   +  H₃O  ⁺ (aq)    [Co(H₂O)₂(OH)₄] ^{2 +} (aq)   +  H₃O  ⁺ (aq) 

[Co(H₂O)₆] ^{2 +} is a polyprotic acid. The acid strength of these complex coordinated ions increases with increasing charge of the metal ion and with decreasing size of such metal ions. The first-step acid ionization equation will yield higher ionization constant compared with the sequential following ionization steps.