Salt Hydrolysis - Ionic Equilibrium | Physical Chemistry PDF Download

Salt Hydrolysis:

1.  Hydrolysis of salt of [SA – SB]:

• Solution is neutral in nature (pH = pOH = 7)

• Ex NaCl, Na₂SO₄, KNO₃ etc.

2.  Hydrolysis of [WA - SB] 

• In this type of salt hydrolysis anion reacts with water therefore called as anionic hydrolysis.

• Solution is basic in nature as [OH^–] increases.

• pH of solution is greater than 7.

1. Relation between k_h, k_w, k_a

CN^– + H₂O  HCN + OH^–

                                                                      …1

For weak acid, HCN  H⁺ + CN^–

                                                                          … 2

For water,    H₂O   H⁺ + OH^–

                        Kw = [H⁺] [OH^–]                                            … 3    

eq (1) x eq. (2) = eq (3)

k_b × k_a= kw

b)  Degree of hydrolysis 

                        CN^– + H₂O  HCN + OH^–

                         C                         0         0

                        C- Ch                Ch     Ch

Since h <<< 1

1 – h ≈ 1

Hence, k_h = Ch²

c)   pH of the solution

            [OH^–]  = Ch

            [OH^–] =  

            [OH^–] =

            Kw = [OH^–] [H⁺]

            [H⁺] =

            [H⁺] =

On taking log both sides

3) Hydrolysis of strong Acid – Weak Base [SA - WB] 

                        NH₄Cl, (NH₄)₂ SO₄ ZnCl₂ etc.

In this type of salt hydrolysis, cation reacts with H₂O, therefore called as cationic hydrolysis.

Solution is acidic in nature as [H⁺] is increased.

pH of the solution is less than 7.

a)   Relation between k_h, k_w and k_b

            NH₄⁺ + H₂O  NH₄OH + H⁺

Hydrolysis constant, k_h =                     …1

For weak base, NH₄OH   NH₄⁺ + OH^–

Kb =                          … 2

For water, H₂O  H⁺ + OH^–

K_w = [H⁺] [OH^–]                                                        … 3

Multiply eq. (1) x (2) = eq. 3

i.e. k_h x k_b = k_w

b) Degree of hydrolysis – represented by h

                        NH₄⁺ + H₂O  NH₄OH + H⁺

                        C                           0            0

                        C – Ch                  Ch         Ch

Since h <<< 1, then (1 – h) ≈ 1

K_h = Ch²

c)   pH of the solution

pH = - log [H⁺]

[H⁺] = Ch =

On taking – log on both sides

4. Hydrolysis of [WA – WB]

ex: NH₄CN, (NH₄)₂ CO₃ etc.

Solution may be neutral, acidic or basic depending upon the nature of acid & base.

a) Relation between Kn₁Kw₁Ka and Kb

            NH₄⁺ + CN^– + H₂O  NH₄OH + HCN

            K_b =                                                                                … 1

For weak base, NH₄OH   NH₄⁺+OH^–

            K_b =                                                                                                       …2

For weak acid, HCN  H⁺ + CN^–

For water, H₂O    H⁺ + OH^–

Kw = [OH^–] [H⁺]

Multiply eq. (1) ×eq (2) ×eq (3) = eq (4)

a) Degree of Hydrolysis 

                                    NH₄⁺ + CN^– + H₂O  NH₄OH + HCN

                                     C        C                            0               0

At equilibrium             C – Ch C – Ch                   Ch              Ch

Since, h <<< 1, then (1 - h) ≈ 1

K_h = h²

b) pH of the solution 

for eq. (3)

[H⁺] =

Since, h <<< 1, then (1-h)  ≈ 1

[H⁺] = ka× h = ka×

Taking log both sides

- log [H⁺] =

Case I: If pk_a = pk_b

Then pH = 7, the solution is neutral

Case II: If pK_a>pk_b

Then pH > 7, then solution is Basic

Case III: pK_a<pK_b

Then, pH < 7, solution is acidic

Example: what is the pH of 0.4 M aqueous solution of NaCN, given that the pk_b (CN^–) is .70.

Solution:         CN^– + H₂O   HCN + OH^–

                        C                        O         O

                        C – Ch               Ch       Ch

pK_b(CN^–) is given means equilibrium constant of this reaction is given. Because this reaction its express the basicity of CN^– ion.

pk_b = - log kb

4.7 = - log kb

K_b = 1.9 × 10^–5

now, k_b

since h <<< 1, the (1 – h) ≈ 1

k_b = Ch²

h = 6.89 × 10^–3

[OH^–] = Ch = 0.4 × 6.89 × 10^–3= 2.76 × 10^–3

pOH = - log [2.76 × 10^–3]

pOH = 2.55

pH = 11.45

Example : Find out the k_h of a centimolar solution of NH₄Cl. If the dissociation constant for NH₄OH is 10^–6 and kw = 10^–14. Find the degree of hydrolysis and pH of solution.

Solution: NH₄Cl is a salt of strong acid and weak base.

                        NH₄⁺ + H₂O  NH₄OH + H⁺

                        C                            0            0

                        C- Ch                   Ch        Ch

Equilibrium constant

K_b = 10^–6

K_h =   =10^–8

K_h = Ch²

10^–8 = Ch²

[H⁺] = Ch = 10^–2× 10^–3 = 10^–5

pH = 5

Dissociation of Polyprotic Acid

H₃A         H⁺        +          H₂A'                            ka₁

C- x                 x+y+z              x – y

H₂A^–        HA^2–    +          H⁺                                          ka₂

x-y                   y – z                x+y+z

HA^2–         A^3–       +          H⁺                                          ka₃

y-z                   z                      x+y+z 

assumption:     ka₁>> ka₂>>> ka₃

hence, x + y + z ≈ x

x – y≈ x

y - z≈ y

So, Ka₁ =

Ka₂ = y

• From eq1., you can calculate the value of x.

• By putting value of x you will get the value of y from eq. 2

• By putting values of x and y, you will get the value of 2 from eq. 3.

Hydrolysis of amphiprotic Anion:

This is one of the case of Anion hydrolysis, cations are not hydrolysed in this case.

NaHCO₃etc. can undergo ionisation to form H⁺ ion and can undergo hydrolysis to form OH^– (Na⁺ ion is not hydrolysed)

                                   Ionization

a)  1)   HCO₃^– + H₂O        CO₃^2– +H₃O⁺

                                                                    acid

                                      Hydrolysis

              HCO₃^– + H₂O       H₂CO₃ + OH^–

                                                                   Base

                        pH (HCO₃^–)    =