Hydrolysis of Salts and pH of their Solutions

Some salts when seen look similar but when they are dissolved in water they will not look similar for example, K₂CO₃, NH₄Cl, NaHCO₃, Na₃PO₄, Na₂HPO_4, NaH₂PO₄ these salts look white in color, but these appear in different colors when dissolved in distilled water this is because of salt hydrolysis. Basic terms related to salt hydrolysis are

• Salt: It is a compound obtained when an acid and base react and this salt gets anion from acid and cation from the base and this process is called neutralization
  Acid + Basic = Salt + Water
• P^H: It is a measure of how acidic or basic a compound is and the PH range generally is 0 to 14.
  PH value of 7 indicates it is a neutral solution and below 7 indicates it as an acidic solution and above 7 indicates it as a basic solution, and PH range is different for different indicators. For example, take litmus paper it has a PH range of 5.5 to 8.3.
•   K (Equilibrium constant): It is the product of the concentration of products raised to their powers divided by the concentration of reactants raised to their powers.

Salt Hydrolysis

The process of cation or anion or both the ions present in a salt react with water to produce acidic, basic, or neutral solution is called salt hydrolysis. There are 3 types of hydrolysis possible. They are Complete hydrolysis, No hydrolysis, Limited hydrolysis.

• Complete hydrolysis: Complete hydrolysis consists of two types. They are acidic salt complete hydrolysis and basic salt complete hydrolysis. Let’s learn about both of these types in detail,

1. Acidic salt complete hydrolysis: If the anion is more acidic than their conjugate base and water is more basic than their conjugate base then the solution is acidic. Here H⁺ is released so it is an acidic solution.
2. Basic salt complete hydrolysis: If the cation is more basic than their conjugate acid and water are more acidic than their conjugate base then the solution is acidic. Here OH^– is released so it is a basic solution.

• No hydrolysis: No hydrolysis consists of two types. They are acidic salt hydrolysis and basic salt hydrolysis. Let’s learn about both of these types in detail,

1. Acidic salt hydrolysis: If the anion is less basic than its conjugate pairs and the water is less acidic than its conjugate pair.
   Example: Cl^– + H₂O ⇢ HCl + OH^–
2. Basic salt hydrolysis: If the cation is less acidic than its conjugate pairs and the water is less basic than its conjugate pair.
   Example: K⁺ + H₂O ⇢ KOH + H⁺

• Limited hydrolysis: When the cations or anions are not so strong compared to their conjugate pairs, hydrolysis relative to strength will take place, and accordingly the solution may be acidic or basic.
  Ions that undergo limited hydrolysis: Cations of weak bases result in the acidic solution. Anions of weak acids yield basic solutions.

There are 4 types of salt possible. Let’s learn about these four types in detail,

• Type 1: Salts of strong acids with strong bases

1. Cations from strong bases: Like Li⁺, Na⁺, k+, Rb+, Cs+.
2. Anions from strong acids: Like Cl-, Br-, I-, NO₃^–, SO^-2₄, ClO^–

NaCl solution in water:

NaCl + H₂O -⇢ Na⁺ + H⁺ + OH^– + Cl^–

Here Na+ and Cl- are strong electrolytes so they will not combine in water but H+ and OH- combine to form H₂O. The reaction is as follows: H⁺+OH^– ⇢ H₂O
This type of salts undergo neither cationic nor anionic hydrolysis and the nature of the solution is neutral. P^H of the solution is 7.

• Type 2: Salts of strong acids with weak bases

1. Cations from weak bases: Like NH₄⁺, Zn⁺², Fe⁺³, Cu⁺², Al⁺³,
2. Anions from strong acids: Like Cl-, Br-, I-, NO₃^–, SO₄^-2, ClO^–

For example let’s consider NH₄Cl in water:

NH₄Cl + H₂O ⇢ NH₄⁺ + H⁺+ OH^– + Cl^–

Here Cl- is a strong electrolyte so it will not combine in water but NH₄⁺ is a weak electrolyte it combines with OH- and forms NH₄OH. The reaction is as follows: NH₄⁺ + OH^– + H⁺ ⇢ NH₄OH + H⁺
Here as H+ is released in water so one can say cationic hydrolysis will occur and one can also say the resulting solution will be acidic. So the resulting P^H will be less than 7. Let the initial concentration of ammonium chloride be C and at equilibrium let ammonium hydroxide concentration be Ch. So, ammonium chloride concentration at equilibrium is C(1 – h).

	NH4Cl ⇢	NH4OH	+ H+
Initially,	C	–	–
At equilibrium,	C(1 – h)	Ch	Ch

Let K_h be the equilibrium constant of the reaction (Here h indicates ammonium chloride undergoes hydrolysis) 

K_h = (NH₄OH) (H⁺)/(NH₄Cl)  [let this be equation 1]

K_h = (Ch)(Ch) / C(1 – h) [Note: ammonium chloride is weak electrolyte so h will be negligible so 1 – h =1]. 

So k_h = Ch² [let this be equation 2]

For ammonium hydroxide, NH₄OH ⇢ NH₄⁺ + OH-

Let K_b be the equilibrium constant of the reaction( Here b indicates ammonium hydroxide as base )

K_b = (NH⁺₄)(OH-)/(NH₄OH) [let this be equation 3]

Now multiply both equation 1 and 3 then K_h × K_b = K_w

So , K_h=K_w/K_b [let this be  equation 4]

Now equate equations 2 and 4,

K_w/K_b = Ch²

h = √(k_h/C) [let this be equation 5]

[H⁺][OH^–] = K_w  => [ OH^–] = K_w/[H⁺]

Since, [H⁺] = Ch => [OH^–] = K_w/Ch

Now substitute equation 5 in this above expression, then   

[OH^–] = Kw/√(k_hC)

Now substitute equation 4 in this expression, then    

[ OH^–] = √(K_wK_b/C)

Now if we multiply with -log on both sides then

-log [ OH^–] = 1/2 [-logk_W – logk_b + logC]

P^OH = 1/2 [P^Kw + P^Kb + logC]

P^H = 14 -1/2 [P^Kw + P^Kb + logC]

(we know that P^Kw = 14)

P^H = 7 – 1/2 [P^Kw + logC]

• TYPE3: Salts of weak acids with strong bases

1. Cations from strong bases: Like Li⁺, Na⁺, k⁺, Rb⁺, Cs⁺
2. Anions from weak acids: Like F^–, CN^–, S^-2, CH3COO^–, CO₃^-2

For example, let’s consider CH₃COOK in water:

CH₃COOK +H₂O⇢CH₃COO^–+H⁺+OH^–+K⁺

[Here K+ is a strong electrolyte so it will not combine in water but CH3COO- is a weak electrolyte it combines with H+ and forms CH₃COOH]. Here reaction is as follows:   

CH₃COO^–+OH^–+H⁺⇢CH₃COOH+OH^–

Here as OH- is released in water so Anionic hydrolysis will occur and say the resulting solution will be base. So the resulting P^H will be more than 7.
Let the initial concentration of potassium acetate be C and at equilibrium let acetic acid concentration be Ch
So, potassium acetate concentration at equilibrium is C(1 – h). 

	CH3COOK⇢	CH3COOH	+OH–
Initially	C	–	–
At equilibrium	C(1 – h)	Ch	Ch

Let K_h  be the equilibrium constant of the reaction (Here h indicates potassium acetate  undergoes hydrolysis) 

K_h = (CH₃COOH)(OH-)/(CH₃COOK) [let this be equation 6]

K_h = (Ch)(Ch)/ C(1 – h)

[Note: Acetic acid is a weak electrolyte so h will be negligible so 1 – h = 1]

So k_h = Ch² [let this be equation 7]

For Acetic acid: CH₃COOH ⇢ CH₃COO^– + H⁺

Let K_a be the equilibrium constant of the reaction (Here a indicates acetic acid as acid)

K_a = (CH₃COO-)(H+)/(CH₃COOH) [let this be equation 8]

NOTE: [H⁺][OH^–] = K_w

Now multiply both equation 6 and 8,

K_h × K_a = K_w

K_h = K_w/K_a  [let this be equation 9]

Now equate equations 7 and 9,

K_w/K_a = Ch²

h = √(k_h/c) [let this be equation 10]

[H⁺][OH^–] = K_w

[H⁺] = K_w/[OH^–]

Since [OH^–] = Ch, 
[H⁺] = K_w/Ch

Now substitute equation 10 in this expression, then      

[H+] = K_w/√(k_hc) 

Now substitute equation 4 in this expression, then  

[H⁺] = √(K_wK_a/C)

Now multiply with -log on both sides then

-log [H⁺] = 1/2 [-logK_w – logk_a + logC]

 P^H = 1/2[P^Kw + P^ka + logC]

Since, P^Kw = 14

P^H = 7 + 1/2 [P^Ka + logC]

For P^OH = 14 – P^H

P^OH = 14 – 1/2 [P^Kw + P^Ka + logC] (since P^Kw = 14)

P^OH = 7 – 1/2 [P^Ka + logC]

• TYPE4: Salts of weak acids and weak bases

1. Cations from weak bases: Like NH⁺₄, Zn⁺², Fe⁺³, Cu⁺², Al⁺³
2. Anions from weak acids: Like F^–, CN-, S^-2, CH3COO-, CO₃^-2

For example, let’s consider ammonium cyanide in water,  

NH₄CN + H₂O ⇢ NH⁺₄ + H⁺ + OH^– + CN^–

Here both NH₄⁺ and CN^– are weak electrolyte so NH₄⁺ combine with OH^– and forms NH₄OH and CN^– combines with H⁺ and forms HCN. The reaction is as follows: NH₄⁺ + OH^– + H⁺ + CN^– ⇢ NH₄OH + HCN.

Here as no  H⁺ and OH- are released in water so we can say both cationic and anionic hydrolysis will occur. Let the initial concentration of ammonium cyanide be C and at equilibrium let ammonium hydroxide and hydrogen cyanide concentration be Ch.
So at an equilibrium concentration of ammonium cyanide be C(1 – h)

	NH4+	+CN–	NH4OH	+HCN
Initially	C	C	–	–
At equilibrium	C(1 – h)	C(1 – h)	Ch	Ch

K for reaction is K = (NH₄OH)(HCN)/(NH₄⁺)(CN^–) [let this be equation 11]

K = (Ch) × (Ch)/ C(1 – h) × C(1 – h) 

Note: Ammonium hydroxide is a weak electrolyte so h will be negligible so 1 – h =1

As, K = h²

So, h = √k [let this be equation 12]

For ammonium hydroxide, NH₄OH ⇢ NH₄⁺ + OH^–

Let K_b be the equilibrium constant of the reaction (Here b indicates ammonium hydroxide as a base)

K_b = (NH₄⁺)(OH^–)/(NH₄OH) [let this be equation 13] 

For hydrogen cyanide, HCN ⇢ H⁺ + CN^–

Let K_a be the equilibrium constant of the reaction (Here a indicates hydrogen cyanide as acid)

K_a = (H⁺)(CN^–)/(HCN) [let this be equation 14]

Multiply equation 11, 13 and 14 we get k × k_a × k_b = [H⁺][OH^–] = K_w

k = k_w/k_ak_b [let this be equation 15] 

Now from equation 14, 

[H⁺] = K_aCh/C(1 – h) 

Note: hydrogen cyanide is a weak electrolyte so h will be negligible so 1 – h =1

So, [H⁺] = K_ah

Now substitute equation 12,

[H⁺] = K_a√k

Now substitute equation 15,

[H⁺] = √k_wk_a/k_b

Now apply -log on both sides, 

P^H = 1/2 P^Kw + 1/2P^Ka – 1/2P^K_b

P^H = 7 + 1/2 (P^Ka – P^Kb)

1. Here, if ka = kb then P^H = 7 It implies the solution is neutral .
2. If k_a is more than k_b then P^H is less than 7. It implies the solution is acidic.
3. If k_a is less than k_b then P^H is more than 7. It implies solution is basic.

Sample Problems 

Question 1: ZnSO₄ undergoes cationic hydrolysis or anionic hydrolysis or both or nor both?

Answer:

We know that salts of strong acid with weak base undergoes cationic hydrolysis. 

If ZnSO₄ solution is dissolved in water:

ZnSO₄ + H₂O ⇢ Zn⁺² + H⁺+ OH^– + SO₄^-2

Here SO₄^-2  is strong electrolyte so it will not combine in water but Zn⁺² is weak electrolyte it combines with OH^– and forms Zn(OH)₂

The reaction is as follows: Zn⁺² + OH^– + H⁺ ⇢ Zn(OH)₂ + H⁺

Here as H⁺ is released in water so we can say cationic hydrolysis will occur and we can also say the resulting solution will be acidic. 

Question 2: Calculate the P^H for 50 ml of 0.1M NaOH + 50ml of 0.1 CH₃COOH. (P^Ka of CH₃COOH=4.8) 

Solution:

	CH3COOH	+NaOH⇢	CH3COONa	+H2O
Initially	5 millimole	5 millimole	–	–
Reacted	5 millimole	5 millimole	5 millimole	5 millimole
Remaining	0 millimole	0 millimole	5 millimole	5 millimole

Salt of strongbase with weak acid  which on water goes anionic hydrolysis.

P^H for a salt of weak acid and strong base

P^H = 7 + 1/2 [PKa + logC]

= 7 + 1/2(4.8 + log10 – 1)

= 8.9        

Question3: Among KCl, CH₃COONa, ZnSO₄, NaOH the following undergoes neither cationic nor anionic hydrolysis. 

Answer:

Salts of strong acid with strong base undergoes neither cationic nor anionic hydrolysis, here in given options only KCl is salt of strong acid with strong base.

KCl solution in water: KCl + H₂O ⇢ K⁺ + H⁺ + OH^– + Cl^–

Here K⁺ and Cl^– are strong electrolytes so they will not combine in water, but H⁺ and OH^– combine to form H₂O. The reaction is as follows: H⁺ + OH^–⇢ H2O

This type of salts undergo neither cationic nor anionic hydrolysis and the nature of solution is neutral, so answer is KCl    

Question4: Salt hydrolysis in water is due to? 

Answer:

Amphiprotic nature: Substance that accepts and donates proton and water can donate both
H⁺ and OH^– and the reaction is as follows: H₂O ⇢ H⁺ + OH^–

Question5: Calculate the extent of hydrolysis and the pH of 0.02M CH₃COONH₄. If [K_b(NH₃) = 1.8 × 10^-5 Ka(CH₃COOH) = 1.8 × 10^-5]

Answer:

CH₃COONH₄ is a salt of weak acid and weak base. 

k = k_w/k_ak_b

= (10 – 10)/(1.8 × 1.8 × 10 – 10) 

= 3.1 × 10^-5

k = h²

h = √k   

h = √(3.1 × 10^-5) = 5.56 × 10^-3

%h = 0.56%

As K_a = K_b hence it will be a neutral solution of P^H = 7 

Question6: The acid ionization constant of  Zn⁺² is 2.0 × 10^-10. What is the pH of 0.001 M solution of ZnCl₂ (log 2 = 0.3).     

Answer:

Zn(OH)₂ + 2HCl ⇢  ZnCl₂

Zn(OH)₂  is weak base and HCl strong acid.

So, ZnCl₂ is a salt of strong acid and weakbase. For salt of strong acid and weakbase,

 K_b × K_a = K_w, so, K_b = K_w/K_a

 K_b = (10^-14)/(2 × 10^-10) 

K_b = 5 × 10^-5

So, P_Kb = 4.3

P^H = 7 – 1/2P_Kb – 1/2logC 

P^H = 7 – 2.15 – 1/2log10 – 3  

P^H = 4.85 + 3/2 = 6.35                                                                                                                

Question7: What is the P^H of a 0.50 M aqueous (NaCN) solution? P_Kb of CN is 4.70. (log 2 = 0.3) 

Answer:

NaCN solution is a salt of strong base and weak acid.

P_Ka + P_Kb = 14, So, P_Ka = 9.3   

P^H = 7 + 1/2 [P_Ka + logC] 

P^H = 7 + 1/2[9.3 + log5 × 10^-1]

= 7 + 4.65 – 0.3/2

= 11.65 – 0.15

= 11.5