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Hydrolysis of Salts and pH of their Solutions

  • Last Updated : 01 Feb, 2022

Some salts when seen look similar but when they are dissolved in water they will not look similar for example, K2CO3, NH4Cl, NaHCO3, Na3PO4, Na2HPO4, 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
  • PH: 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. PH 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 PH 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)ChCh

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

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

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

So kh = Ch2 [let this be equation 2]

For ammonium hydroxide, NH4OH ⇢ NH4+ + OH-

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

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

Now multiply both equation 1 and 3 then Kh × Kb = Kw

So , Kh=Kw/Kb [let this be  equation 4]

Now equate equations 2 and 4,

Kw/Kb = Ch2 

h = √(kh/C) [let this be equation 5]

[H+][OH] = Kw  => [ OH] = Kw/[H+]

Since, [H+] = Ch => [OH] = Kw/Ch

Now substitute equation 5 in this above expression, then   

[OH] = Kw/√(khC)

Now substitute equation 4 in this expression, then    

[ OH] = √(KwKb/C)

Now if we multiply with -log on both sides then

-log [ OH] = 1/2 [-logkW – logkb + logC]

POH = 1/2 [PKw + PKb + logC]

PH = 14 -1/2 [PKw + PKb + logC]

(we know that PKw = 14)

PH = 7 – 1/2 [PKw + 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 PH 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 equilibriumC(1 – h)         Ch  Ch  

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

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

Kh = (Ch)(Ch)/ C(1 – h)

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

So kh = Ch2 [let this be equation 7]

For Acetic acid: CH3COOH ⇢ CH3COO+ H+

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

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

NOTE: [H+][OH] = Kw

Now multiply both equation 6 and 8,

Kh × Ka = Kw

Kh = Kw/Ka  [let this be equation 9]

Now equate equations 7 and 9,

Kw/Ka = Ch2

h = √(kh/c) [let this be equation 10]

[H+][OH] = Kw

[H+] = Kw/[OH]

Since [OH] = Ch, 
[H+] = Kw/Ch

Now substitute equation 10 in this expression, then      

[H+] = Kw/√(khc) 

Now substitute equation 4 in this expression, then  

[H+] = √(KwKa/C)

Now multiply with -log on both sides then

-log [H+] = 1/2 [-logKw – logka + logC]

 PH = 1/2[PKw + Pka + logC]

Since, PKw = 14

PH = 7 + 1/2 [PKa + logC]

For POH = 14 – PH

POH = 14 – 1/2 [PKw + PKa + logC] (since PKw = 14)

POH = 7 – 1/2 [PKa + 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 CNcombines 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++CNNH4OH  +HCN
Initially       CC
At equilibriumC(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 Kb be the equilibrium constant of the reaction (Here b indicates ammonium hydroxide as a base)

Kb = (NH4+)(OH)/(NH4OH) [let this be equation 13] 

For hydrogen cyanide, HCN ⇢ H+ + CN

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

Ka = (H+)(CN)/(HCN) [let this be equation 14]

Multiply equation 11, 13 and 14 we get k × ka × kb = [H+][OH] = Kw 

k = kw/kakb [let this be equation 15] 

Now from equation 14, 

[H+] = KaCh/C(1 – h) 

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

So, [H+] = Kah

Now substitute equation 12,

[H+] = Ka√k

Now substitute equation 15,

[H+] = √kwka/kb

Now apply -log on both sides, 

PH = 1/2 PKw + 1/2PKa – 1/2PKb

PH = 7 + 1/2 (PKa – PKb)

  1. Here, if ka = kb then PH = 7 It implies the solution is neutral .
  2. If ka is more than kb then PH is less than 7. It implies the solution is acidic.
  3. If ka is less than kb then PH 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 PH for 50 ml of 0.1M NaOH + 50ml of 0.1 CH3COOH. (PKa of CH3COOH=4.8) 

Solution:

 CH3COOH +NaOH⇢CH3COONa+H2O
Initially 5 millimole  5 millimole  
Reacted 5 millimole5 millimole   5 millimole   5 millimole   
Remaining   0 millimole 0 millimole 5 millimole5 millimole

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

PH for a salt of weak acid and strong base

PH = 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 OHcombine 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 OHand the reaction is as follows: H2O ⇢ H+ + OH 

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

Answer:

CH3COONH4 is a salt of weak acid and weak base. 

k = kw/kakb

= (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 Ka = Kb hence it will be a neutral solution of PH = 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) is weak base and HCl strong acid.

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

 Kb × Ka = Kw, so, Kb = Kw/Ka 

 Kb = (10-14)/(2 × 10-10

Kb = 5 × 10-5

So, PKb = 4.3

PH = 7 – 1/2PKb – 1/2logC 

PH = 7 – 2.15 – 1/2log10 – 3  

PH = 4.85 + 3/2 = 6.35                                                                                                                

Question7: What is the PH of a 0.50 M aqueous (NaCN) solution? PKb of CN is 4.70. (log 2 = 0.3) 

Answer:

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

PKa + PKb = 14, So, PKa = 9.3   

PH = 7 + 1/2 [PKa + logC] 

PH = 7 + 1/2[9.3 + log5 × 10-1]

= 7 + 4.65 – 0.3/2

= 11.65 – 0.15

= 11.5 


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