[3] CATIONIC AS WELL AS ANIONIC HYDROLYSIS:

Take a salt (CH3COONH4) of the weak acid (CH3COOH) and the weak base (NH4OH) . and dissolve in water, therefore, the salt completely dissociates as given below.

The ions get hydrolysed according to the reaction.

Such salts undergoes hydrolysis because ,the aqueous solution contains unionised acid as well as  base molecules .

The nature of aqueous solution of such salt depends on the equilibrium constant for cationic or anionic hydrolysis.

Multiplying and dividing by H⁺ & OH^– and rearranging,

There is an important issue that needs clarification before we move on further. In this case,

 we can see that both the ions (i.e., cation and anion) get hydrolyzed to produce a weak acid and a weak base (hence, we can’t predict whether the solution is acidic, basic or neutral). We have considered the degree of hydrolysis of both the ions to be the same. Now we present an explanation as to why this is incorrect and then state reasons for the validity of this assumption

n.

 Actually the hydrolysis reaction given earlier, 

Now, we calculate the pH of the solution as:

If the reaction for hydrolysis is in equilibrium then all the reversible processes occurring in water must be in equilibrium .

The H+ or OH- ions may be calculated from the dissociation constant of acid or base , here calculation of H+ from acid is given as below .

We know that at 25° Pkw of water is 14 .

Hence 

             pH = 7+ 1/2[Pka~pkb]

If Kh1<Kh2 then ka>kb. and  pKa <pKb

as results Solution become acidic 

If Kh1>Kh2 then ka<kB and  pKa>pKb

as results  solution become  basic 

ILLUSTRATIVE EXAMPLE (1): calculate the pH of 0.2 M NH4CN Solution. ( Given Ka HCN is 3x10-10 and kb NH4OH is 2.0x10-5)

(Ans-pH 9.5 )

ILLUSTRATIVE EXAMPLE (2):

Calculate the DOD and pH of 0.2M NaCN Solution (Given Ka of HCN is 2.0x10-10)

(Ans- DOD = √2×10-10 and pH =11.5)

ILLUSTRATIVE EXAMPLE (3):

Calculate the DOD (h) and pH of 0.2 M C6CH5NH3Cl Solution (Given Ka C6CH5NH3Cl is =4.0×10-8)

(Ans- DOD =√20×10-4  and pH is 6.6)

ILLUSTRATIVE EXAMPLE (4):

ILLUSTRATIVE EXAMPLE (5):

[2] CATIONIC SALT HYDROLYSIS:

(2) CATIONIC HYDROLYSIS OR ACIDIC SALTS HYDROLYSIS:

          (Salt of a Weak Base and a Strong Acid)

      Let the acid be HCl and the base be NH₄OH. Therefore the salt would be NH₄Cl.

      NH₄Cl completely dissociates into NH₄⁺and Cl^– ions.

      HCl being a strong acid dissociates completely to give H⁺ ions and Cl^– ions.

In this hydrolysis, NH₄OH and H⁺ are being produced. This implies that the solution is acidic

To calculate pH,

Multiplying and dividing by OH^– and rearranging,

Now, substituting the concentrations,

ILLUSTRATIVE EXAMPLE (1):

ILLUSTRATIVE EXAMPLE (2):

ILLUSTRATIVE EXAMPLE (3):

ILLUSTRATIVE EXAMPLE (4):

ILLUSTRATIVE EXAMPLE (5):

     

[1] ANIONIC SALT HYDROLYSIS:

When a salt is dissolved in a solvent, it first dissociates into its constituent ions. This process is called dissolution. Now, if these ions chemically react with water, the process is called hydrolysis.

Salt hydrolysis is may be consider as the reverse of process of neutralization

We can also say that “combination of any of the ion furnished by the salt with water molecules is called hydrolysis “

Cationic hydrolysis will make the solution acidic but anionic hydrolysis will make the solution basic

(1) NEUTRAL SALTS: (Salts of strong acids and strong bases)

  A salts formed by complete neutralization of strong acid and strong base are called neutral salt such salts will not undergoes hydrolysis so aqueous solution of such salts be must neutral.

The salts that undergo hydrolysis after dissolution are
(2) ALKALINE SALTS:(Salts of weak acids and strong bases)
(3) ACIDIC SALTS:(Salts of weak bases and strong acids)
(4)  Salts of weak acids and weak bases

(1) ANIONIC HYDROLYSIS OR ALKALINE SALTS HYDROLYSIS:

                  (Salt of a Weak Acid and Strong Base) 

      Let us take a certain amount of weak acid (CH₃COOH) and add to it the same amount of a strong base (NaOH). They will react to produce CH₃COONa. 

      CH₃COONa being a strong electrolyte, completely dissociates into its constituent ions.

      Now, the ions produced would react with H₂O. This process is called hydrolysis

We know that NaOH is a strong base and therefore it would be completely dissociated to give Na⁺ and OH^– ions.

      Canceling Na⁺ on both the sides,

We can note here that ions coming from strong bases do not get hydrolysed. We should note here that the solution will be basic. This is because the amount of CH₃COOH produced and OH^– produced are equal. But CH₃COOH will not completely dissociate to give H⁺ ions. Therefore [OH^–] ions will be greater than [H⁺] ions.

      Since the reaction is at equilibrium,

This equilibrium constant K_c is given a new symbol, K_h

If we multiply and divide the above equation by [H⁺] of the solution, then

CASE (1): If a is very much less than 1, then 1-a= 1 this approximation valid when C/Kh is greater than 100”

CASE (2): If C/Kh is lower than 100 than calculate h by formation of quadratic equation.

ILLUSTRATIVE EXAMPLE (1): A 0.0258 M solution of the sodium salt, NaH of the weak monoprotic acid, HA has a pH of 9.65. Calculate Ka of the acid AH.

ILLUSTRATIVE EXAMPLE (2): What is the pH of 0.10 M CH₃COONa solution. Hydrolysis constant of sodium acetate is 5.6 × 10^-10   ?

SOLUTION: Hydrolysis of the salt may be represented as

ILLUSTRATIVE EXAMPLE (3): Calculate pH of 1.0 x 10^-3 M Sodium phenolate (Na⁺O^-C₆H₅ ) Ka for C₆H₅OH is 1.0 x10^{-10 .}

SOLUTION:  (Ans-  pH=10.43)

(2) CATIONIC HYDROLYSIS    OR  ACIDIC SALTS HYDROLYSIS:

         (Salt of a Weak Base and a Strong Acid)    .......

SALTS AND SALT HYDROLYSIS

(1) Salts are the compounds formed by neutralization between acids and bases which are containing at least one positive part (Cation) other than H⁺ and also at least  one Negative part (Anion) other than OH^-.

(2) Salts may taste salty, bitter, a stringer or sweet or tasteless

(3) Solution of salts may be acidic, basic or neutral.

(4) Fused salts and their aqueous solutions conduct electricity and undergo electrolysis

(5)The salts are generally crystalline solids

CLASSIFICATION OF SALTS:

These salts may be classified into four categories.

(A) SIMPLE SALTS:

The salts formed by the neutralization process between acid and base. These are of three types.

(1) Normal salt:     

The salt formed by the loss of all possible protons (replaceable H⁺ ions)

For example NaCl, NaNO₃, K₂SO₄, Ca₃ (PO4)₂, Na₃BO₃, Na₂HPO, NaH₂PO₂ etc.

(2) Acidic salts:

Salts formed by incomplete neutralization of polybasic acid. Such salts contain one or more replaceable H atom.

For examples NaHCO3, NaHSO₄, NaH₂PO₄, Na₂HPO₄ etc.

Above salts when neutralised by base form normal salts.

(3) Basic salts:

Salts formed by incomplete neutralization of poly acidic bases are called basic salts. These salt contain one or more hydroxyl groups.basic salt when neutralised by acids form normal salts.

Ex. Zn(OH)Cl,  Mg(OH)Cl,  Fe(OH)₂Cl, Bi(OH)₂ etc.

(B) DOUBLE SALTS:

(1)The addition compounds formed by the combination of two simple salts are termed as double salts.

(2) Double salts are stable in solid state only.

(3) When dissolved in water, it furnishes all the ions present in the simple salt form which it has been constituted.

(4)The solution of double salt shows the properties of the samples salts from which it has been constituted

For examples

Mohar’s salt-FeSO₄ (NH₄)₂SO₄ .6H₂O (Ferrous ammonium sulphate)

Alum’s- K₂SO₄Al₂ (SO4)₃. 24H₂O (Potassium ammonium sulphate)

Karnalite- KCl.MgCl_2.6 (H₂O)

Dolomite- CaCO₃.MgCO₃ or CaMg (CO₃)₂

(C) COMPLEX SALTS:

(1) Complex salts are formed by combination of two simple salts or molecular compounds.

      For examples   K₄Fe (CN)₆, Co(NH₃)₆ SO₄ etc.

(2) Complex salts are stable in solid states as well as solutions

(3) Complex salts On dissolving in water, if furnishes a complex ion.

(4) The properties of the solution are different from the properties of the substance from which it has been constituted.

(D) MIXED SALTS:

(1)The salt which furnishes more than one cation or more than one anion when dissolved in water is Called mixed salt.

FOR EXAMPLE - CaOCl₂, NaKSO₄, NaNH₄HPO4 etc.

SALT HYDROLYSIS:

[1] ANIONIC SALT HYDROLYSIS:
[2] CATIONIC SALT HYDROLYSIS:
[3] CATIONIC AS WELL AS ANIONIC HYDROLYSIS:

[4] Amphoteric salt hydrolysis:

BUFFER SOLUTION

Topics cover :
(1) DEFINITION
(2)TYPE OF BUFFER SOLUTION
(3) pH OF ACIDIC BUFFER
(4) pH OF BASIC BUFFER
(5) DILUTION OF BUFFER
(6) BUFFER CAPACITY OR BUFFER  INDEX
(7) IDEAL BUFFER SOLUTION

(1) DEFINITION:
The aqueous electrolyte solution which resist the any change in pH even after addition of small amount of strong acid or strong base  called  buffer Solution.

(2)TYPE OF BUFFER SOLUTION:
(1) Simple Buffer Solution or Neutral buffer
(2) Mixed Buffer Solution

(3) pH OF ACIDIC BUFFER:

(4) pH OF BASIC BUFFER:

(5) DILUTION OF BUFFER:

(6) BUFFER CAPACITY OR BUFFER  INDEX:

(7) IDEAL BUFFER SOLUTION:

ILLUSTRATIVE EXAMPLE (1):
Pkb(NH4OH) is 5 and a buffer solution contains 0.1M NH4OH and 0.1M (NH4)SO4 calculate pH of this buffer solution ?
(Ans-5.3 )

ILLUSTRATIVE EXAMPLE (2):
Pls of HX is 4.7 , (1) find the pH of Solution having 0.5 M HX and KX 0.25M ? (2) pH of this solution if it is diluted 10000 times ?
(Ans- (1) 4.4 (2) )

ILLUSTRATIVE EXAMPLE (3):
Calculate pH of acidic buffer mixture containing 1.0 M HA (Ka=1.5x10-1) and 0.1M NaH .
(Ans- 0.824)

ILLUSTRATIVE EXAMPLE (4):
Calculate the mass of NH3 and NH4Cl required to prepared a buffer solution of pH  =9.0 when total  concentration of buffering reagents is 0.6 mole /litre .( Pkb for NH3 is 4.7, log2 is 0.3010)
(Ans- a=0.2 mole ,b=0.4 mole)

ILLUSTRATIVE EXAMPLE (5):
One liter buffer solution is prepared by mixing of 1.0 mole HCOOH (formic acid) and 1 mole HCOONa . (Given Pka HCOOH =4.0) then calculate
(i) pH of buffer
(ii) pH of Solution if 1/3 mole of HCl is added
(iii) pH of Solution if 1/3 mole of NaCl is added
(iv) pH of Solution if it diluted to 10 liter
(v) pH of Solution if it diluted to 1000 liter
(Ans- i-4.0 , ii- 3.7 , iii-4.3010 , vi- 4.0 v- 4.07)

ILLUSTRATIVE EXAMPLE (6):
Calculate the pH of an aqueous solution originally containing 0.4 M acetic acid and 0.2 M sodium acetate (ka CH3COOH= 1.8x10-5 ).
(Ans- 4.4)

ILLUSTRATIVE EXAMPLE (7):
Calculate pH of Solution Originally having 0.2 M (NH4)SO4 and 0.4 M NH4OH (given Kb NH4OH =1.8x10-5)
(Ans- 9.26)
ILLUSTRATIVE EXAMPLE (8):
In 100 ml of 0.4M C6H6COOH Solution,0.1M C6H6COONa is added , calculate pH of resulting solution (given Ka C6H6COOH is 4x10-5)
(Ans-4.1)

ILLUSTRATIVE EXAMPLE (9):
A solution contains 0.2 mole acetic acid and 0.10 mole of sodium acetate ,made up to 10 liter volume , calculate the pH of Solution ( given Ka CH3COOH is 1.8x10-5)
(Ans-4.44)
ILLUSTRATIVE EXAMPLE (10):
What mass of , in gram ,of NaNO2 must added to 700 ml of 0.165 M HNO2 to produce a Solution with pH of 3.50 ? ( Ka HNO2 is 6.0x10-4)
(Ans- 15.1gm)

ILLUSTRATIVE EXAMPLE (11):
In 500 ml of a buffer solution containing 0.8 M CH3COOH and 0.6 M CH3COONa , 0.2 M HCl is added. Calculate the pH of Solution before and after adding HCl. (Ka CH3COOH is 1.8x10-5) .
(Ans- i - 4.62  ii- 3.96)

ILLUSTRATIVE EXAMPLE (12):
A mixture of 0.2 mole RNH2 and 0.4 mole RNH3Cl is  mixed .the volume of Solution prepared is 10 liter  (given Kb RNH2 is 10-5) calculate.
(i) pH of resulting solution
(ii) pH of Solution if diluted to 1000 litres
(iii) pH of Solution if 200 ml buffer is mixed with 2 milimoles of H+
(iv) pH of Solution if 200 ml buffer is mixed with 2 milimoles of OH-
(Ans- i-  5.30 , ii- 5.31 , ii- 8.3 , iv -9.0)