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Showing posts with label SALT HYDROLYSIS:. Show all posts
Showing posts with label SALT HYDROLYSIS:. Show all posts

Saturday, September 26, 2020

What are DOUBLE SALTS ?

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.MgCl2.6 (H2O)
Dolomite- CaCO3.MgCO3 or CaMg (CO3)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

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.


Sunday, November 4, 2018

[5] POLYVALENT ION HYDROLYSIS:

(1) POLY ANION HYDROLYSIS:(Poly acidic base):
For example PO4-3 (n=3) , C2O4-2 (n=2), S-2 (n=2) , N-3 (n=3) etc.
Na2S, NH3PO4, Na3PO4   (Hydrolysis is study just like hydrolysis of polybasic acids / polyacidic base) we know that dissociation constant of phosphoric acid is given as:

DISSOCIATION CONSTANT OF PHOSPHORIC ACID:

Step-1
Step-2
Step-3
HYDROLYSIS OF ANION (PO4-3):
Step wise illustration of hydrolysis of poly basic acids and polyacidic base is given as - 
Step-1
Experimentally we know that   Ka1>>Ka2>>Ka3, hence x >>y>>Z   so y and z can be neglected with respect to x 
Special Case (1):  
Special Case (2)
          then x can be calculate by quadratic equation. 
Step-2
Experimentally we know that   Ka1>>Ka2>>Ka hence   x>>y>>Z    So y and z can be neglected with respect to x and the x present in denominator and numerator both  are cancelled .
Step-3
Experimentally we know that   Ka1>>Ka2>>Ka hence   x>> y>>Z  So y and z present in numerator can be neglected with respect to x and the z present in denominator is also neglected with respect to y.
Finally -  
And concentration of different species  
 (2) POLY CATION HYDROLYSIS: (Poly basic acid):
For example Ca+2 , Fe+3 , Al+3 , Mn+2 , Cd+2  , Zn+3  etc
ILLUSTRATIVE EXAMPLE:
Calculate pH and concentration of all species in 0.1 M solution of FeCl3 given Fe(OH)3 have  Kb1=10-3 , Kb2=10-7 and Kb3=10-12.

SOLUTION:
  Fe+3 àundergo hydrolysis while, Cl-  à do not under goes hydrolysis.

   We know that 
Experimentally know that Kb1>>Kb2>>Kb3      so x>>Y and Y>>Z
Hence y and z neglected with respect to x
HYDROLYSIS OF FeCl3:
Experimentally we know that   Ka1>>Ka2>>Ka3 ,   hence  x>>y>>Z    So  y and  z can be neglected with respect to x

TITRATION BASED POLYVALENT ION HYDROLYSIS:

(1) TITRATION OF Na2CO3 Vs HCl:
(2) TITRATION OF NA3PO4 Vs HCl:

(1) TITRATION OF Na2CO3 Vs HCl:

ILLUSTRATIVE EXAMPLE:
The following volume of 0.1M HCl Solution is added to the in20 ml 0.1M Na2CO3 Solution the pH of resulting solution in each case.
(H2CO3, Ka1= 4×10-6, Ka2= 5×10-11)
(1) 0.0 ml (No HCl added)
(2) 10 ml HCl is added
(3) 20 ml HCl is added
(4) 30 ml HCl is added
(5) 40 ml HCl is added 
                                                            

Saturday, November 3, 2018

Amphoteric salts hydrolysis:

Example of amphoteric salts  NaHS, NaHCO3, Na2HPO4, NaH2PO4

(A) HCO3- act as conjugate acid as well as conjugate base:


Both reaction will support each other extent of hydrolysis and extent of dissociation is same.

(B) Here H2PO4- and HPO4-2 are amphoteric anions. The pH of amphoteric salts anions is independent of concentration of salts.

Here HPO4-2 is conjugate base of H2PO4- and H3PO4 is conjugate acid of H2PO4-Similarly PO4-3 is conjugate base of HPO4-2 and H2PO4-1 is conjugate  acid of HPO4-2

When these salts are dissolved in water [H3O+] concentration can be determined as;

ILLUSTRATIVE EXAMPLE: Calculate pH of solution of

(1) 100 ml 0.1M H3PO4 + 100 ml 0.1M NaOH.

(2) 100 ml 0.1M H3PO4 + 200 ml 0.1M NaOH.

(3) 100 ml 0.1M H3PO4 + 300 ml 0.1M NaOH.

(4) 100 ml 0.1M H3PO4 + 400 ml 0.1M NaOH.

                                                     

Wednesday, October 31, 2018

[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, thereforethe 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 Kc is given a new symbol, Kh
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)    .......