Welcome to ChemZipper !!!: October 2018

Search This Blog

[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)    .......

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.MgCl2.6 (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:

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)

BRONSTED LOWERY ACID-BASE CONCEPT

According to Bronsted theory the species which donate protons (H+) in any medium is consider as acid and the species which accept proton is consider as base.
Acid and base characters are realised in the presence of each other.
For example
CONJUGATE ACID-BASE PAIRS
(1) conjugate pair is acid-base pair differing in single proton (H+)
(2) conjugate acid is written by adding  H+ and conjugate is written by removing H+
.
(3) Strong acid has weak conjugate base and vice versa. Similarly strong base has weak conjugate and vice versa.

(4) Equilibrium always moves from strong acid to weak acid and strong base to weak base.


(5) Conjugate acid - base pair differ by only one proton. Reaction will always proceed from strong acid to weak acid or from strong base to weak base.
MERITES OF BRONSTED CONCEPT:
(1) the role of solvent clearly defined.
(2) the acidic and basic  character may be observed in non aqueous medium also .
(3) the acidic ,basic or Amphoteric nature of most of the substance may be defined.
(4) the acid having greater tendency to donate protons are stronger acid and base  having greater tendency to accept protons are stronger base .
(5)In conjugate pair ,if one is strong then other must be weak .
The weak acid or base are normally determined by comparing the the stability of different acid or base
DEMERITES OF BRONSTED CONCEPT:
(1) Proton is a nuclear particle hence reaction should not explained in term of proton.
(2) the neutralized process becomes multiples step process.
(3) Most of the Amphoteric solvent become Amphoteric.


AMPHOTERIC SPECIES (Amphiprotic):
The species which have a tendency to donate proton as well as accept proton (H+) such species are known as Amphoteric species.
For example H2
O,NH3 HS- ,HPO3- ,HC2OO4- , H2O4 etc

ILLUSTRATIVE EXAMPLES:
(1)The conjugate base of HCO3 is –
  (A) H2CO3        (B) CO2             (C) H2O      (D) CO3 
(2) The conjugate acid of HSO3- is -
  (A) SO32-          (B) SO42-                  (C) H2SO4    (D) H2SO3

(Ans: 1-D 2-D)
                                                                 @@@

ARRHENIUS ACID-BASE CONCEPT

CHARACTERS OF ACIDS:
(1) acids convert blue litmus to red and methyl orange  indicator to  red .
(2) Sour in taste.
(3) It liberate hydrogen gas with active metals.
(4) Acids neutralised the effect of base 
(5) acids increases the conduction of water.
CHARACTERS OF BASES:
(1) bases convert the red litmus to blue and methyl orange indicator to yellow.
(2) Phenylphthlene indicator (white) to pink.
(3) Bitter in taste and soapy in touch.
(4) Bases neutralised the effect of acid.
(5) They increases conductance of water.
ACID BASE CONCEPT:
(1) ARRHENIUS ACID-BASE CONCEPT
(2) BRONSTED LOWERY ACID-BASE CONCEPT 

(3) LEWIS ACID-BASE CONCEPT

(1) ARRHENIUS ACID-BASE CONCEPT
the substance which produces H+ in aqueous solution is consider as acid and the substance which give produces OH- in aqueous solution is consider as base
Arrhenius theory depend upon dissociation of water.







Type of Arrhenius acids:

Type of Arrhenius Bases :
         

Feature of Arrhenius theory:

(1) OH¯ ion is present also in hydrated form of H3O2¯, H7O4¯, H5O3¯
      H+ ion is present also in hydrated form of H3O+, H5O2+, H7O3+, H9O4+
(2) Neutralisation reaction can be easily explain by the Arrhenius theory .acids furnishing H+ ion  in water to large extent are strong acid.

Strength of acid or base:

(3) If Ka increases then concentration of H+ increases hence acidic strength is increases.
Similarly the base furnishing OH- ions to the large extent are strong base
Kb is dissociation constant for bases, if Kb is increases OH- increases, hence basic strength of base

(4)The term strong is used only for those acids or bases or bases which dissolved almost completely in water.

LIMITATIONS OF ARRHENIUS THEORY
(1) This theory explain nature of a substance only aqueous medium . It cannot be applied for non aqueous solution.
(2) It could not explain formation of hydronium ions like H3O- , H5O2- , and H7O3- .
(3) the nature of aqueous solution of AlCl3, CuSO4 ,BF3, B(OH)3 etc are acidic and aqueous solution of NH3 ,NaCO3 ,RNH2 R2NH,  R3N , C2H5N  etc are basic in nature cannot be explain by Arrhenius Concept.
(4) there are many  Amphoteric hydroxide Zn(OH)2  Al(OH)3 ,Pb(OH)2 , which cannot be  explain  by Arrhenius Concept.
(5) Arrhenius explain only when H+ is released it cannot explain when H+ is taken.
                                                        @@@@

SPONTANEOUS AND NON SPONTANEOUS PROCESS

SPONTANEOUS PROCESS: Those process which have natural tendency to take place ,they may or may not require initiation.
For example
(1) Flow of water from higher level to lower level .
(2) Flow of heat from high temperature to lower temperature.
(3) Radioactivity
(4) Cooling of cup of tea .
(5) Evaporation
(6) Condensation 
(7) Sublimation
(8) Burning of candle
(9) Dissolution of salt and sugar in water
(10) Burning of fuel
(11) melting of ice at room temperature.
NON SPONTANEOUS PROCESS: All spontaneous processes are non spontaneous processes in reverse direction it requires spot of external energy for their progress.
CRITERIA FOR SPONTANEITY:
(1) If dS universe is greater than Zero  then process is spontaneous (reversible)
(2) If dS is Zero then process is in reversible state of equilibrium
(3) If dS is lower than Zero then  process is non spontaneous
For  a spontaneous process entropy of univers is greater than Zero . In order to use entropy has a sole criteria , we need to have information about system as well as surrounding.
dS(system) + dS(surr) is greater than  Zero
.
.
.
.
In order to explain the spontaneous behaviour by using the parameters of system we can established  two criteria.
(1) Randomness
(2) Criteria of energy
A spontaneous process is one in which energy ( enthalpy) decreasing and Randomness of system increasing.
For spontaneous proceDH=-ve and dS=+ve

LEWIS ACID-BASE CONCEPT

According to Lewis acid base theory- acids are electron pair acceptor (electron deficient) and bases are electron pair donor and the combination of Lewis acid and Lewis base occur through coordinate bond formation.
For example
Other examples of Lewis acid-base neutralisation.
LEWIS ACIDS:
(1) Cations are act as Lewis acids example, H+ , Ag+ , Cu+2 , Na+ , Fe+2 , Hg+2  etc.
(2) electron deficient compounds are act as lewise acids for example , BF3 ,ACl3 ,BCl3 ,FeCl3 etc
(3) compounds with available vacant d orbital act as Lewis acids for example , SiCl4 ,PCl5 , SOF4 ,etc
(4) compounds containing multiple bonds for example  CO2.
LEWIS BASES:
(1) Anions are act as Lewis base H- ,CH3- ,NH2- , OH- , X- etc.
     All the anions are not Lewis base for example (PCl5)
(2) Molecules having lone pairs act as Lewis base example NH3 ,H2O , R-OH ,R-O-R , etc
(3) compounds with non polar multiples bond  are also act as Lewis base, example C2H4 , C2H2    (Ligand in coordination compounds)
MERITS OF LEWIS CONCEPT: 
(1) The acid/base nature of substance may also defined without any solvent.
(2) It is highly useful to explain coordination compounds in which central metal atom into behave as Lewis acid and ligands behave as Lewis base.
(3) the acid having greater tendency to accept lone pair bare stronger acid and base have greater tendency to donate lone pair are stronger base.
DEMERITES OF LEWIS CONCEPT:
(1) It is extremaly generalized Concept , almost all the compounds either become acid or base  by this Concept.

                                                                       @@@


LEVELLING EFFECT OF WATER

When strong acids like HClO4, HCl, HBr, HI, HNO3 etc are dissolved in water they are equally ionised (100%) it means they are equally strong in water this is called levelling effect of water .
Similarly for strong bases also , NaOH, KOH  Ca(OH)2 etc are in water behave the same . hence relative strength of strong acids or base can not be compare in water .
(1) Water does no show levelling effect for weak acids or base because they are ionised upto different extent in water.
(2) All the acids are stronger than H3O+ ion, consider as strong acids and weak than H3O+ ion are consider as weak acids.
(3) All the bases stronger than OH- are consider as strong base and weaker than OH- , then they are consider  as weak base.

DEFINITION: If more than one acids or bases are showing same acidic or basic strength in same solvent it is called labelling effect and the solvent is called labelling solvent.
For example HClO4 and HI equally dessociation in aqueous medium (99.99%)  it means they have equal strength.
Similarly HF and HCl are equally dessociation in NaOH (100 %).
If their strength differ in same solvent ,it is called differential effect and the solvent is called differential solvent

For example HClO4 dessociate 99.99 % in acetic acid while HI dissociate 99.8% in acetic acid, it means acidic strength of HClO4  is more than HI in acetic acid.
The reason behind labelling and diffential  effect is ability of solvent to donate or accept protons .
For acids, acidic solvents are differential and basic solvent are labelling.
 

VOLUME STRENGTH OF H2O2(aq)

 Hydrogen peroxide is a colourless liquid used as bleaching agent in some food and act as oxidising as well as reducing agent.The most common use of hydrogen peroxide is in cosmetic including personal care, hair dye, bleach shampoos conditioners and first aid  antiseptic.
Hydrogen peroxide also used as propellant for torpedoes on account of its exothermic decomposition.
The strength of Hydrogen peroxide expressed several ways like Volume strength, W/V % (% strength), Molarity, Normality etc.

COMMERCIAL LABELLING OF H2O2 AS ‘X’ VOLUME :  
Commercially H2O2 labelled as 'X' volume H2O2 (at P=1 ATM and T=273 K)
Where 'X' may be  10V, 20V, 30V, 40 V etc. 

Here  'X' Volume H2O2 means that 1ml of such solution of hydrogen peroxide on decomposition  produce 'X' ml of oxygen (at P=1 ATM and T=273 K).
For example 10 Volume H2O2 means that 1ml of such solution of hydrogen peroxide on decomposition  produce '10' ml of oxygen (at P=1 ATM and T=273 K).

W/V % or PERCENTAGE STRENGTH OF  H2O2SOLUTION: 
The amount of H2O2 in gram present in 100 ml of aqueous solution of H2O2. Called (w/v) or percentage strength.

Illustration:  taken 10 V (Vol) H2O2:

By Stoichiometric calculations:

Thus 1ml, 10 V labelled H2O2 (aq) solution contains 0.0303 gm H2O2.
Similarly 100 ml, 10 V labelled H2O2 (aq) solution contains 0.0303x100=3.03 gm of H2O2. This called W/V % or percentage strength of H2O2 solution.

MOLARITY AND NORMALITY OF H2O:
 And 1000 ml, 10 V labelled H2O2 (aq) solution contains 0.0303x1000=30.30 gm H2O2. It means 30.30 gm H2O2 dissolved in 1000 ml (1L) Solution.

  Hence Molarity of 10 V H2O2 Solution is = 0.89 M
  Normality of 10V H2O2 (aq) Solution: 
Hence Normality of 10 V H2O2 Solution is =1.79 N
 We also know that NORMALITY= MOLARITY X n-FACTOR
                              Normality= O.89x2 =1.79 N

ALTERNATIVE MATHAD TO DTERMINED MOLARITY AND NORMALITY:
Volume of O2 provided by 1unit Volume of H2O2 aq Solution
ILLUSTRATIVE EXAMPLES:
EXAMPLE: (1) A solution of H2O2 has Normality N/1.7 calculate w/V strength of this solution?.
EXAMPLE: (2) Convert 20 V H2O2 into % strength or w/v strength?
EXAMPLE: (3) Convert 6.8 % strength of H2O2 into Volume strength?
EXAMPLE: (4) 500 ml of a H2O2 solution on decomposition produce 2 mole of H2O. Calculate the Volume strength of of H2O2 of solution? (Given Volume of O2 measured at 1atm and 273 K)  (Ans - 44.8 V)
EXAMPLE: (5) What is the percentage strength of "15" volume of H2O2?
EXAMPLE: (6) What volume of H2O2 solution of "11.2 volume strength is required to liberate 2240 ml of O2 at NTP?
EXAMPLE: (7) 30 gm Ba(MnO4)2 sample containing inert impurity completely reacting  with 100 ml "28" volume strength of H2O2 in acidic medium then what will be the percentage purity of Ba(MnO4)2 in the sample ?. (Ba=137, Mn = 55, O =16)
EXAMPLE: (8) the strength of "20 Volume" H2O2 is equal? 
EXAMPLE: (9) What is weight of available oxygen from a solution of H2O2, if 20 ml of this solution need 25 ml, N/20 KMnO4 for complete oxidation? (Ans- 0.25)
EXAMPLE :( 10) calculate molarity of 44.8 V strength of 
H2O2 aq Solution?.
EXAMPLE: (11) calculate the molarity of H2O2 if 11.2 ml H2O2 require 30 ml of 0.5 M K2Cr2O7 for its complete oxidation; also calculate the Volume strength of H2O2? (Ans M=45/11.2 and volume strength=45)

Top Search Topics