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PERCENTAGE (%) AVAILABLE CHLORINE IN BLEACHING POWDER:

Bleaching powder is not a true compound but it is a mixed salt of calcium hypochlorite [Ca (OCl) 2].3H2O and basic calcium chloride [CaCl2Ca (OH) 2H2O]. [Simply it is represented by CaOCl2.
The bleaching action of [CaOCl2] is due to liberation of oxygen with limited quantity of dilute acid.                                                                                                        

Whereas disinfectant action available of Cl2 on reaction with excess of acid.
The Chlorine liberates is called available Chlorine, and calculation of available Chlorine is called IODOMETRY.

IODOMETRIC-CALCULATION OF AVAILABLE % CHLORINE: Take a W (gm) sample of bleaching powder; and  when bleaching powder treated with dilute acid or water then liberates chlorine gas.
The chlorine produced in above reaction is titrated with KI solution which produced (I2) Iodine which is further completely titrated with hypo solution.                                          

If number of millimoles of hypo solution consume is M V then millimoles of %available chlorine is calculated as ;
                            Millimoles of Iodine is =1/2 Millimoles Hypo solution
                            Millimoles of Iodine is = Millimoles chlorine
Weight of available chlorine is = Number of moles x molecular Wt of Chlorine 
ILLUSTRATIVE EXAMPLE(1): If trace of chlorine are not remove from pulp used for paper manufacturing , then on the long standing it weaken the paper and makes it yellowish , which of following antichlor can be used to remove chlorine from pulp ?.
(A) Na2S2O3        (B) conc. HCl            (C) NaHCO3           (D) Both (A) and (B)
SOLUTION:
An antichlor  is a substance used to decompose residual hypochlorite or chlorine after use of chlorine based bleaching , in order to prevent ongoing reactions .example s of antichlor – Sodiumbisulphite (NaHSO3), Pottassiumbisulphite (KHSO3) ,  Sodiummetasulphite (Na2S2O3), Sodiumthiosulphate  (Na2S2O3) and hydrogen peroxide (H2O2).
ILLUSTRATIVE EXAMPLE (2): Available chlorine in sample of bleaching powder can be calculated as per the following reactions.
If 4 gm of bleaching powder dissolved to give 100 ml solution 25 ml of it react with excess of CH3COOH and KI .The Iodine liberation required 10 ml of 0.125 N Hypo solutions. Calculate % of available chlorine in the sample.
(A)  21%                       (B) 35 %                     (C) 45 %             (D) 4.4%
SOLUTION:
ILLUSTRATIVE EXAMPLE (3): Calculate the % of available chlorine in the sample in a sample of 3.35 gm of bleaching powder which was dissolved to 100 ml water. 25 ml of this solution on treatment react with KI and dilute acid. Required  20 ml 0.125 N Hypo solution (Sodiumthiosulphate- Na2S2O3). 
SOLUTION:
ILLUSTRATIVE EXAMPLE (4): 25 ml of household bleach solution was mixed with 30 ml of 0.50 M KI and 10 ml of 4.0N acetic acid. In this titration of the liberated iodine 48 ml of 0.25 N Na2S2O3 was used to reach the end point. The Molarity of household bleach solution is?
SOLUTION:

PH OF POLYPROTIC ACIDS:

The acids which are capable to furnish more than one protons to water are called polyprotic acids or polyprotic acids are those acids which have more than one acidic hydrogen.
For examples

Polyprotic  acids are dissociate in steps wise with different value of dissociation constant for each step for example carbonic acid is dissociate in two steps as-
Step-1 

Step-2 

Over all reaction 

Experimentally known that for typical Polyprotic acids Ka1>> Ka2 as result all the H+ is due to the first acid ionization.
These dissociation constant also illustrate simultaneous equilibria in the acid dissociation of poly acids both are happening at same time , so both anions HCO3-1 and  CO3-2 are present in equilibrium mixture in solution.
PH CALCULATION OF POLYPROTIC ACIDS:

Consider a general triprotic acid (H3A) and dissociate in three steps
Step-1 
Experimentally we know that   Ka1>>Ka2>>Ka3, hence x >>y>>Z   so y and z can be neglected with respect to x 
                   This is the quadratic equation solve by following formulae
Step-2   
Experimentally we know that   Ka1>>Ka2>>Ka3, 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:
In such cases following assumption are applicable. In generally Ka1 >> Ka2 >> Ka3
x >> y >> z     [H+] = x, [H2A¯ ] = x,  [HA2–] = y,  [A3–] = z, [H3A] = c – x
  
ILUUSTRATIVE EXAMPLE (1):     0.1M H3X,
 find the concentration of (X-3 ) ?.
 SOLUTION:  We know that:  
Calculate the value of (z) by solving these three equations (z) =10-19 
ILUUSTRATIVE EXAMPLE (2): Calculate the concentration of all species of significant contribution present in 0.1M H3PO4 Solution.(Given ka1 7.5×10-3,Ka2 6.2×10-8 and Ka3 3.6×10-13)
ILUUSTRATIVE EXAMPLE (3): Calculate concentration of (a) H+ (B) HCO3-1 (C) CO3-2 (D) DOD of HCO3-1 in 0.1M aqueous solution.(Given for Ka1=1.6x10-7  Ka2=4x10-11).
ILUUSTRATIVE EXAMPLE (4): Find concentration of (a) H+ (B) HCO3-1 (C) CO3-2 in 0.01M aqueous  solution of carbonic acid if pH of the solution is 4.18, .(Given for Ka1=4.45x10-7  Ka2=4.69x10-11).
ILUUSTRATIVE EXAMPLE (5): Consider the acid dissociation of 0.25 M H2CO3. What are the concentration of all species in equilibrium in each stage and pH?                                                       
Answer key
(2) (H+=HPO3-1) = 0.0273 M, H3PO4= 0.1- x= 0.0727M, PO4-3=8.26x10-19
(3) H+= 4.0x10-5 M, HCO3-1= x-y=4.0x10-5 M   DOD= 10-6
(4) H+=6.6x10-5 x=2.1095x10-5, y =CO3 -2=ka2=4.69x10-11 M
(5) H+=x=3.24x10-4 M , HCO3-1=x= 3.24 x10-4 M
                                               @@@

POLARITY AND FAJAN'S RULE:

IONIC CHARACTER IN COVALENT COMPOUNDS:
(1) Pauling used dipole moment for depicting percentage of polarity and ionic character of the bond. According to Pauling, a bond can never be 100% ionic.
ILLUSTRATIVE EXAMPLE (1): Calculate the % of ionic character of a bond having length = 0.83 Å and 1.82 D as its observed dipole moment?

SOLUTION: To calculate μ considering 100% ionic bond
                      = 4.8 × 10–10 × 0.83 × 10–8esu cm
                      = 4.8 × 0.83 × 10–18 esu cm = 3.984 D
                          % ionic character = 1.82/3.984 × 100 = 45.68%

(2) When electronegativity difference between two atoms is 2.1, there is 50% ionic character in the bond.
(3) When electronegativity difference is zero (identical atoms), the bond will be 100% covalent.
HANNEY AND SMITH EQUATION:
According to Haney and Smith, the percentage of ionic character of a polar covalent bond can be calculated with help of the following expression.
%ionic character:

POULING EQUATION:

ILLUSTRATIVE EXAMPLE (2) : Calculate the & ionic character in H-X bond of
Halogen acids. Given that electronegativity of  H ,F, Cl, Br. and I  are 2.1, 4.0, 3.0, 2.8,
and 2.5 respectively.



FAJAN'S RULE:

Consider an ionic bond formed between two oppositely charged ions of unequal size of A+  and
B-. In this bond cation and anion remains bonded by electrostatic force of attraction by the valence electrons of larger anion (B-) are attracted by small cation (A+) and so the shape of valence shell is deformed and electron cloud of valence shell of anion remains no longer symmetrical. This process is called Polarisation of anion by cation. The ability of cation to polarise anion is called polarising power and the tendency of anion to get polasied, is called polarisability.

Due to polarisation of anion, an ionic bond acquires some partial covalent character and
greater is the extent of polarisation, more is the covalent character incorporated in ionic bond.
The magnitude of polarization depends upon following factors :

(1)SIZE OF THE CATION: Polarization of the anion increases as the size of the cation decreases i.e., the electrovalent compounds having smaller cations show more of the covalent nature. For example, in the case of halides of alkaline earth metals, the covalent character decreases as we move down the group. Hence melting point increases in the order of

                  BeCl2 < MgCl2 < CaCl2 < SrCl2 < BaCl2 
  
(2)SIZE OF ANION: The larger the size of the anion, more easily it will be polarized by the cation i.e., as the size of the anion increases for a given cation, the covalent character increases. For example, in the case of halides of calcium, the covalent character increases from F anion to I anion i.e. increasing covalent character
                               CaF2 < CaCl2 <CaBr2 < CaI2
Similarly, in case of trihalides of aluminum, the covalent character increases with increase in size of halide anion i.e.
                                  AlF3 <AlCl3< AlBr3 < AlI3
Covalent character increases as the size of halide ion increases

(3) CHARGE ON EITHER OF THE IONS: As the charge on the cation increases, its tendency to polarize the anion increases. This brings more and more covalent nature in the electrovalent compound. Whereas with the increasing charge of anion, its ability to get polarized, by the cation, also increases. For example, in the case of NaCl, MgCl2 and AlCl3 the polarization increases, thereby covalent character becomes more and more as the charge on the cation increases. Similarly, lead forms two chlorides PbCl2 and PbCl4 having charges +2 and +4 respectively. PbCl4 shows covalent nature. Similarly among NaCl, Na2S, Na3P, the charge of the anions are increasing, therefore the increasing order of
Covalent character.  NaCl < Na2S < Na3P

(4) NATURE OF THE CATION: Cations with 18 electrons (s2 p6 d10) in outermost shell polarize an anion more strongly than cations of 8 electrons (s2 p6) type. The d electrons of the 18 electron shell screen the nuclear charge of the cation less effectively than the s and p electrons of the 18- electron shell. Hence the 18-electron cations behave as if they had a greater charge. Copper (I) and Silver (I) halides are more covalent in nature compared with the corresponding sodium and potassium halides although charge on the ions is the same and the sizes of the corresponding ions are similar. This illustrates the effect of 18- electron configuration of Cu+ (3s2 3p6 3d10) and Ag+ (4s2, p6, d10) ions.
ILLUSTRATIVE EXAMPLE:
The decomposition temperature of Li2CO3 is less than that of Na2CO3. Explain.
SOLUTION: As Li+ ion is smaller than Na+ ion, thus small cation (Li+) will favour more covalent character in Li2CO3 and hence it has lower decomposition temperature than that of Na2CO3.
                                                                  @@@

TITRATION OF TRIPROTIC ACID WITH STRONG BASE:

TRIPROTIC Vs NaOH: [H3PO4 Vs NaOH]
ILLUSTRATIVE EXAMPLE: Phosphoric acid (H3PO4), is a Triprotic acid with Ka1= 10-3 , Ka2=10-8 and Ka3= 10-12 .Consider the titration of 0.10M 100 ml H3PO4 with 0.1M NaOH Solution and answers the following questions.
(A) Write out the reactions associated with Ka1, Ka2 and Ka3.
(B) Calculate the pH after the following titration points:
(1) 100 ml of 0.1M H3PO4 + 0.0 ml of 0.1 ml NaOH
(2) 100 ml of 0.1M H3PO4 + 50 ml of 0.1 ml NaOH
(3) 100 ml of 0.1M H3PO4 + 100 ml of 0.1 ml NaOH
(4) 100 ml of 0.1M H3PO4 + 150 ml of 0.1 ml NaOH
(5) 100 ml of 0.1M H3PO4 + 200 ml of 0.1 ml NaOH
(6) 100 ml of 0.1M H3PO4 + 250 ml of 0.1 ml NaOH
(7) 100 ml of 0.1M H3PO4 + 300 ml of 0.1 ml NaOH 
(C) Sketch the titration curve for this titration.
(D) What weak acid and what conjugate base make the best phosphate Buffer at pH ~7.0?

(A) Write out the reactions associated with Ka1, Ka2 and Ka3.

 SOLUTION:
Step-1    
Step-2  
 Step-3    
(B)  Calculate the pH after the following titration points;

(1) 100 ml of 0.1M H3PO4 + 0.0 ml of 0.1 ml NaOH 
SOLUTION:
By approximation we know that if

(2) 100 ml of 0.1M H3PO4 + 50 ml of 0.1 ml NaOH 
SOLUTION:
Given mill moles (M x V) of acid= 0.1x100= 10 and base 0.1x50 = 5.0
Here H3PO4/NaH2PO4 remain same in solution and which are act acidic buffer
(3) 100 ml of 0.1M H3PO4 + 100 ml of 0.1 ml NaOH 
SOLUTION:
Given mill moles (M x V) of acid= 0.1x100 = 10 and base 0.1x100 = 10
Here only NaH2PO4 remain same in solution and which are under go amphoteric hydrolysis. PH of amphoteric salts is independent of concentration of salt.

(4) 100 ml of 0.1M H3PO4 + 150 ml of 0.1 ml NaOH
SOLUTION:
Given millimoles (M x V) of acid= 0.1x100 = 10 and base 0.1x150 = 15
Here only NaOH (5 millimoles) and NaH2PO4 both are remain present in solution hence which are further go titration.
Here NaH2PO4 and Na2HPO4 remain same in solution and which are act acidic buffer

(5) 100 ml of 0.1M H3PO4 + 200 ml of 0.1 ml NaOH
SOLUTION:
Given mill moles (M x V) of acid= 0.1x100 = 10 and base 0.1x 200 = 20
Here only NaOH (10 millimoles) and NaH2PO4 (10 millimoles) both are remain present in solution hence which are further go titration.

Here only Na2HPO4 remain same in solution and which are under go amphoteric hydrolysis. pH of amphoteric salts is independent of concentration of salt.
(6) 100 ml of 0.1M H3PO4 + 250 ml of 0.1 ml NaOH 
SOLUTION:
Given mill moles (M x V) of acid= 0.1x100 = 10 and base 0.1x 250 = 25
Here NaOH (15 millimoles) and NaH2PO4 (10 millimoles) both are remain present in solution hence which are further go titration.
Here Na2HPO4 (10 millimoles) and (5.0millimoles) both remain same in solution and which are further go titration.
Here Na2HPO4 and Na3PO4 remain same in solution and which are act acidic buffer

(7) 100 ml of 0.1M H3PO4 + 300 ml of 0.1 ml NaOH 
SOLUTION:
Given millimoles (M x V) of acid= 0.1x100 = 10 and base 0.1x 300 = 30
Here NaOH (20 millimoles) and NaH2PO4 (10 millimoles) both are remain present in solution hence which are further go titration.
Here NaOH (10 millimoles) and Na2HPO4 (10 millimoles) both remain same in solution and which are further go titration.
Finally 10 millimoles (Molarity=10/400) N3PO4 is formed which undergo polyvalent salt hydrolysis 

HYDROLYSIS OF ANION (PO4-3):

Step wise illustration of hydrolysis of poly basic acids and polyacidic base is given as -
Na+ ion do not under goes hydrolysis while PO4-3 undergoes step wise hydrolysis
Experimentally we know that   Ka1>>Ka2>>Ka3, hence x >>y>>Z   so y and z can be neglected with respect to x   it mean total OHions count from only first step OH- ions coming from 2nd and 3rd  hydrolysis is ignored 

Special Case (1):  
Special Case (2) 
This is the quadratic equation solve by following formulae
So approximation not valid hence follows 2nd case

(C) Sketch the titration curve for this titration:

SOLUTION:    

S.N.
Given condition
comments
PH
1
100 ml of 0.1M H3PO4 + 0.0 ml of 0.1 ml NaOH 
H3PO4 Polyprotic acid
02.02
2
100 ml of 0.1M H3PO4 + 50 ml of 0.1 ml NaOH 
Acidic Best buffer
PH= Pka1 (1st half E-Point)
03.00
3
100 ml of 0.1M H3PO4 + 100 ml of 0.1 ml NaOH 
Amphoteric salt Hydrolysis
PH=1/2(Pka1+Pka2)
05.50
4
100 ml of 0.1M H3PO4 + 150 ml of 0.1 ml NaOH
Acidic Best buffer
PH= Pka2 (2st half E-Point)
08.00
5
100 ml of 0.1M H3PO4 + 200 ml of 0.1 ml NaOH
Amphoteric salt Hydrolysis
PH=1/2(Pka2+Pka3)
10.00
6
100 ml of 0.1M H3PO4 + 250 ml of 0.1 ml NaOH 
Acidic Best buffer
PH= Pka3 (3st half E-Point)
12.00
7
100 ml of 0.1M H3PO4 + 300 ml of 0.1 ml NaOH 
Poly anionic salt Hydrolysis
3rd Equivalent point
12.07
 Graphical representation of titration:



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