ACID-BASE TITRATION:

THEORY OF INDICATOR'S:

It is method to determine concentration of any Solution with the help of a solution of known concentration.
EQUIVALENT POINT:
When the complete reaction occurs between the solutions, it is called equivalent point of titration.
END POINT: 
When sudden change in colour of solution occurs, it is called end point of titration.
INDICATORS:
(1) They are very weak organic acids or bases that show different colours at different pH value.
(2) They exist in ionized and unionized form which have different colours .
(3) The point at which Number of equivalent of acid and base become equal is known as equivalent point / Stoichiometric point / neutralization point.
(4) We assure that indicators do affect the pH of Solution.
(5) Around the equivalent point pH change the drastic change.
(6) For the successful titration end point should be as close as possible to the equivalent point.
TYPE OF INDICATOR'S:
(1) ACIDIC INDICATOR'S
(2) BASIC INDICATOR'S

(1) ACIDIC INDICATOR'S:

Phenophthalene (HPh): It is weak acid and can represented as [HPh] If it ionised gives [H⁺ ] and [Ph^- ] 

Addition of [H⁺]: In addition of an acid the ionization of HPh is practically negligible and as the equilibrium shift left hand side due to high concentration of [H⁺] ion thus solution would remain colourless.

Addition of [OH^- ]: In the presence of a base H⁺]^[ ions  are removed by [OH^-] ions in the in the form of water molecules and the above equilibrium shift to right hand side .Thus the concentration of Ph- ions increases in solution and they impart a pink colours to the solution.

INDICATOR THEORY: Let consider acidic indicators [HPh] Phenophthalene. An indicator has two colouring parts.

(1) Unionized part of indicators 

(2) Ionized part of indicators

The relative concentration of these species will depend upon PH of medium .

Take negative log both sides

CASE (1): It is observed that 

Then colour of indicator decided by concentration of [Ph^-)]

CASE (2): It is observed that 

Then colour of indicator decided by concentration of [HPh]

PH RANGE OF INDICATORS:

The given indicators work between a pH range i.e.

Then working of indicators is best;

ILLUSTRATIVE EXAMPLE (1):  For an acidic indicator, dissociation constant, K_in is 2×10^-6 Calculate pH range of indicator.

SOLUTION:  

ILLUSTRATIVE EXAMPLE (2): The pH range of a basic indicator is 4 to 6.5 Calculate the dissociation constant of indicator?.

SOLUTION:  pK_In must be midpoint of pH range for acidic indicators and pOH range for basic indicators

The pH range = 4 to 6.5 so pOH range is 10 to 7.5

Hence Pk_In =   (10+7.5) /2 = 8.75 

ILLUSTRATIVE EXAMPLE (3):  For an indicator pKa is 6 Calculate pH of Solution having this indicator such that 40% indicator molecules remain in ionised form.

SOLUTION:  We know that  

(2) BASIC INDICATOR'S

Methyl orange (MeOH): It is weak base and can represented an MeOH If it ionised gives [ Me⁺] and[ OH^- ]  and methyl orange is an intensely coloured indicator that is red below pH 3.1 and orange-yellow above pH 4.4

The red (acid) form has an [H+] attached to one of the N atoms and the yellow (basic) form has lost the [H+]

Addition of [H⁺ ] ^: In the presence of an acid, OH^- ions are removed in the form of water molecules and the above equilibrium shift to right hand side .This effect Me⁺ ions are produced which impart red coloure to the Solution.

Addition of [OH^- ]^: In addition of alkali the concentration of OH^- increases in the solution and equilibrium shift left hand side i.e. the ionization of MeOH is practically negligible .thus solution acquired the colour of unionised methyl orange molecules i.e. yellow

DEGREE OF DISSOCISTION OF INDICATOR'S (DOD): Consider a general indicator dissociation;

ILLUSTRATIVE EXAMPLE (4):  For acidic indicator, pH range is 3 to 4.6 calculate the ratio of[ In^- ] and H⁺ for the appearance of solution in a single colour.

SOLUTION:

Given pH range 3.0 to 4.6 so pK_In= (3.0+ 4.6)/2 =3.6

ILLUSTRATIVE EXAMPLE (5): An indicator with Ka = 10^-5 is solution with pH = 6 Calculate % of indicator in ionised form?

SOLUTION:

ILLUSTRATIVE EXAMPLE (6): The pH of at which an acid indicator with Ka is 10^-15 changes colour when indicator concentration 1×10^-5 M is? 

SOLUTION:

TYPE OF TITRATION:

(A) ACID-BASE TITRATION: 

(A-1): Strong acid Vs Strong base:

(A-2): Weak acid Vs Strong base:
(A-3): Weak base Vs Strong base:
(A-4): Weak acid Vs Weak base: weak acid and weak base titration can not be carried out because due to very low PH change , their is no suitable indicator for this titration.
(A-5); Salt of SB and WA Vs Strong acid:

(A-6); Double Indicators Titration:

(A-7): Back Titration:
(B):TITRATION OF POLYPROTIC ACIDS:
(1) Titration of diprotic acid with strong base:
(2) Titration of triprotic acid with strong base:

(C) REDOX TITRATION:

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>>Ka_3, 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>>Ka_3, 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>>Ka₃   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,  [HA^2–] = y,  [A^3–] = 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

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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 AlC_l3 the polarization increases, thereby covalent character becomes more and more as the charge on the cation increases. Similarly, lead forms two chlorides PbC_l2 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.

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