TITRATIONS OF DIPROTIC ACID WITH STRONG BASE:

DIPROTIC ACID  Vs NaOH:  [H₂CO₃ and Oxalic acid H₂A]

ILLUSTRATIVE EXAMPLE: Give the answers of following questions when 100 ml of Malonic acid is titrated with 0.10 M NaOH the (Given that Ka₁=1.5×10^-3 and Ka₂=2.0×10^-6 for Malonic acid HOOC-CH₂-COOH represented as H₂A ).

(A) Write out the reactions and equilibrium expression associated with Ka₁ and Ka₂.

(B) Calculate the PH when:

(1)100 ml of 0.10M H₂A + 0.0 ml of 0.10 ml NaOH 

(2) 100 ml of 0.10M H₂A+ 50 ml of 0.10 ml NaOH
(3) 100 ml of 0.10M H₂A+ 100 ml of 0.10 ml NaOH
(4) 100 ml of 0.10M H₂A+ 150 ml of 0.10 ml NaOH
(5) 100 ml of 0.10M H₂A + 200 ml of 0.10 ml NaOH

(C) Sketch the titration curve for this titration.

(A) Write out the reactions and equilibrium expression associated with Ka₁ and Ka₂
SOLUTION:

(B) Calculate the PH when:

SOLUTION:

 (1) At Point-(1):100 ml of 0.10 M H₂A + 0.0 ml of 0.10 ml NaOH 

Note: Ignore the amount of [H+] coming from 2^nd dissociation of acid 

(2) At Point-(2): 100 ml of 0.10 M H₂A + 50 ml of 0.10 ml NaOH 

At Point -2- is the half of the first equivalent point   where [H₂A=HA^-]   hence pH of solution is due to best buffer

(3) At Point-(3):100 ml of 0.10M H₂A + 100 ml of 0.10 ml NaOH]

At point-3- 1^st equivalent point occurs and here major species is HA^- which is act as both acid and base (Amphoteric species).

(4) At Point-(4): 100 ml of 0.10 M H₂A + 150 ml of 0.10 ml NaOH

At Point -4- is the half of the 2^nd equivalent point   where [HA^- = A^-2]   hence pH of solution is due to best buffer .

(5) At Point-(5):100 ml of 0.10 M H₂A + 200 ml of 0.10 ml NaOH

At point-5- It is 2^nd equivalent point there is only A^-2 present which act as weak base and undergo polyvalent anion  hydrolysis;

(C) Sketch the titration curve for the titration point of question (B) titration.

S.N.	Given condition	comments	PH
1	100 ml of 0.1M H2A + 0.0 ml of 0.1 ml NaOH	H2A Diprotic acid	1.94
2	100 ml of 0.1M H2A + 50 ml of 0.1 ml NaOH	Acidic Best buffer PH= Pka1 (1st half E-Point)	2.82
3	100 ml of 0.1M H2A + 100 ml of 0.1 ml NaOH	Amphoteric salt Hydrolysis PH=1/2(Pka1+Pka2)	4.26
4	100 ml of 0.1M H2A + 150 ml of 0.1 ml NaOH	Acidic Best buffer PH= Pka2 (2st half E-Point)	5.70
5	100 ml of 0.1M H2A + 200 ml of 0.1 ml NaOH	Anionic salt Hydrolysis	9.11

Graphical representation: 

TITRATION OF TRIPROTIC ACID WITH STRONG BASE:

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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