Volumetric Analysis - Redox Reactions | Physical Chemistry PDF Download

7.  VOLUMETRIC ANALYSIS 

Now, we have developed enough platforms to understand the law of equivalents and volumetric analysis.

The volumetric analysis is an analytical method of estimating the concentration of a substance in a solution by adding exactly same number of equivalents of another substance present in the solution of known concentration .

This is the basic principle of titration. Volumetric analysis is also known as titrimetric analysis. The substance whose solution is employed to estimate the concentration of unknown solution is called titrant and the substance whose concentration is to be determined is called titrate.

The Volumetric Analysis is divided into following types:

1. Simple Titrations
2. Back Titrations
3. Double Titrations

7.1 SIMPLE TITRATIONS

The aim of simple titration is to find the concentration of an unknown solution with the help of the known concentration of another solution.

Let us take a solution of substance ‘A’ of unknown concentration. We are provided with solution of another substance ‘B’ whose concentration is known (N₁). We take a certain known volume
(V₂litres) of ‘A’ in a flask and start adding ‘B’ from burette to ‘A’ slowly till all the ‘A’ is consumed by ‘B’. This can be known with the aid of suitable indicator, which shows color change after the complete consumption of ‘A’. Let the volume of B consumed is V₁ litre. According to the law of equivalents, the number of equivalents of ‘A’ would be equal to the number of equivalents of ‘B’.

N₁V₁ = N₂ V₂, where N₂ is the concentration of ‘A’.

Thus using this equation, the value of N₂ can be calculated.

There are four types of simple titrations, namely

(A) Acid-Base titrations

(B) Redox titrations

(C) Precipitation titrations

(D) Complexometric titrations

7.1.1    ACID-BASE TITRATIONS

In this type of titration , the concentration of an acid in a solution is estimated by adding a solution of standard base and vice-versa. The equivalence point is detected by adding a few drops of suitable indicator to the solution whose concentration is to be estimated. An acid-base indicator gives different colors with acids and bases . the choice of an indicator in a particular titration depends on the pH-range of the indicator and the pH change near the equivalence point. For example,

I. STRONG ACID –STRONG BASE TITRATION :

In the titration of HCl Vs NaOH, the equivalence point lies in pH-range of 4-10. Thus, methyl red ( pH-range 4.2 to 6.3 ) , methyl orange ( pH-range 3.1 to 4.4 )and phenolphthalein (pH-range 8.3 to 10) are suitable indicators.

II.WEAK ACID – STRONG BASE TITRATION :

In the titration of CH₃COOH and NaOH , the equivalence point lies in pH-range of 7.5 to 10. Thus, phenolphthalein is the suitable indicator.

III.WEAK BASE – STRONG ACID TITRATION :

In the titration of NH₄OH and HCl , the equivalence point lies in pH-range of 4 to 6.5. Thus, methyl orange and methyl red are the suitable indicators.

IV.WEAK ACID – WEAK BASE TITRATION :

In the titration of CH₃COOH and NH₄OH , , the equivalence point lies in

pH-range of 6.5 to 7.5 and the pH change is not sharp at the equivalence point. Thus, no simple indicator can be employed to detect the equivalence point.

7.1.2  REDOX TITRATIONS

In a redox titration, an oxidant is estimated by adding reductant or vice-versa. For example, Fe²⁺ ions can be estimated by titration against acidified KMnO₄ solution when Fe²⁺ ions can be estimated by titration against acidified KMnO₄ solution when Fe²⁺ ions are oxidized to Fe³⁺ ions and KMnO₄ is reduced to Mn²⁺ in the presence of acidic medium. KMnO₄ functions as self-indicator as its purple color is discharged at the equivalence point.

In addition to acidified KMnO₄ , acidified K₂Cr₂O₇ can also be employed. Other redox titrations are iodimetry etc.

(a)  Iodimetry – This titration involves free iodine. Such direct estimation of iodine is called iodimetry. This involves the titration of iodine solution with known sodium thiosulphate solution, whose normality is N. Let the volume of sodium thiosulphate used be V litre.

I₂ + 2Na₂S₂O₃ → 2NaI + Na₂S₄O₆

Equivalents of I₂ = Equivalents of Na₂S₂O₃ used = N x V

Moles of I₂ = 

Mass of free I₂ in the solution =

(b) Iodometry – This is an indirect method  of estimation of iodine. An oxidizing agent is made to react with excess of solid KI . The oxidizing agent oxidizes I^- to I₂. This liberated iodine is then made to react with Na₂S₂O₃  solution of normality N. Let the volume of thiosulphate solution required be V litre.

Equivalents of ‘A’ = Equivalents of I₂ = Equivalents of Na₂S₂O₃ used = N x V

Equivalents of I₂ liberated from KI =  N x V

Equivalents of ‘A’ = N x V

Let the n-factor of ‘A’ in its reaction with KI be x, then

Mass of ‘A’ consumed =  M_A( where M_A is the molar mass of A )

7.1.3  PRECIPITATION TITRATIONS

In a titration of this kind, cations and anions combine to form a compound of very low solubility. Thus, a solid residue separates out. For example

AgNO₃ + NaCl → AgCl↓(white) +NaNO₃

BaCl₂ + H₂SO₄ → BaSO₄↓(white ) + 2HCl

7.1.4   COMPLEXOMETRIC TITRATIONS 

In this type of titration, the titrate combines with the titrant to form complex salts . The complex salts may or may not be soluble. For example,

AgNO₃ + 2KCN  →  K[Ag(CN)₂] + KNO₃

CuSO₄ + 4NH₄OH  →  [Cu(NH₃)₄]SO₄ + 4H₂O

7.2 BACK TITRATIONS

Let us assume that we have an impure solid substance ‘C’, weighing ‘w’ g and we are required to calculate the % purity of ‘C’ in the sample. We are also provided with two solutions ‘A’ and ‘B’, where the concentration of ‘B’ is known (N₁) and that of ‘A’ is unknown. For the back titration to work, following conditions are to be satisfied (a) Compounds ‘A’, ‘B’ and ‘C’ should be such that ‘A’ and ‘B’ react with each other. (b) ‘A’ and pure ‘C’ also react with each other but the impurity present in ‘C’ does not react with ‘A’ . (c) Also the product of ‘A’ and ‘C’ should not react with ‘B’.

Now we take out certain volume of ‘A’ in flask (the equivalents of ‘A’ taken should be ≥ equivalents of pure ‘C’ in the sample ) and perform a simple titration using ‘B’ . Let us assume that the volume of ‘B’ used be V₁ litre.

Equivalents of ‘B’ reacted with ‘A’ = N₁V₁

Equivalents of ‘A’ initially = N₁V₁

In another flask, we again take some volume of ‘A’ but now ‘C’ is added to this flask. Pure part of ‘C’ reacts with ‘A’ and excess of ‘A’ is back titrated with ‘B’. Let the volume of ‘B’ consumed is V₂ litre .

Equivalents of ‘B’ reacted with excess of ‘A’ = N₁V₂

Equivalents of ‘A’ in excess = N₁V₂ 

Equivalents of ‘A’ reacted with pure ‘C’ =( N₁V₁ - N₁V₂ )

Equivalents of pure ‘C’ = ( N₁V₁ - N₁V₂ )

Let the n-factor of ‘C’ in its reaction with ‘A’ be x, then the moles of pure ‘C’ = ( N₁V₁ - N₁V₂ )/x

Mass of pure

% purity of ‘C’ =

7.3 DOUBLE TITRATIONS

The purpose of double titration is to determine the percentage composition of an alkali mixture or an acid mixture. In the present case, we will find the percentage composition of an alkali mixture. Let us consider a solid mixture of NaOH, Na₂CO₃ and some inert impurities, weighing ‘w’ g. We are required to find the % composition of this alkali mixture. We are also given an acid reagent (HCl) of known concentration M₁ that can react with the alkali sample.

We first dissolve this mixture in water to make an alkaline solution and then we add two indicators, namely phenolphthalein and methyl orange to the solution. Now, we titrate this alkaline solution with standard HCl.

NaOH is a strong base while Na₂CO₃ is a weak base. So it is obvious that NaOH reacts first with HCl completely and Na₂CO₃ reacts only after complete NaOH is neutralized.

NaOH + HCl  → NaCl + H₂O     ….. (i)

Once NaOH has reacted completely, then Na₂CO₃ starts reacting with HCl in two steps:

Na₂CO₃ + HCl  → NaHCO₃+ NaCl ….. (ii)

NaHCO₃ + HCl → NaCl + CO₂ + H₂O    ……. (iii)

It is clear that when we add HCl to the alkaline solution, alkali is neutralized and the pH of the solution decreases. Initially, the pH decrease would be rapid as strong base (NaOH) is neutralized completely. When Na₂CO₃ is converted into NaHCO₃ completely, the solution is still, weakly basic due to the presence of NaHCO₃ (which is weaker as compared to Na₂CO₃). At this point, phenolphthalein changes color since it requires this weakly basic solution to show its color change. When HCl is further added, the pH again decreases and when all the NaHCO₃ reacts to form NaCl, CO₂ and H₂O the solution becomes weakly acidic due to the presence of weak acid (H₂CO₃). At this point, methyl orange changes color as it requires this weakly acidic solution to show its color change.

Thus in general, phenolphthalein shows color change when the solution contains weakly basic NaHCO₃ along with other neutral substances while methyl orange shows color change when solution contains weakly acidic H₂CO₃ along with other neutral substances .

Let the volume of HCl used up for the first and second reaction be V₁ litre (this is volume of HCl used from the beginning of titration up to the point when phenolphthalein shows color change) and the volume of HCl required for the third reaction be V₂ litre (this is the volume of HCl used from the point where phenolphthalein changes color upto the point when methyl orange shows color change. Then,

Moles of HCl consumed by NaHCO₃ = Moles of NaHCO₃ reacted = M₁V₂

Moles Of NaHCO₃ formed from Na₂CO₃ = M₁V₂

Moles of Na₂CO₃ in the mixture = M₁V₂

Mass of Na₂CO₃ in the mixture = M₁V₂  x 106

% of Na₂CO₃ in the mixture = (M₁V₂ x 106)/2 x 100

Moles of HCl used in the reaction (i) and (ii) = M₁V₁

Moles of HCl used in reaction (ii) = M₁V₂

Moles of HCl used in reaction (i) = (M₁V₁ - M₁V₂)

Moles of NaOH = (M₁V₁ - M₁V₂)

Mass of NaOH = (M₁V₁ - M₁V₂) x 40

% of NaOH in the mixture = [(M₁V₁ - M₁V₂)x40]/w  x 100

Here, we have determined the % composition of the mixture using mole concept, as the balanced reactions were available. If we were to solve this by equivalent concept, then the procedure adopted would be

NaOH   +    HCl  →    NaCl    +   H₂O(i)

Na₂CO₃  +  HCl    →   NaHCO₃ + NaCl  (ii)

(n=1)  (n=1)   (n=1)

Phenolphthalein shows end point after reaction (ii)

NaHCO₃   +   HCl →  NaCl   +  CO₂  +  H₂O

(n=1)         (n=1)

And methyl orange shows end point after reaction (iii).

At phenolphthalein end point,

(equivalents of HCl )₁ =equivalents of NaOH +equivalents of Na₂CO₃ (n=1)……..(iv)

= M₁V₁

At methyl orange end point,

(equivalents of HCl )₂ = equivalents of NaHCO₃ reacted ( n=1 )

= equivalents of NaHCO₃ produced ( n=1 ) = equivalents of Na₂CO₃ ( n=1 )

M₁V₁ = Equivalents of Na₂CO₃ ( n=1 ) = ½ equivalents of Na₂CO₃ ( n=2 )  …… (v)

Subtracting the equation (v) from (iv),]

Equivalents of NaOH = (M₁V₁ - M₁V₂)

Moles of NaOH = (M₁V₁ - M₁V₂)          (Since n-factor of NaOH is 1 )

% of NaOH in the mixture = [ (M₁V₁ - M₁V₂) x 40 ]/w  x 100

Equivalents of Na₂CO₃ ( n=1 ) = M₁V₂

And moles of Na₂CO₃ = M₁V₂

% of Na₂CO₃ in the mixture =( M₁V₂ x106 )/w  x 100

In the above case, we have taken alkali mixture of NaOH and Na₂CO₃. But other alkali mixtures can also be taken. For example,

(i) If a mixture of NaOH and NaHCO₃ has been taken 

(a) Equivalents of NaOH = Equivalents of HCl required upto phenolphthalein end point

(b) Equivalents of NaHCO₃ = Equivalents of HCl required from phenolphthalein end point to methyl orange end point.

(ii) If a mixture of Na₂CO₃ and NaHCO₃ has been taken

(a)  ½ equivalents of Na₂CO₃ ( n=2 ) = Equivalents of HCl required upto phenolphthalein end point.

(b)  ½ equivalents of  Na₂CO₃ ( n=2 ) = Equivalents of NaHCO₃ initially ( n=1 ).

= Equivalents of HCl required from phenolphthalein end point to methyl orange end point.

(iii) If a mixture of NaOH, Na₂CO₃ and NaHCO₃ has been taken 

(a)  Equivalents of NaOH + ½ Equivalents of Na₂CO₃ (n=2)

= Equivalents of HCl required up to phenolphthalein end point.

(b)  ½ equivalents of  Na₂CO₃ ( n=2 ) = Equivalents of NaHCO₃ initially ( n=1 ).

=Equivalents of HCl required from phenolphthalein end point to methyl orange end point.

8.   VOLUME STRENGTH OF H₂O₂ SOLUTION

When a solution of H₂O₂ is labeled as ‘x volumes’ , it means that  1 volume ( 1ml or 1 L ) of H₂O₂ solution would liberate x volumes ( x ml or x L) of O₂ at STP on complete decomposition.

If a H₂O₂ solution  (acting as reducing agent ) has normality N and it is to be reacted with KMnO₄ solution ( acting as oxidizing agent ), we can say that there are N equivalents of H₂O₂ present in 1 L of this H₂O₂ solution. 1 ml of H₂O₂ `of this solution would contain  equivalents.

Moles of H₂O₂ in 1 ml of this solution =  [from equation (ii) ] When these many moles of H₂O₂ in 1 ml of solution are allowed to decompose according to the reaction, the volume of O₂ released ( in ml ) by them at STP will give the volume trength of H₂O₂ solution. Moles of O₂ given by 1 ml of this solution = [from eq. (i)]

Volume of O₂ at STP given by 1 ml of this solution = 

Volume strength of H₂O₂ = 5.6 x Normality of H₂O₂ 

9 . PERCENTAGE LABELING OF OLEUM

Oleum contains H₂SO₄ and SO₃ only. When Oleum is diluted ( by adding water ) , SO₃ reacts with H₂O to form H₂SO₄ , thus increasing the mass of the solution.

The total mass of H₂SO₄ obtained by diluting 100 g of oleum sample with required amount of water, is equal to the percentage labeling of oleum.                                                                                          

Percentage labeling of oleum = Total mass of H₂SO₄ present in oleum after dilution.

=  mass of H₂SO₄ initially present + mass of H₂SO₄  produced on dilution.

If we have a sample of oleum labeled as 109%, this means that 100g of oleum on dilution gives 109g of H₂SO_4. Let us calculate the composition of oleum, which is labeled as 109%.                                              

Let the mass of SO₃ in the sample be x g, then the mass of H₂SO₄ would be ( 100 – x)g. On dilution,

                                                                                                                                                   
Moles of SO₃ in oleum = x/80 = Moles of H₂SO₄ formed on dilution.

Mass of H₂SO₄ formed on dilution =  Total mass of H₂SO₄ present in oleum after dilution =   + (100-x) = 109. Thus, x=40 . Thus, oleum contained 40% SO₃ and 60% H₂SO₄.

Atlernatively, let the mass of oleum sample be 100g, which on dilution becomes 109g. This implies that 9g of H₂O was added. 

Moles of H₂O added = 9/18 = Moles of SO₃ present in oleum sample.                                                   

Mass of SO₃ in oleum = (9/18) x 80 = 40g.                                                                                                         

Thus oleum sample contained 40% SO₃ of and 60% of H₂SO₄

10. MODIFICATIONS IN LAW OF EQUIVALENCE

There are certain modifications required in order to use law of equivalence. These modification are

(i) The equivalents of a substance produced and reacted may not necessarily be same . If the n-

factor of the substance, in the reaction in which it is produced were different than the n-

factorof the same substance,when it is reacting, then the equivalent of the substance

produced and reacted would be different.

(ii) The equivalents of the same substances can be added or subtracted only when they are having the same n-factor.

(iii) In a reaction, the equivalents of oxidizing agents would always be equal to the equivalents of reducing agents, irrespective of the number of agents used in the reaction.