Law of Equivalence

Before heading for the law of equivalents, let us first discuss certain definitions.

Molarity (M)

It is defined as the number of moles of solute present in one litre of solution.

Molarity (M) =  

Let the weight of solute be w g, molar mass of solute be M₁g/mol and the volume of solution be V litre.

Number of moles of solute = 

Hence M 

Hence Number of moles of solute

Normality (N)

It is defined as the number of equivalents of a solute present in one litre of solution. Equivalent is also the term used for amount of substance like mole with the difference that one equivalent of a substance in different reactions may be different as well as the one equivalent of each substance is also different.

Normality (N) = 

Let the weight of solute be w g, equivalent mass of solute be E g/eqv. and the volume of solution be V litre.

Number of equivalents of solute = 

Hence  N  

Hence Number of equivalents of solute =  = N × V (in litre)

Equivalent Mass

Equivalents mass = 

Hence Number of equivalents of solute  

Hence Number of equivalents of solute = n × number of moles of solute

Also, 

N = M × n

Hence

Where n is n- factor or valency factor here.

Dilution Effects

When a solution is diluted, the moles and equivalents of solute do not change but molarity and normality changes while on taking out a small volume of solution from a larger volume, the molarity and normality of solution do not change but moles and equivalents change proportionately.

In stoichiometry, the biggest problem is that for solving a problem we need to know a balanced chemical reaction. Since the number of chemical reactions are too many, it is not possible to remember all those chemical reactions. So, there is need to develop an approach which does not require the use of balanced chemical reaction. This approach makes use of a law called law of equivalence.

The law of equivalence provide, us the molar ratio of reactants and products without knowing the complete balanced reaction, which is as good as having a balanced chemical reaction. The molar ratio of reactants and products can be known by knowing the n-factor of relevant species.

According to the law of equivalence, whenever two substances react, the equivalents of one will be equal to the equivalents of other and the equivalents of any product will also be equal to that of the reactant.

Let us suppose we have a reaction, A + B → C + D.

In this reaction, the number of moles of electrons lost by 1 mole of A are x and the number of mole of electrons gained by 1 mole of B are y. Since, the number of mole of electrons lost and gained are not sane, the molar ratio which A & B react cannot be 1 : 1. 

Thus, if we take y moles of A, then the total moles of electrons lost by y moles of A would be (x × y).

Similarly, if x moles of B are taken, then the total mole of electrons gained by x moles of B would be (y × x).

Thus, the number of electrons lost by A and number of electrons gained by B becomes equal. For reactant A, its n-factor is x and the number of moles used are y. 

So, The equivalents of A reacting = moles of A reacting × n-factor of A = y × x.

Similarly, for reactant B, its n-factor is y and the number of moles used are x, So,

The equivalents of B reacting = moles of B reacting × n-factor of B = x × y

Thus, the equivalents of A reacting would be equal to the equivalent of B reacting. Thus, the balancing coefficients of the reactant would be as

The n-factor of A & B are in the ratio of x : y, and their molar ratio is y : x. Thus, molar ratio is inverse of the n-factor ratio.

In general, whenever two substances react with their n-factors in the ratio of a : b, then their molar ratio in a balanced chemical reaction would be b : a.

To get the equivalents of a substance, its n-factor is to be known.

Let the weight of the substance used in the reaction be w g. Then, equivalents of substance reacted would be   (where E and M₁ are the equivalent mass and molar mass of the substance). Thus, in order to calculate the equivalents of substance, knowledge of n-factor is a must.

Volume Strength of H₂O₂ Solution

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

H₂O₂  H₂O + ½O₂  …….. (i)

If a H₂O₂ solution (acting as reducing agent) has normality N and its is to be reacted with KMnO₄ solution (acting as oxidizing agent). Our task is to determine its volume strength. We can say that there are N equivalents of H₂O₂ present in 1 litre of this H₂O₂ solution.

1 ml of H₂O₂ of this solution would contain   equivalents.

H₂O₂ + KMnO₄  O₂ + Mn⁺²             ……..(ii)

 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, H₂O₂  H₂O + ½O₂, the volume of O₂ released (in ml) by them at STP will give the volume strength of H₂O₂ solution.

Moles of O₂ given by 1 ml of this solution =  [from equation (i)]

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

Volume strength of H₂O₂ = 5.6 × Normality of H₂O₂

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.

SO₃ + H₂O →  H₂SO₄

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 100 g of oleum and dilution gives 109 g of H₂SO₄.

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,

SO₃ + H₂O →  H₂SO₄

Moles of SO₃ in oleum =  = Moles of H₂SO₃ formed on dilution

Mass of H₂SO₄ formed on dilution = 

Total mass of H₂SO₄ present in oleum after dilution =  + (100 – x) = 109

x = 40.

Thus, oleum contained 40% SO₃ & 60% H₂SO₄.

Alternatively,

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

SO₃ + H₂O →  H₂SO₄

Moles of H₂O added =  Moles of SO₃ present in oleum sample.

Mass of SO₃ in oleum =  × 80 = 40 g

Thus, oleum sample contained 40% SO₃ and 60% H₂SO₄.
 

Question 1: What is the mass of HCl present in 500 ml of 1 molar aqueous solution of HCl ?

a. 1 g

b. 36 g

c. 18 g

d. 22 g

Question 2: Which of the following things not change which a solution is diluted?

a. Molarity

b. Molality

c. Normality

d. Equivalents of solute

Question 3: 

What would be molaity of a solution in which 5 moles of a solute is present in 2 L of the solution?

a. 5 M

b. 2 M

c.2.5 M

d. 1.5 M

Question 4: In general, whenever two substances react with their n-factors in the ratio of a : b, then their molar ratio in a balanced chemical reaction would be

a. b : a.

b. a : b

c. 2a : b

d.  2b : a.

Related Resources

• Look here for past year papers of IIT JEE  

• Click here to refer syllabus of chemistry for  IIT JEE

• You can also have a look at gravimetric analysis