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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) = \frac{\textup{Number\ of\ moles\ of\ solute}}{\textup{Volume\ of\ solution\ in\ litres}} 

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

Number of moles of solute = \frac{\textup{Weight\ \ of\ solute}}{\textup{Molar\ mass\ of\ solute}}=\frac{w}{M_1}

Hence M =\frac{w}{M_1}\times \frac{1}{V(in\ litres)}

Hence Number of moles of solute =\frac{w}{M_1} = \textup{M} \times \textup{V}

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) = 

\frac{\textup{Number\ of\ equivalents\ of\ solute}}{\textup{Volume\ of\ solution\ in\ litres}}

Normality (N)  Is defined as the number of equivalents of a solute present in one litre of solutionLet 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 = \frac{\textup{Weight\ \ of\ solute}}{\textup{Equivalent\ mass\ of\ solute}}=\frac{w}{E}

Hence  N  =\frac{w}{E}\times \frac{1}{V(in\ litres)}

Hence Number of equivalents of solute = \frac{w}{E} = N × V (in litre)

Equivalent Mass

Equivalents mass = \frac{\textup{Atomic\ or\ molecular\ mass}}{\textup{'n'\ factor}} = \frac{M_1}{n}

Hence Number of equivalents of solute  =\frac{w}{E}=\frac{w}{M_1/n}=\frac{w\times n}{M_1}

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

N = \frac{w}{M_1/n}\times \frac{1}{V (in\ litre)}=\frac{w}{M_1}\times \frac{1}{V(in\ litre )}\times n

Also, 

N = M × n

Hence

Normality of solution = n × molarity of solution

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.

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

 yA                   +          xB       →        C         +          D

(n-factor = x)              (n-factor = y)

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  \frac{w}{E}=\frac{w}{M_1/n}=\frac{w\times n}{M_1} (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   \frac{\textup{N}}{1000}equivalents.

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

 Moles of H2O2 in 1 ml of this solution =\frac{1}{2}\times \frac{\textup{N}}{1000}          [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 =\frac{1}{2}\times \frac{1}{2}\times \frac{\textup{N}}{1000}  [from equation (i)]

Volume of O2 at STP given by 1ml of this solution =\frac{1}{2}\times \frac{1}{2}\times \frac{\textup{N}}{1000}\times 22400 = 5.6 \times\textup{N}

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 = \frac{\textup{x}}{\textup{80}} = Moles of H2SO3 formed on dilution

Mass of H2SO4 formed on dilution =\frac{x\times 98}{80} 

Total mass of H2SO4 present in oleum after dilution = \frac{x\times 98}{80} + (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.

 Modifications in law of equivalence

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

  • 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-factor of the same substance, when it is reacting, then the equivalent of the substance produced and reacted would be different.

  • The equivalents of the same substances can be added or subtracted only when they are of the similar n-factor.

  • 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.

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.

Q.1

Q.2

Q.3

Q.4

c

d

c

a

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