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How to find Equivalent Weight in Chemistry ? Chemical Formula

Equivalent Weight

The number obtained by dividing the atomic mass of an element by its valency is called the Equivalent Weight of that element.

Equivalent weight of Element = Atom weight of Element / Valency

The atomic mass of hydrogen is 1 (1.008 to be precise). The valency of hydrogen is always 1. Therefore, the equivalent weight of hydrogen is 1. In most of the reactions the valency of oxygen is 2.

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Therefore, in most of the reactions, the equivalent weight of oxygen is 16/2 = 8. In most reactions of chlorine has 1 valency. Therefore, in most of the reactions, the equivalent weight of chlorine is 35.5/1 = 35.5.

Equivalent Weight Definition

The equivalent weight of other substances is defined as –

The equivalent weight of a substance is a number that tells how many parts of it combine with the weight of 1 part of hydrogen, 8 parts of oxygen, 355 parts of chlorine or the equivalent weight of any other substance, or combine them with other substances.

Equivalent Weight

Example:

2H2 + O2 → 2H2O

Since 4 grams of hydrogen react with 32 grams of oxygen, so 1 gram of hydrogen will react with 8 grams of oxygen. So the equivalent weight of oxygen is 8.

H2 + Cl2 → 2HCl

Since 2 grams of hydrogen react with 71 grams of chlorine, the equivalent weight of chlorine is 35.5.

Mg + 2HCl → MgCl2 + H2

Since 24 grams of magnesium is equal to 2 grams of hydrogen, the equivalent weight of magnesium is 12.

NaOH + HCl → NaCl + H2O

According to the above example the equivalent weight of HCl is 36.5. In the above reaction 36.5 g of HCl react with 40 g of NaOH. Hence, the equivalent weight of NaOH in the above reaction is 40.

4Al + 3O2 → 2Al2O3

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The equivalent weight of oxygen is 8. 96 grams of oxygen combine with 108 grams of aluminum. So 8 grams of oxygen will combine with 9 grams of aluminum. So the equivalent weight of aluminum is 9.

Some definitions and principles used in Volumetric Analysis

Volumetric Analysis:

The branch of chemical analysis in which the quantities of substances in solutions are determined by measuring the volumes of the solutions of the reacting substances. It is called Volumetric Analysis.

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In volumetric analysis, a solution of a known concentration or strength is treated with a solution of unknown strength. After the reaction is completed, the volume of both the solutions is determined and by calculation the strength of the solution having known strength is determined.

Titration:

In volumetric analysis, the solution of one substance is mixed with the solution of another substance in a fixed volume until the reaction is completed. This process of mixing solutions is called Titration.

Single and double titrations

When a solution of a substance is treated with a solution of known strength of another substance to determine the strength, then this process is called Single Titration.

When the strength of a solution of a substance is determined with the help of a solution of known strength of the same substance, then two types of Titration have to be done –

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(i) Titration of a solution of known strength of that substance with that of a solution of unknown strength of another substance.

(ii) Titration of a solution of unknown strength of that substance with a solution of another substance.

This process of doing two types of titration is called Double Titration.

Titration

In double titration, in addition to the known and unknown strength solutions of a substance, a solution of another substance with an unknown strength is used, it is called an intermediate solution.

End point:

In the process of titration, when another solution is mixed in one solution, then a situation comes when both the solutions have completely reacted with each other.

This position is called the end point and it is usually known by the change of color by the indicator.

Indicator:

In double titration, in addition to the known and unknown strength solutions of a substance, a solution of another substance with an unknown strength is used, it is called an intermediate solution.

Internal Indicator

These are used to find the end point by putting it inside any one solution in the titration method. Titration is done by adding one drop of phenolthelen to the alkali-acid reaction. Phenolthelen is an internal indicator.

External Indicator

In the titration method, the indicator which is used by keeping it out of the mixture of the reacting solutions, is called external indicator.

Self Indicator

If any one of the substances reacting in the titration method acts as an indicator, then this substance is called self indicator.

Due to the presence of that substance at the end point, the color changes. KMnO4 solution acts as a self indicator in KMnO4/oxylic acid oxidation-reduction titration. Light pink color comes at the end point.

Phenolphthalein(C₂₀H₁₄O₄), methyl orange and litmus are some useful indicators. Phenolphthalein is pink in alkaline medium and colorless in neutral or acidic medium.

Methyl orange is yellow in alkaline medium, pink in acidic medium and orange in neutral medium. Litmus is red in acidic solution, blue in basic solution and purple in neutral solution.

Classification of Titrations

There are mainly four types of titration based on the nature of the reaction taking place in the process of titration –

  1. Acid – base Titrations
  2. Precipitation Titrations
  3. Oxidation – reduction Titrations or Redox Titrations
  4. Iodine Titrations

In Acid-base Titrations, neutralization reaction of acid and base takes place and in Oxidation-Reduction Titrations the action of an oxidising substance with a reducing substance takes place.

In Acid-Base Titrations, if the strength of a solution of an acid is determined, it is called Acidimetry and if the strength of a solution of an alkali is determined, it is called alkalimetry.

Equivalent Weight in chemistry

(1) Like atomic weight and molecular weight, equivalent weight is also a number, it does not have any unit.

(2) The weights of the substances that take part in a chemical reaction and the substances formed from it are in the ratio of their equivalent weights. This is called the law of chemical equilibrium.

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Let us assume that the reaction of two substances A and B forms two other substances C and D.

A + B → C + D

So according to the law of chemical equilibrium :-

Weight of A/Weight of B = Equivalent weight of A/ Equivalent weight of B

Same with

Weight of A/ Weight of C = equivalent weight of A/ Equivalent weight of C

Similarly, the ratio of the weights of A and B, B and C, B and D and C and D can also be expressed.

(3) All substances have a fixed atomic mass or molecular weight. The equivalent weight of all substances is not fixed.

The equivalent weight of a substance depends on the chemical reaction in which it is participating. Many substances exhibit variable equivalent weight.

Example: The equivalent weight of KMnO4 is 31.6 in acidic medium, 158 in alkaline medium and 52.7 in neutral medium.

Gram Equivalent Weight

When the equivalent weight of a substance is expressed in grams, then it is called gram equivalent weight. For example: If the equivalent weight of oxylic acid is 63, then 63 grams becomes its gram equivalent weight.

Number of Equivalents

A gram equivalent of a substance has the same quantity as its equivalent weight.

Therefore

Number of gram equivalent = weight of substance in gram / equivalent weight of substance

The number of gram equivalent is also called number of equivalents.

Therefore

number of equivalents = weight of matter in grams / equivalent weight of padarth

There are 1000 milli-equivalents in 1 equivalent of a substance.

Therefore

number of equivalents = number of milli wquivalents / 1000

According to the Mole concept:

number of moles = mass of the substance in grams / molecular mass of the substance

Number of moles x molecular weight = number of equivalents x equivalent weight

This relationship is very important. With its help, number of equivalents can be found from number of moles and number of moles can be found from number of equivalents.

Method to find equivalent weight of acids:

Knowing the molecular weight of an acid and its insolubility, its equivalent weight can be found –

equivalent weight of acid = molecular mass of acid/ phosphorescence of acid

The basicity of an acid is the number of replaceable protons (H+) present in a molecule of it.

Example: The basicity of HCl, H2SO4 and H3PO4 are 1, 2 and 3 respectively.

Acids whose basicities are greater than 1, show the variable equivalent weight.

Example: The basicities of H3PO4 in the following reactions are 1, 2 and 3 respectively.

H3PO4 + NaOH → NaH2PO4 + H2O

H3PO4 + 2NaOH → Na2HPO4 + 2H2O

H3PO4 + 3NaOH → Na3PO4 + 3H2O

Hence, in the above reactions, the equivalent weight of H3PO4 is molecular weight/1, molecular weight/2 and molecular weight/3 respectively.

Method to find Bases equivalent weight:

Knowing the molecular weight of a base or base and its acidity, its equivalent weight can be found –

equivalent weight of base = molecular weight of base/ acidity of base

The acidity of a base is the presence of replaceable hydroxyle ions (OH) in one of its molecules.

It is the number of protons (H+) that can be accepted by a molecule.

Example: Acidities of NaoH, Ba(OH)2, Al(OH)3 and NH3 are 1, 2, 3 and 1 respectively.

 Like polyatomic acids, multi-acidic bases also exhibit variable equivalent weight.

Example: The acidity of Ca(OH)2 in the following reactions are 1 and 2 respectively –

Ca(OH)2 + HCl  → Ca(OH)Cl + H2O

Ca(OH)2 + 2HCl  → CaCl2 + H2O

Hence, in the above reactions, the equivalent weight of Ca(OH)2 is molecular weight/1 and molecular weight/2 respectively.

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Method to find the equivalent weight of Ions:

To find the equivalent weight of an ion, divide its ion weight by the amount of charge present on it.

Equivalent weight of ion = ion weight / Amount of charge

Example: The ion mass of CO32- is 12 + 3 x 16 = 60. Hence its equivalent weight = 60 / 2 = 30.

Method to find the equivalent weight of Salts:

To find the equivalent weight of a salt, divide its molecular weight by the numerical value of some oxidation number of its acidic or basic radical.

Example: The molecular weight of BaCl2 is 208 and the numerical value of the oxidation number of Ba2++ is 2.

Therefore

Equivalent weight of BaCl2 = 208/2 = 104

The equivalent weight of a salt is equal to the sum of the equivalent weights of the cation and anion present in it.

equivalent weight of salt = equivalent weight of cation + equivalent weight of anion

Example:

Equivalent weight of Ca3(PO4)2 = equivalent weight of Ca2++ + equivalent weight of PO43-

Like acids and bases, salts and ions also show variable equivalent weights. The above definitions of equivalent weights of salt and ion are valid only when a complete reaction of an acid and a base forms a salt or an ion or a salt or ion undergoes a complete reaction with an acid, base, salt or ion.

In other cases, the equivalent weight of the salt or ion cannot be determined on the basis of the above definitions.

Na2CO3 + HCl  → NaHCO3 + NaCl

Na2CO3 + 2HCl  → 2NaCl + H2O + CO2

Example:

In the above reactions, the equivalent weight of Na2CO3 is molecular weight/1 and molecular weight/2 respectively.

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