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`) = `text(number of equivalents of solute)/text(volume of solution in litres)`

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

Number of equivalents of soIute = `text(weight of solute)/text(equivalent mass of solute) =w/E`

`N = (w/E) xx (1/text(V in litres))`

`(w/(M_w//n)) xx(1/text(V in litres))`

Hence , Normality of solution = `n` factor `xx` molarity of solution

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 equivalents of the reactant.

For a reaction `nA + mB -> pC + qD` (It is not necessary to balance the equation)

Equivalents of `A` = Equivalents of `B` = Equivalents of `C` = Equivalents of `D`

Acids are the species which furnish `H^+` ions when dissolved in a solvent. For acids, n-factor is defined as the number of `H^+` ions replaced by `1` mole of acid in a reaction. Note that the n-factor for acid is not equal to its basicity; i.e. the number of moles of replaceable `H^+` atoms present in one mole of acid.

For example, n-factor of `HCl` = `1`,

n-factor of `HN_3` = `1`,

n-factor of `H_2SO_4` = `1` or `2`, depending upon extent of reaction it undergoes.

`H_2SO_4 + NaOH -> NaHSO_4 + H_2O`

Although one mole of `H_2SO_4` has `2` replaceable `H` atoms but in this reaction `H_2SO_4` has given only one `H^+` ion, so its n-factor would be `1`.

`H_2SO_4 + 2NaOH -> Na_2SO_4 + 2H_2O`

The n-factor of `H_2SO_4` in this reaction would be 2.

Similarly, n-factor of `H_2SO_3` = `1` or `2`

n-factor of `H_2CO_3` = `1` or `2`
n-factor of `H_3PO_4` = `1` or `2` or `3`
n-factor of `H_3PO_3` = `1` or `2` because one of the `H` is not replaceable in `H_3PO_3`.

Bases are the species, which furnish `OH^-` ions when dissolved in a solvent. For bases, n-factor is defined as the number of `OH^-` ions replaced by `1` mole of base in a reaction. Note that n-factor is not equal to its acidity i.e. the number of moles of replaceable `OH^-` ions present in `1` mole of base.

For example, n-factor of `NaOH` = `1`
n-factor of `Zn(OH)_2` = `1` or `2`
n-factor of `Ca(OH)_2` = `1` or `2`
n-factor of `Al(OH)_3` = `1` or `2` or `3`
n-factor of `NH_4(OH)` = `1`

The n - factor for such salts is defined as the total moles of cationic /anionic charge replaced in 1 mole of the salt. For the reaction,

`Na_3PO_4+BaCl_2-> NaCl+Ba_3(PO_4)_2`

To get one mole of `Ba_3(PO_4)_2`, two moles of `Na_3PO_4` are required, which means six moles of `Na^+` are completely replaced by 3 moles of `Ba^(2+)` ions. So, six moles of cationic charge is replaced by 2 moles of `Na_3PO_4`, thus each mole of `Na_3PO_4` replaces 3 moles of cationic charge. Hence, `n` - factor of `Na_3PO_4` in this reaction is `3`.

The n-factor of such salts is defined as the number of moles of electrons exchanged (lost or gained) by one mole of the salt.
Let us have a salt `A_aB_b` in which oxidation state of `A` is `+ x`. It changes to a compound, which has atom `D` in it. The oxidation state of `A` in `A_cD` be `+y`.

`A_a^(+x) B_b -> A_c^(+y)D`

The n-factor of `A_aB_b` is calculated as `n =| ax - ay |`

To calculate n-factor of a salt of such type, we take one mole of the reactant and find the number of mole of the element whose oxidation state is changing. This is multiplied with the oxidation state of the element in the reactant, which gives us the total oxidation state of the element in the reactant. Now, we calculate the total oxidation state of the same element in the product for the same number of mole of atoms of that element in the reactant. Remember that the total oxidation state of the same element in the product is not calculated for the number of mole of atoms of that element in the product.

For example, let us calculate the n-factor `KMnO_4` for the given chemical change.

`KMnO_4 overset(H^+) -> Mn^(+2)`

In this reaction, oxidation state of `Mn` changes from `+ 7` to `+2`. Thus, `KMnO_4` is acting as oxidising agent, since it is reduced.

n-factor of `KMnO_4 =|1 xx (+7) - 1 (+2)| =5`

Similarly,

(a) `KMnO_4 overset(H_2O) -> Mn^(+4)`

n-factor of `KMnO_4` = `|1xx (+7) -1xx(+4)|` =`3`

(b) `KMnO_4 overset(OH^-) -> Mn^(+6)`

n-factor of `KMnO_4` = `|1xx (+7) -1xx(+6)|` =`1`

It can be seen that in all the above chemical changes, `KMnO_4` is acting as oxidising agent, yet its n- factor is not same in all reactions. Thus, the n-factor of a compound is not fixed, it depends on the type and the extent of reaction it undergoes.

Let us have a salt `A_aB_b` in which oxidation state of `A` is `x` . It undergoes a reaction such that element `A` changes its oxidation state and goes in more than one (two) products with the same oxidation state (but different oxidation state than in the reactant). In such case, the n-factor is calculated in the same manner as in the previous case.

Let the chemical change be

`A_a^(+x)B_b -> A_c^(+y)D + A_e^(+y) F`

In such cases, the number of products in which element `A` is present is of no significance since the oxidation state of `A` in both the products is same. The point of importance is not the number of products containing that element which undergoes change in oxidation state but the oxidation state of the element is of importance. The n-factor of `A_aB_b` is calculated in the same way as in the previous case.

n-factor of `A_aB_b = |a x -ay|`

For example, let us calculate the n- factor of `K_2Cr_2O_7` for the given chemical change

`Cr_2O_7^(2-) -> Cr^(+3) + Cr^(+3)`

In this reaction, oxidation state of `Cr` changes from `+6` to `+3` in both products.

n-factor of `K_2Cr_2O_7 = |2 xx (+6)-2 xx (+3)| = 6`

Let the chemical change be

`A_a^(+x)B_b -> A_c^(+y)D + A_e^(+z) F`

In such cases, n-factor calculation is not possible until we know that how much of `A` has changed its oxidation state to `+y` and how much of `A` has changed its oxidation state. to `+z`. This is because the number of moles of electrons lost or gained by one mole of `A_aB_b` would depend on the fact that how much of `A` underwent change to oxidation state `+y` and how much of `A` underwent change to oxidation state `+z`. This is possible only by knowing the balanced chemical reaction. If we know the balanced chemical reaction, then the n-factor calculation is of no use because problem can be solved using mole concept. But nevertheless, n-factor
calculation in such cases can be done as follows.

Let us take a chemical change, `2Mn^(+ 7) -> Mn^(+4) + Mn^(+2)` out of the two moles of `Mn^(+7)`, one mole `Mn^(+7)` changes to `Mn^(+4)` by gaining 3 mole of electrons and the other mole of `Mn^(+7)` changes to `Mn^(+2)` by gaining 5 mole of electrons, so in all 8 mole of electrons are gained by 2 mole of `Mn^(+7)`. So each mole of `Mn^(+7)` has gained `8//2 = 4` mole of electrons. Thus, 4 would be the n-factor of `Mn^(+7)` in this reaction.

If the reaction would have been

`3Mn^(+7) -> 2Mn^(+2) + Mn^(+4)`

Out of `3` moles of `Mn^(+1)`, two moles of `Mn^(+1)` changes to `Mn^(+1)` by gaining `10` mole of electrons and one mole of `Mn^(+1)`
changes to `Mn^(+1)` by gained `3` mole of electrons. Thus each mole of `Mn^(+1)` have gained `13//3` mole of electrons. Therefore, the n-factor of `Mn^(+1)` in this reaction would be `13//3`.

Note that n-factor can be a fraction because it is not the number of electrons exchanged but it is the number of moles of electrons exchanged which can be a fraction.

Now, if the reaction would have been `3Mn^(+7) -> Mn^(+2) + 2Mn^(+4)` . Thus, each mole of `Mn^(+2)` have gained `11//3` mole of electron. Therefore, n-factor of `Mn^(+1)` in this reaction would be `11 //3`.

Let the reaction be

`A_a^(+x) B_b -> A_e^(+x)F + A_c^(+y)D`

For such reactions also, the n-factor calculation is not possible without the knowledge of balanced chemical reaction because n-factor of `A_aB_b` would depend on the fact that how much of `A` underwent change to oxidation state `+y` and how much of `A` remained in the same oxidation state `+x`.

For example, if we have a chemical change as

`2Mn^(+7) -> Mn^(+7) + Mn^(+2)`

(the compounds containing `Mn` in `+ 7` state in reactant and product are different.)

In this reaction, `5` moles of electrons are gained by `2` moles of `Mn^(+7)`, so each mole of `Mn^(+7)` takes up `5//2` mole
of electrons. Therefore, n-factor of `Mn^(+7)` in this reaction would be `5//2`.

Let the change be represented as

`A_a^(+x)B_b -> A_c^(+y)D + E_fB^(+z)`

In this reaction, both `A` and `B` are changing their oxidation states and both of them are either getting oxidised or reduced. In such cases, the n-factor of the compound would be the sum of individual n-factors of `A` and `B`.

n-factor of `A= |a x - ay|`

n-factor of `B = |- a x -bz|` because the total oxidation state of `'b'` `B's` in the reactant is - `a x` (as the total oxidation state of `'a'` `A's` in the reactant is `+a x`) and the total oxidation state of `y` `B's` in the product is `bz`.

n-factor of `A_aB_b` = `|ax - ay| + |-ax-bz|`

In general, the n-factor of the salt will be the total number of mole of electrons lost or gained by one mole of the salt.

For example, we have a reaction,

`Cu_2^(+1) S^(-2) -> Cu^(+2) + overset(+4)SO_2`

in which `Cu^+` and `S^(2-)` both are getting oxidised to `Cu^(+2)` and `S^(+4)` respectively.

n-factor of `Cu_2S` = `|2xx(+1) -2xx(+2)| + |1xx(-2) -1xx(+4)|` = `8`

If we have a salt which react in a fashion that atoms of one of the element are getting oxidised and the atoms of another element are getting reduced and no other element on the reactant side is getting oxidised or reduced, than the n-factor of such a salt can be calculated either by taking the total number of moles of electrons lost or total number of mole of electrons gained by one mole of the salt.

For example, decomposition reaction of `KClO_3` is represented as

`Koverset(+5)(Cl) overset(-2)O_3 -> Koverset(-1)Cl + overset(0)O_2`

In this reaction, `O^(2-)` is getting oxidised to `O_2` and `Cl^(+5)` is getting reduced to `Cl^(-1)`. In each case, `6` mole of electrons are exchanged whether we consider oxidation or reduction.

n-factor of `KClO_3` considering oxidation = `|3xx(- 2}- 3(0)| =6`

or n - factor of `KClO_3` considering reduction = `|1 xx(+5)- 1 xx( - 1)| = 6`

Disproportionation reactions are the reactions in which oxidising and reducing agents are same or the same element from the same compound is getting oxidised as well as reduced. n-factor of a disproportionation reaction can only be calculated using a balanced chemical reaction. We will categorize disproportionation reactions into two types.

(a) Disproportionation reactions in which moles of compound getting oxidised and reduced are same i.e. moles of oxidising agent and reducing agent are same. The n-factor for such compounds is calculated by either the number of mole of electrons lost or gained by
one mole of the compound because in such a case, n-factor of the compound acting as oxidizing agent or as reducing agent would be same.
For example, `2H_2O_2 -> 2H_2O + O_2`

Out of the `2` mole of `H_2O_2` used in reaction, one mole of `H_2O_2` gets oxidised to `O_2` (oxidation state of O changes from `- 1` to `0`) while the other mole of `H_2O_2` gets reduced to `H_2O` (oxidation state of `O` changes from `-1` to `-2`). When `1` mole of `H_2O_2` gets oxidised to `O_2` the half-reaction would be

`O_2^(-2) -> O_2^0 + 2e^(-)`

and when `1` mole of `H_2O_2` gets reduced to `H_2O` , the half-reaction would be `O_2^(2-) + 2e^(-) -> 2O^(-2)`

Thus, it is evident that one mole of `H_2O_2` (which is either getting oxidised or reduced) will lose or gain `2` mole of electrons. Therefore, n-factor of `H_2O_2` as oxidizing as well as reducing agent in this reaction is `2`. Thus,

`underset(n=2)(undersettext(reducing agent)(H_2O_2)) + underset(n=2)(undersettext(oxidising agent)(H_2O_2)) -> underset(n=1)(2H_2O) + underset(n=2)(O_2)`

Or when the reaction is written as

`2H_2O_2 -> 2H_2O + O_2`

where `H_2O_2` ia not distinguished as how much of it functions as oxidizing agent and how much as reducing agent, then n-factor calculation can be done in the following manner. Find the number of electrons exchanged (lost or gained) using the balanced equation and divide it by the number of moles of `H_2O_2` involved in the reaction. Thus, the n-factor of `H_2O_2` when the reaction is written without segregating oxidising and reducing agent is `2/2 =1`

`underset(n=1)(2H_2O_2) -> underset(n=1)(2H_2O) + underset(n=2)(O_2)`

(b) Disproportionation reactions in which moles of compound getting oxidised and reduced are not same i.e. moles of oxidising agent and reducing agent are not same.

For example,

`6Br_2 + 12 OH^(-) -> 10Br^(-) + 2BrO_3^(-) + 6H_2O`

In this reaction, the mole of electrons lost by the oxidation of some of the moles of `Br_2` are same as the number of mole of electrons gained by the reduction of rest of the moles of `Br_2`. Of the `6` moles of `Br_2` used, one mole is getting oxidized, losing `10` electrons (as reducing agent) and `5` moles of `Br_2` are getting reduced and accepts `10` moles of electron (as oxidizing agent).

`Br_2 -> 2 Br^(+5) + 10 e^-`
`5Br_2 + 10e^(-) -> 10 Br^-`

Total reaction is : `underset(n=10)(undersettext(reducing agent)(Br_2)) + underset(n=2)(undersettext(oxidising agent)(5Br_2)) -> underset(n=1)(10Br^(-)) + underset(n=5)(2Br^(+5))`

Thus, n-factor of `Br_2` acting as oxidizing agent is `2` and that `Br_2` acting as reducing agent has n-factor `10`.

Or when the reaction is written as
`6Br_2 -> 10Br^( -) + 2Br^(+5)`

where, `Br_2` is not distinguished as how much of it functions as oxidizing agent and how much as reducing agent, then for calculating n-factor of compound in such reactions, first find the total number of mole of electrons exchanged (lost or gained) using the balanced equation and divide it with the number of mole of `Br_2` involved in the reaction to get the number of mole of electrons exchanged by one mole of `Br_2`. In the overall reaction, the number of mole of electrons exchanged (lost or gained) is `10` and the moles of `Br_2` used in the reaction are `6`. Thus, each mole of `Br_2` has exchanged `10//6` or `5//3` mole of electrons. Therefore, the n-factor of `Br_2` when the reaction is written without segregating oxidising and reducing agent is `5//3`.

`underset(n=5//3)(6Br_2) -> underset(n=1)(10Br^(-)) + underset(n=5)(2Br^(+5))`

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 a 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 estimated is called titrate.

The volumetric analysis is divided into following types:
(A) Simple titrations
(B) Back titrations
(C) Double 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 a substance `'A'` of unknown concentration (say `N_1`) . We are provided with solution of another substance `'B'` whose concentration is known (`N_2`). We take a certain known volume (`V _1` litre) 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 colour change after the complete consumption of `'A'`. Let the volume of `B` consumed is `V_2` litre.

According to the law of equivalents, the number of equivalents of `'A'` would be equal to the number of equivalents of `'B'`.
`N_1V_1=N_2V_2,` where `N_1` is the concentration of `'A'`.

Thus using this equation, the value of `N_1` can be calculated. For example, in a redox titration, an oxidant is estimated by adding reductant or vice- versa. For example, `Fe^(2+)` ions can be estimated by titration against acidified `KMnO_4` solution when `Fe^(2+)` ions are oxidised to `Fe^(3+)` ions and `KMnO_4` is reduced to `Mn^(2+)` in the presence of acidic medium. `KMnO_4` functions as self - indicator as its purple colour is discharged at the equivalence point.

`MnO_4^(-) + 8H^+ + 5e^(-) -> Mn^(+2) + 4H_2O`
`[Fe^(+2) -> Fe^(3+) + e^(-)] xx5`

`underset(n=5)(MnO_4^(-)) + 8H^+ + underset(n=1)(5Fe^(+2)) -> Mn^(+2) + 5Fe^(3+) + 4H_2O`

In addition to acidified `KMnO_4`, acidified `K_2Cr_2O_7` can also be employed. Other redox titrations are iodimetry, iodometry etc.