n- Factor Calculation

Acids

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 HCI = 1,

n-factor of HNO₃ = 1,

n-factor of H₂SO₄ = 1 or 2, depending upon extent of reaction it undergoes.

H₂SO₄ + NaOH →  NaHSO₄ + H₂O

Although one mole of H2SO4 ahs 2 replaceable H atoms but in this reaction H2SO4 has given only one H+ ion, so its n-factor would be 1.

 H₂SO₄ + 2NaOH →  Na₂SO₄ + 2H₂O

The n-factor of H2SO4 in this reaction would be 2.

Similarly,

• n-factor of H₂SO₃ = 1 or 2

• n-factor of H₂CO₃ = 1 or 2

• n-factor of H₂PO₄ = 1 or 2 or 3

• n-factor of H₃PO₃ = 1 or 2 because one of the H is not replaceable in H₃PO₃. This can be seen using its structure .
  The H atoms which are linked to oxygen are replaceable while the H atom linked directly to central atom (P) is nonreplaceable.

• n-factor of H₃BO₃ = 1
  In H₃BO₃, although all three H are linked to oxygen, yet all 3 H are not replaceable. Here, boron atom is electron deficient, so it acts as a Lewis acid. When H₃BO₃ is added to water, then oxygen atom of H₂O through its lone pair attack the boron atom, as follows

The net reaction is H₃BO₃ + 2H₂O →  [B(OH)₄]^– + H₃O⁺.
Thus, one mole of H₃BO₃ in solution gives only one mole of H⁺, so its n-factor is 1.

Bases

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 AI(OH)₃ = 1 or 2 or 3

• n factor of NH₄(OH) = 1

Salts which react such that no atom undergoes change in oxidation state

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

2Na₃PO₄ + 3BaCI₂ →  6NaCI + Ba₃(PO₄)₂

To get one mole of Ba₃(PO₄₎₂, two moles of Na₃PO₄ are required, which means six moles of Na⁺ are completely replaced by 3 moles of Ba²⁺ ions. So, six moles of cationic charge is replaced by 2 moles of Na₃PO₄, thus each mole of Na₃PO₄ replaces 3 moles of cationic charge. Hence, n-factor of Na₃PO₄ in this reaction is 3.

Salts which react in a manner that only one atom undergoes change in oxidation state and goes only in one product

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^+x_a B_b → A^+y_cD

The n-factor of A_aB_b is calculated as

      n = ½ax – ay½

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 KMnO₄ for the given chemical change.

KMnO₄  Mn⁺².

In this reaction, oxidation state of Mn changes from +7 to +2. Thus, KMnO₄ is acting as oxidizing agent, since it is reduced.

Hence n-factor of KMnO₄ = ½1 × (+7) – 1 × (+2)½ = 5

Similarly,

KMnO₄  Mn⁺⁴

            n-factor of KMnO₄ = ½1 × (+7) – 1 × (+4)½ = 3

KMnO₄  Mn⁺⁶

           n-factor of KMnO₄ = ½1 × (+7) – 1 × (+6)½ = 1

It can be seen that in all the above chemical changes, KMnO₄ is acting as oxidizing 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.

Salts that react in a manner that only one atom undergo change in oxidation state but goes in two products with the same oxidation state.
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 it 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 case 4.

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 case 4.
Hence n-factor of A_aB_b = ½ax – ay½
For example, let us calculate the n-factor of K₂Cr₂O₇ for the given chemical change.
Cr₂O₇^2– → Cr³⁺ + Cr³⁺
In this reaction, oxidation state of Cr changes from +6 to +3 in both products.
Hence n-factor of K₂Cr₂O₇ = ½2 × (+6) –2 × (+3)½ = 6

Salts which react in a manner that only one atom undergoes change in oxidation state but goes in two products with different oxidation state (different than in the reactant) as a result of either oxidation or reduction

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+7, two moles of Mn+7 changes to Mn+2 by gaining 10 mole of electrons and one mole of Mn+7 changes to Mn+4 bygained 3 mole of electrons.
Thus each mole of Mn+7 have gained 13/3 mole of electrons. Therefore, the n-factor of Mn+7 in this reaction would be 13/3.Note that n-factor can be 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.

Thus, each mole of Mn⁺⁷ have gained 11/3 mole of electron. Therefore, n-factor of Mn⁺⁷ in this reaction would be 11/3.

Salts which react in a fashion that only one atom undergoes change in oxidation state but goes in two products with different oxidation state (in one product with same oxidation state and in other with different oxidation state than in the reactant)

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.
Calculate n-factor of Mn⁺⁷, if the reaction would have been 3Mn⁺⁷ → Mn⁺⁷ + 2Mn⁺².

Salts that react in a manner that two type of atoms in the salt undergo change in oxidation state (both the atoms are either getting oxidised or reduced).

In this reaction, both A and B are changing their oxidation states and both of them are either getting oxidized or reduced. In such case, the n-factor of the compound would be the sum of individual n-factors of A and B.
n-factor of A = ½ax – ay½
n-factor of B = ½– ax – bx½ because the total oxidation state of ‘b’ B’s in the reactant is –ax (as the total oxidation state of ‘a’ A’s in the reactant is +ax) and the total oxidation state of y B’s in the product is bz.
Hence n-factor of A_aB_b = ½ax – ay½ + ½– ax – bz½
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⁺¹₂S^–2 →  Cu²⁺ + SO₂ in which Cu⁺ and S^2– both are getting oxidized to Cu²⁺ and S⁺⁴ respectively.
Hence n-factor of Cu2S = ½2 × (+1) –2 (+2)½ + ½1 × (–2) –1 × (+4)½ = 8

Refer to the following video for a solved problem on normality and n– factor calculation.

Salts that react in a manner that two atoms in the salt undergoes change in oxidation state (on atom is getting oxidised and the other is getting reduced).
If we have a salt which react in a fashion that atoms of one of the elements are getting oxidized and the atoms of another element are getting reduced and no other element on the reactant side is getting oxidized 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 KCIO₃ is represented as
KCIO₃ → KCI + O₂
In this reaction, O^2– is getting oxidized to O₂ and CI⁺⁵ is getting reduced to CI^–1. In each case, 6 mole of electrons are exchanged whether we consider oxidation or reduction.
n-factor of KCIO₃ considering oxidation = ½3(–2)–3(0)½ = 6
or n-factor of KCIO₃ considering reduction = ½1 × (+5) –1 × (–1)½ = 6

Slats or compounds which undergoes disproportionation reaction.
Disproportionation reactions are the reactions in which oxidizing and reducing agents are same or the same element from the same compound is getting oxidized 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.

Disproportionation reactions in which moles of compound getting oxidized and reduced are same i.e. moles of oxidizing 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, 2H₂O₂ → 2H₂O + O₂
Out of the 2 mole of H₂O₂ used in reaction, one mole of H₂O₂ gets oxidized to O₂ (oxidation state of O changes from –1 to 0) while the other mole of H₂O₂ gets reduced to H₂O (oxidation state of O changes from –1 to –2). When 1 mole of H₂O₂ gets oxidized to O₂, the half-reaction would be  →  + 2e^– and when 1 mole of H2O2 gets reduced to H₂O, the half-reaction would be  → 2e^– → 2O^2–.
Thus, it is evident that one mole of H₂O₂ (which is either getting oxidized or reduced) will lose or gain 2 mole of electrons. Therefore, n-factor of H₂O₂ as oxidizing as well as reducing agent in this reaction is 2. 
where, H₂O₂ is 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 H₂O₂ involved in the reaction. Thus, the n-factor of H₂O₂ when the reaction is written without segregating oxidizing and reducing agent is  = 1.

                        2H₂O₂ → 2H₂O + O₂

                                ^{(n=1)        (n=1)          (n=2)}

Disproportionation reactions in which moles of compound getting oxidized and reduced are not same i.e. moles of oxidizing agent and reducing agent are not same.
6Br₂ + 12 OH^– → 10Br^– + 2BrO^–₃ + 6H₂O
In this reaction, the mole of electrons lost by the oxidation of some of the moles of Br₂ are same as the number of moles of electrons gained by the reduction of rest of the moles of Br₂. Of the 6 moles of Br₂ used, one mole is getting oxidized, loosing 10 electrons (as reducing agent) and 5 moles of Br₂ are getting reduced and accepts 10 moles of electrons (as oxidizing agent).
where Br₂ is not distinguished as how much of it functions as oxidizing agent and how much as reducing agent, then for calculating n-factor of compounds 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 Br₂ involved in the reaction to get the number of mole of electrons exchanged by one mole of Br₂.
In the overall reaction, the number of mole of electrons exchanged (lost or gained) is 10 and the moles of Br₂ used in the reaction are 6. Thus, each mole of Br₂ has exchanged 10/6 or 5/3 mole of electrons.
Therefore, the n-factor of Br₂ when the reaction is written without segregating oxidizing and reducing agent is 5/3.
6Br₂ → 10Br– + 2Br⁺⁵
(n=5/3)     (n=1)     (n=5)

Question 1: n-factor for  HNO₃ is

a. 1

b. 2

c. 3

d. 4

Question 2:  What of the following alternatives does not represents the possible value of n-factor of H₂PO₄ ?

a. 1

b. 2

c. 3

d. 4

Question 3:  In the reaction 2Na₃PO₄ + 3BaCI₂ → 6NaCI + Ba₃(PO₄)₂

n-factor of Na₃PO₄ in this reaction is..

a. 1

b. 2

c. 3

d. 4

Question 4:  In the reaction 6Br₂ → 10Br– + 2Br⁺⁵  the n-factor of Br₂ is

a. 1

b. 2

c. 3

d. None of the above

Related Resources

• Click here for past year papers of IIT JEE  

• Have a look  syllabus of chemistry for  IIT JEE

• You can also refer to volumetric analysis