Loading....
Coupon Accepted Successfully!

 

Acids

Acids are the species which furnish H+ ions when dissolved in a solvent. For acids, the n-factor is defined as the number of H+ ions replaced by 1 mol of acid in a reaction.

 

Note the n-factor for acid is not equal to its basicity, i.e., the number of moles of replaceable H+ atoms present in 1 mol of acid.

Bases

Bases are the species which furnish OH ions when dissolved in a solvent. For bases, the n-factor is defined as the number of OH ions replaced by 1 mol of base in a reaction.

 

Note the n-factor is not equal to its acidity, i.e., the number of moles of replaceable OH- ions present in 1 mol of base.

Salts

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 cationic/anionic charge replaced in 1 mol of the salt. Cosider the reaction:
 
Na3PO4 + BaCl2 NaCl + Ba3(PO4)2
 
To get 1 mol of Ba3(PO4)2, 2 mol of Na3PO4 are required, which means 6 mol of Na+ are completely replaced by 3 mol of Ba2+ ions. So, 6 mol of cationic charge is replaced by 2 mol of Na3PO4; thus each mole of Na3PO4 replaces 3 mol of cationic charge. Hence, the 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 gives in only one product
 
The n-factor of such salts is defined as the number of moles of electrons exchanged (lost or gained) by 1 mol of the salt.
 
Let us have a salt AaBb 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 AcD is +y.
 
Description: 11778.png
 
The n-factor of AnBb is calculated as
n = |axay|
 
To calculate the n-factor of a salt of such type, we take 1 mol of the reactant and find the number of moles 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 moles 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 moles of atoms of that element in the product.
 
For example, let us calculate the n-factor of KMnO4 for the given chemical change:
 
Description: 11786.png
 
In this reaction, oxidation state of Mn changes from +7 to +2. Thus, KMnO4 is acting as an oxidizing agent, since it is reduced.
 
n-factor of KMnO4 = |1 × (+7) – 1 × (+2)| = 5
 
Salts that react in a manner that two types of atoms in the salt undergo change in oxidation state (both the atoms are either getting oxidized or reduced)
 
Let the change be represented as:
 
Description: 11795.png
 
In this reaction, both A and B are changing their oxidation states and both of them are getting either oxidized 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 = |axay|
 
n-factor of B = |axbz|
 
Because the total oxidation state of “bB’s in the reactant is ax (as the total oxidation state of “aA’s in the reactant is +ax) and the total oxidation state of y B’s in the product is bz. Therefore,
 
n-factor of AaBb = |axay| + |axbz|
 
In general, the n-factor of the salt will be the total number of moles of electrons lost or gained by 1 mol of the salt. For example, we have a reaction Description: 11808.pngin which both Cu+ and S2– are getting oxidized to Cu2+ and S+4, respectively. Therefore,
 
n-factor of Cu2S = |2 × (+1) 2 × (+2)| + |1 × (2) 1 × (+4)| = 8
 
Salts which react in a manner that two atoms in the salt undergoes change in oxidation state (one atom is getting oxidized and the other is getting reduced)
 
If we have a salt which react in a fashion that atoms of one of the element 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, then 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 moles of electrons gained by one mole of the salt.
 
For example, decomposition reaction of KClO3 is represented as
 
Description: 11819.png
In this reaction, O2 is getting oxidized to O2 and Cl+5 is getting reduced to Cl1.
 
In each case, 6 mol of electrons are exchanged whether we consider oxidation or reduction.
 
n-factor of KClO3 considering oxidation = |3(2)3(0)| = 6
or n-factor of KClO3 considering reduction = |1 × (+5) 1 × (1)| = 6




Test Your Skills Now!
Take a Quiz now
Reviewer Name