Question

how to calculate n-factor of all acid ,base ?

Answer 1

n factor is atomic weight or molecular weight by equivalent weight. n factorfor acids is its basicity, ie, number of replaceable hydrogen ion per molecule. For exampleg H C l has n factor 1 n H X 2 S O X 4 has 2 H X 3 P O X 3 has n factor 2 because when H X 3 P O X 3 dissociates to form H+ and HPO3^-2.

Answer 2

For 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 has 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.

 

 

 

For 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

 

 

 

For 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 mole of the salt. For the reaction,

 

 

 

2Na3PO4 + 3 BaCI2 →  6 NaCI + 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 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 be +y.

 

Aa+x Bb → Ac+yD

 

The n-factor of AaBb is calculated as

 

     n = |ax – ay|