2.1 $n$-factor

  Solutions

• 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 per mole of acid in a reaction. Let us take some examples.
  
  1. $HCl$ consists of one $H^+$ ion which means only one $H^+$ can be displaced. So, the $n$-factor is $1$.
  
  2. $H_2SO_4$ consists of two $H^+$ ions which means maximum of two $H^+$ ions can be displaced depending upon the type of reaction. So, the $n$-factor of $H_2SO_4$ is $1$ or $2$ depending on the number of $H^+$ ions displaced as shown in reactions below.
  $$\begin{equation} \begin{aligned} {H_2}S{O_4} + NaOH \to NaHS{O_4} + {H_2}O\;\left( {n - {\text{factor }}1} \right) \\ {H_2}S{O_4} + 2NaOH \to N{a_2}S{O_4} + 2{H_2}O\;\left( {n - {\text{factor 2}}} \right) \\\end{aligned} \end{equation} $$
  
  3. $H_3PO_3$, from the chemical formulae it looks like that three $H^+$ ions can be displaced but from the structure shown, one $H^+$ ion is directly attached to the central atom phosphorus $(P)$ which is non-replacable. Therefore, only two $H^+$ ions can be displaced. So, $n$-factor is $1$ or $2$ depending on the number of $H^+$ ions displaced.
  

• For Base: 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 displaced by $1$ mole of base in a reaction. For example, $n$-factor of $NaOH$ is $1$, $Ca{\left( {OH} \right)_2}$ is $1$ or $2$.

• For Salts: The $n$-factor for salts in which no atom undergoes change in oxidation state is defined as the total moles of cationic/anionic charge replaced in $1$ mole of the salt.
  
  For example. In the reaction shown $$2N{a_3}P{O_4} + 3BaC{l_2} \to 6NaCl + B{a_3}{\left( {P{O_4}} \right)_2}$$ The product of the reaction is one mole $Ba_3(PO_4)_2$, for which two moles of $Na_3PO_4$ are required. As we know that sodium exists as $Na^+$ ions and Barium as $Ba^{2+}$, we can say that $6$ moles of $Na^+$ are completely replaced by $3$ moles of $Ba^{2+}$ ions.
  
  Therefore, we can conclude that $6$ moles of cationic charge is replaced by $2$ moles of $Na_3PO_4$, therefore, using unitary method we can say that $1$ mole of $Na_3PO_4$ replaces $3$ moles of cationic charge.
  
  Hence, $n$-factor of $Na_3PO_4$ in this reaction is $3$.

• For oxidising/reducing agents: The $n$-factor is defined as the number of electrons involved in oxidation/reduction half reactions per mole of oxidising agent/reducing agent.
  
  For example, in the reaction of $KMnO_4$ shown below, the oxidation state of $Mn$ is changed from $+7$ to $+2$ in which $5$ electrons are involved. $$5{e^ - } + 8{H^ + } + Mn{O_4}^ - \to M{n^{2 + }} + {H_2}O$$
  Therefore, the $n$-factor of $KMnO_4$ is $+5$.