Stoichiometry
1.0 The Mole
2.0 The Limiting Reagent
3.0 Gravimetric Analysis
4.0 Volumetric Analysis
5.0 Calculation of n-factor
6.0 Redox Reactions
7.0 Titration
7.1 Simple Titration
7.2 Double Titration
7.3 Method
7.4 Titration of the solution containing both $N{a_2}C{O_3}$ and $NaHC{O_3}$
7.5 Titration of the solution containing both $NaOH$ and $N{a_2}C{O_3}$
7.6 Back Titration
8.0 Iodimetric and Iodometric Titrations
9.0 Volume strength peroxide solution
10.0 Percentage Labeling of Oleum
11.0 Hardness of Water
5.1 Acid-Base Reaction
7.2 Double Titration
7.3 Method
7.4 Titration of the solution containing both $N{a_2}C{O_3}$ and $NaHC{O_3}$
7.5 Titration of the solution containing both $NaOH$ and $N{a_2}C{O_3}$
7.6 Back Titration
According, to the Arrhenius, an acid is a substance that furnishes ${H^ + }$ ion(s) in solution, a base is a substance that furnishes $OH–$ ion(s) in solution and neutralization is a reaction in which ${H^ + }$ ion furnished by acid combines with $OH–$ ions furnished by base. The number of ${H^ + }$ ion(s) furnished per molecule of the acid is its n-factor also called basicity. Similarly the number of $OH–$ ion(s) furnished by the base per molecule is its n-factor, also called acidity.
\[\mathop {HCl}\limits_{(n = 1)} \xrightarrow{{}}{H^ + } + C{l^ - }\]
\[\mathop {{H_2}S{O_4}}\limits_{(n = 1)} \xrightarrow{{}}{H^ + } + HSO_4^ - \]
\[\mathop {{H_2}S{O_4}}\limits_{(n = 2)} \xrightarrow{{}}2{H^ + } + SO_4^{2 - }\]
\[\mathop {{H_3}P{O_4}}\limits_{(n = 1)} \xrightarrow{{}}{H^ + } + {H_2}PO_4^ - \]
\[\mathop {{H_3}P{O_4}}\limits_{(n = 3)} \xrightarrow{{}}3{H^ + } + PO_4^{3 - }\]
\[\mathop {{H_3}P{O_3}}\limits_{(n = 1)} \xrightarrow{{}}{H^ + } + {H_2}PO_3^ - \]
\[\mathop {{H_3}P{O_3}}\limits_{(n = 2)} \xrightarrow{{}}2{H^ + } + HPO_3^ - \]
The n-factor of ${H_3}P{O_3}$ cannot be 3 as it has only two dissociable ${H^ + }$ ions. So, its n-factor or dissociable protons is 1 or 2 as one of the $H-$ atoms is linked with P atom directly.
Similarly,
\[\mathop {C{H_3}COOH}\limits_{(n = 1)} \xrightarrow{{}}C{H_3}CO{O^ - } + {H^ + }\]
n-factor of $C{H_3}COOH$ is 1, because it contains only one dissociable ${H^ + }$ ion.
Now, we will consider the n-factor of some bases.
\[\mathop {NaOH}\limits_{(n = 1)} \xrightarrow{{}}N{a^ + } + O{H^ - }\]
\[\mathop {Ba{{(OH)}_2}}\limits_{(n = 1)} \xrightarrow{{}}{[Ba(OH)]^ - } + O{H^ - }\]
\[\mathop {Ba{{(OH)}_2}}\limits_{(n = 2)} \xrightarrow{{}}B{a^{2 + }} + 2O{H^ - }\]
\[\mathop {Al{{(OH)}_3}}\limits_{(n = 3)} \xrightarrow{{}}A{l^{3 + }} + 3O{H^ - }\]
Similarly, n-factor of $Al{\left( {OH} \right)_3}$ can also be 1 or 2 or 3, depending upon the number of $OH–$ released.