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
7.5 Titration of the solution containing both $NaOH$ and $N{a_2}C{O_3}$
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
A given volume of the aqueous solution is titrated with an acid of known normality using phenophthalein indicator. Suppose $‘a’$ milli-equivalents of acid are used in the first end point then,
milli equivalent of $NaOH$ $+\frac{1}{2}$ milli equivalent of $N{a_{2}}C{O_3}=$ milli equivalent of acid $=b$
Now in the same already titrated solution methyl orange indicator is added and again titrated to the end point.
Suppose $‘b’$ milli equivalents of the acid are used at the second end point.
$\frac{1}{2}$ milli equivalents of $N{a_2}C{O_3}$ = milli equivalents of acid $= b$
From equation $(1)$ and $(2)$
Milli equivalents of acid used by $N{a_2}C{O_3} = 2b$$=$$ \equiv milli\ equivalents\ of\ N{a_2}C{O_3}$$
Milli equivalents of acid used by $NaOH= a – b = milli\ equivalent\ of\ NaOH$
Knowing the milli equivalents of $N{a_2}C{O_3}$ or $NaOH$ and the volume of the solution titrated, their normality can be calculated.