7.4 Titration of the solution containing both $N{a_2}C{O_3}$ and $NaHC{O_3}$

  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
    8.0 Iodimetric and Iodometric Titrations
    9.0 Volume strength peroxide solution
    10.0 Percentage Labeling of Oleum
    11.0 Hardness of Water

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7.4 Titration of the solution containing both $N{a_2}C{O_3}$ and $NaHC{O_3}$

Given volume of the solution is titrated by an acid using phenolphthalein indicator.

Suppose $‘a’$ milli equivalents of acid are used in the first end point. Then

$\frac{1}{2}$ milli equivalent of $N{a_2}C{O_3}$ = milli equivalents of acid $= a\ \ \ …(1)$

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. Then

$\frac{1}{2}$ milli equivalents of $N{a_2}C{O_3}$ $+$ milli equivalents of $NaHC{O_3}$ $=$ milli equivalents of acid $ = b\;\;\;..(2)$

From equation $(1)$ and $(2)$

Milli equivalents of acid used by $N{a_2}C{O_3}= 2a$

$ \equiv $ milli equivalents of $N{a_2}C{O_3}$

Milli equivalents of acid used by $NaHC{O_3} = b - a = $ milli equivalent of $NaHC{O_3}$

Knowing the milli equivalents of the base and volume of the solution titrated, the normality (strength) of the bases can be calculated.