Redox and Equivalent ConceptHard
Question
In the mixture of NaHCO3 and Na2CO3, the volume of a given HCl required is x ml with phenolphthalein indicator and further y ml is required with methyl orange indicator. Hence, the volume of HCl for complete reaction of NaHCO3 present in the original mixture is
Options
A.2x
B.y
C.x/2
D.(y − x)
Solution
Let NaHCO3 = a mole, Na2CO3 = 6 mole.
In the presence of phenolphthalein, neqHCl = neqNa2CO3
Or $\frac{x \times N}{1000} = b \times 1(1)$
In the presence of methyl orange, $n_{eq}HCl = n_{eq}HCO_{3}\left( \text{Original} \right) + n_{eq}NaHCO_{3}\left( \text{formed} \right)$
Or $\frac{y \times N}{1000} = a \times 1 + b \times 1(2)$
From (1) and (2), we get: VHCl only forms NaHCO3 (original)$= (y - x)\text{ ml}$.
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