Question

A mixture of ${ (n }_{ 1 }$ moles each of ${ Na }_{ 2 }{ C }_{ 2 }{ O }_{ 4 }$ and ${ NaH }{ C }_{ 2 }{ O }_{ 4 }$ is titrated separately with ${ H }_{ 2 }{ O }_{ 2 }$ and KOH to reach the equivalence point. Which of the following statements is/are correct?
This question has multiple correct options
a.) Moles of ${ H }_{ 2 }{ O }_{ 2 }$ and KOH are ${ (n }_{ 1 }{ +n }_{ 2 }{ ) }$ and ${ (n }_{ 2 }$ respectively.
b.) Moles of ${ H }_{ 2 }{ O }_{ 2 }$ and KOH are ${ n }_{ 1 }{ + }\dfrac { n_{ 2 } }{ 2 }$ and ${ (n }_{ 1 }$ respectively
c.) n-factor of ${ NaH }{ C }_{ 2 }{ O }_{ 4 }$ with KOH and ${ H }_{ 2 }{ O }_{ 2 }$ respectively are 1 and 2.
d.) n-factor of ${ Na }_{ 2 }{ C }_{ 2 }{ O }_{ 4 }$ with ${ H }_{ 2 }{ O }_{ 2 }$ and KOH respectively are 2 and 1.

Answer 1

n-factor can be simply defined as the change in the oxidation state of a particular ion. This will be equal to the number of electrons gained or lost by that species during the reaction.

Complete Solution :
${ Na }_{ 2 }{ C }_{ 2 }{ O }_{ 4 }$ and ${ NaH }{ C }_{ 2 }{ O }_{ 4 }$ both react with ${ H }_{ 2 }{ O }_{ 2 }$ as a reducing agent only. (n-factor for both is 2).
Equivalents of ${ H }_{ 2 }{ O }_{ 2 }$ = Equivalents of ${ Na }_{ 2 }{ C }_{ 2 }{ O }_{ 4 }$ + Equivalents of ${ NaH }{ C }_{ 2 }{ O }_{ 4 }$
${ 2\times moles\quad of\quad H }_{ 2 }{ O }_{ 2 }$ = ${ n }_{ 1 }{ \times 2+n }_{ 2 }{ \times 2 }$
Therefore, moles of ${ H }_{ 2 }{ O }_{ 2 }$ = $\dfrac { { n }_{ 1 }{ \times 2+n }_{ 2 }{ \times 2 } }{ 2 }$
Hence, moles of ${ H }_{ 2 }{ O }_{ 2 }$ = ${ (n }_{ 1 }{ +n }_{ 2 }{ ) }$

- As we know that only ${ NaH }{ C }_{ 2 }{ O }_{ 4 }$ reacts with KOH as acid-base titration. Its n-factor is 1 as it has only ${ H }^{ + }$ ion.
Equivalents of KOH = Equivalents of ${ NaH }{ C }_{ 2 }{ O }_{ 4 }$
${ 1\times }$ moles of KOH = ${ n }_{ 2 }{ \times 1 }$
Therefore, moles of KOH = ${ (n }_{ 2 }$
Hence, moles of ${ H }_{ 2 }{ O }_{ 2 }$ and KOH are ${ (n }_{ 1 }{ +n }_{ 2 }{ ) }$ and ${ (n }_{ 2 }$ respectively.
Therefore, the n-factor of ${ NaH }{ C }_{ 2 }{ O }_{ 4 }$ with KOH and ${ H }_{ 2 }{ O }_{ 2 }$ are 1 and 2 respectively.
and n-factor of ${ Na }_{ 2 }{ C }_{ 2 }{ O }_{ 4 }$ with ${ H }_{ 2 }{ O }_{ 2 }$ and KOH are 2 and 2 respectively.
So, the correct answer is “Option A and C”.

Additional Information:
- For Acid: Acids are the species which furnish ${ H }^{ + }$ ions when dissolved in a solvent. For acids, the n-factor is defined as the number of ${ H }^{ + }$ ions replaced by 1 mole of acid in a reaction. Note that the n-factor for acid is not equivalent to its basicity; i.e. the number of moles of replaceable ${ H }^{ + }$ atoms present in one mole of acid.
- For Bases: Bases are the species, which furnish hydroxide ions when dissolved in a solvent. For bases, the n-factor is defined as the number of hydroxide ions replaced by 1 mole of the base in a reaction. Note that n-factor is not equivalent to its acidity i.e. the number of moles of replaceable hydroxide ions present in 1 mole of the base.

Note: The possibility to make a mistake is that you may choose option D. But the n-factor of ${ Na }_{ 2 }{ C }_{ 2 }{ O }_{ 4 }$ with ${ H }_{ 2 }{ O }_{ 2 }$ is 2 as it gives two ${ H }^{ + }$ ions not 1 same with KOH the n-factor is 2.