Question

Given, ${{H}_{2}}{{C}_{2}}{{O}_{4}}$ and $NaH{{C}_{2}}{{O}_{4}}$ behave as acids as well as reducing agents. Which of the following statements are correct?
A.Equivalent weights of ${{H}_{2}}{{C}_{2}}{{O}_{4}}$ and $NaH{{C}_{2}}{{O}_{4}}$are equal to their molecular weights when acting as reducing agents
B.Equivalent weights of ${{H}_{2}}{{C}_{2}}{{O}_{4}}$and $NaH{{C}_{2}}{{O}_{4}}$are equal to half their molecular weights when acting as reducing agents
C.100 mL of 1 M solution of each is neutralized by equal volumes of 1 N $Ca{{(OH)}_{2}}$
D.100 mL of 1 M solution of each is neutralized by equal volumes of 1 M $KMn{{O}_{4}}$

Answer 1

The equivalent weight of an element is equal to gram atomic weight divided by its valence. For example equivalent weights of few elements, silver (Ag) = 107.868 g, magnesium (Mg) = $\dfrac{24.312}{2}$ g, aluminum (Al) = $\dfrac{26.312}{3}$ g and sulfur (S) = $\dfrac{32.064}{2}$ g.
If we are going to add an acid to base a neutral compound is going to form, the product is called salt and the reaction is called neutralization reaction.

Complete answer:
In the question it is given that oxalic acid (${{H}_{2}}{{C}_{2}}{{O}_{4}}$) and sodium oxalate ($NaH{{C}_{2}}{{O}_{4}}$) behaves as acids as well as reducing agents.
The given statement is correct.
Now we have to discuss the options given.
Coming to option A, Equivalent weight of ${{H}_{2}}{{C}_{2}}{{O}_{4}}$ and $NaH{{C}_{2}}{{O}_{4}}$are equal to their molecular weights when acting as reducing agents.
The n factor for ${{H}_{2}}{{C}_{2}}{{O}_{4}}$an acid is n = 2 (it releases two protons easily)
The n factor for ${{H}_{2}}{{C}_{2}}{{O}_{4}}$a reducing agent is n = 2 (it donates two hydrogens very easily)
Coming to$NaH{{C}_{2}}{{O}_{4}}$ the n factor as an acid n = 1 (it releases one proton only)
The n factor for $NaH{{C}_{2}}{{O}_{4}}$a reducing agent is n = 2 (it forms${{C}_{2}}O_{4}^{2-}$).
Therefore n-factor for${{H}_{2}}{{C}_{2}}{{O}_{4}}$, $NaH{{C}_{2}}{{O}_{4}}$as an acid is not the same, so their equivalent weights are not equal to their molecular weights.
So, option A is wrong.
Coming to option B, Equivalent weights of ${{H}_{2}}{{C}_{2}}{{O}_{4}}$and $NaH{{C}_{2}}{{O}_{4}}$ are equal to half their molecular weights when acting as reducing agents. Yes it is correct because both the molecules will convert into ${{C}_{2}}O_{4}^{2-}$by giving two ions (as per option A discussion). So, option B is correct.
Coming to option C, 100 mL of 1 M of each solution ​ is neutralized by equal volumes of 1 M$Ca{{(OH)}_{2}}$​. It is wrong because 100 ml of 1 M solution of $NaH{{C}_{2}}{{O}_{4}}$ is neutralized by 50 ml of$Ca{{(OH)}_{2}}$.
Coming to option D, 100 mL of 1M solution of each is neutralized by equal volumes of 1 M of $KMn{{O}_{4}}$ ​. It is correct because n-factor for both the chemicals is 2 and
mEq of $KMn{{O}_{4}}$= mEq of ${{H}_{2}}{{C}_{2}}{{O}_{4}}$
mEq of $KMn{{O}_{4}}$ ​= mEq of $NaH{{C}_{2}}{{O}_{4}}$
here mEq = milliequivalents.
So, option B and Option D are correct.

Note: Don’t be confused with the terms molecular weight and equivalent weight. Both are not the same.
Molecular weight: The sum of the atomic masses of all atoms present in a given molecule is called molecular weight.
Equivalent weight: It is a ratio of atomic masses of all atoms present in the given molecule to the charge of the ion produced during its solubility.