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

Hint: 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) = $${\displaystyle \frac{24.312}{2}}$$ g, aluminum (Al) = $${\displaystyle \frac{26.312}{3}}$$ g and sulfur (S) = $${\displaystyle \frac{32.064}{2}}$$ 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.