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) = \[\dfrac{24.312}{2}\] 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.