For the reaction, H 3 PO 2 + NaOH → NaH 2 PO 2 + H 2 O What is the equivalent weight of H 3 P O 2 ? (mol. wt. is M)

Equivalent weight of H 3 PO 2 = n − factor M

For the reaction, H 3 PO 2 + NaOH → NaH 2 PO 2 + H 2 O What is the equ | TestHub

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$Eq u i v a l e n tw e i g h t = \frac{\frac{M o l ec u l a r w e i g h t}{A t o mi c w e i g h t}}{n - f a c t or}$

The n-factor is very important in both redox and non-redox reactions. With the help of the n-factor, we can predict the molar ratio of the reactant species taking part in reactions. The reciprocal of the n-factor's ratio of the reactants is the molar ratio of the reactants.

In general, the n-factor of an acid/base is the number of moles of H⁺/OH⁻ furnished per mole of acid/base. The n-factor of a reactant is the number of moles of electrons lost or gained per mole of reactant.

Example 1:

(1) In acidic medium: $KMn O_{4}$ (n = 5) → $M n^{2 +}$

(2) In basic medium: $KMn O_{4}$ (n = 3) → $M n^{2 +}$

(3) In neutral medium: $KMn O_{4}$ (n = 1) → $M n^{6 +}$

Example 2:

$Fe C_{2} O_{4} \to F e^{3 +} + 2C O_{2}$

Total number of moles of electrons lost by 1 mole of $Fe C_{2} O_{4}$ is $1 + (1 \times 2) = 3$ .

Therefore, the n-factor of $Fe C_{2} O_{4}$ is 3.

For the reaction, $H_{3} PO_{2} + NaOH \to NaH_{2} PO_{2} + H_{2} O$

What is the equivalent weight of $H_{3} P O_{2} ?$

(mol. wt. is M)

Options:

Answer:

A

Solution:

Equivalent weight of $H_{3} PO_{2} = \frac{M}{n - factor}$

Stream:JEESubject:ChemistryTopic:Redox ReactionsSubtopic:Equivalent Weight and n-factor

⏱ 2mℹ️ Source: QB

Chemistry - Redox Reactions Question with Solution | TestHub

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