Question

Calculate the equivalent weight of $S{O_2}$ in the following reactions.
a) $S{O_2} + 2{H_2}S = 3S + 2{H_2}O$
b)$5S{O_2} + 2KMn{O_4} + 2{H_2}O = {K_2}S{O_4} + 2MnS{O_4} + 2{H_2}S{O_4}$

Answer 1

Hint:We define the equivalent weight of the substance as the ratio of the molecular weight or mass of the compound to the n-factor or the acidity or basicity. We can calculate the n- factor by determining the change in oxidation state.

Complete step by step answer:
In the question we have to calculate the equivalent weight of the sulfur dioxide from the given reaction. We know the molecular weight of the sulfur dioxide is $64\,g/mol.$
Now, we calculate the n-factor. First, calculate the change in the oxidation state of sulfur in the reactions.
The given reactions are,
a) $S{O_2} + 2{H_2}S = 3S + 2{H_2}O$
In the above reaction, the oxidation state of sulfur is changed from $ + 4$ to zero. Thus the n-factor is $4.$ The equivalent weight of sulfur can be calculated as,
The equivalent weight of sulfur$ = \dfrac{{64}}{4} = 16\,g$.
b)$5S{O_2} + 2KMn{O_4} + 2{H_2}O = {K_2}S{O_4} + 2MnS{O_4} + 2{H_2}S{O_4}$
In the above, the oxidation state of sulfur is changed from $ + 4$ to\[ + 6\]. Thus the n-factor is$2$. Now we can calculate the equivalent weight of sulfur is calculated as,
The equivalent weight of sulfur$ = \dfrac{{64}}{2} = 32\,g$
Thus, the equivalent weight of sulfur in reaction (a) is $16\,g$ and the equivalent weight of sulfur in reaction (b) is $32g$.

Note:
 We can know, 'n' factor of an acid is the number of ions replaced by one mole of acid. The n-factor for acid isn't the same as its basicity; i.e. the number of moles of usable Hydrogen atoms present in one mole of acid.
The oxidation state of the single element is zero but if it has any charge present on it then it is considered as n-factor while basicity for the acidic substance and the acidity is defined for the basic substance.