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Question: \(28NO_3^ - + 3A{s_2}{S_3} + 4{H_2}O \to 6AsO_4^{3 - } + 28NO + 9SO_4^{2 - } + 8{H^ + }\) What wil...

28NO3−+3As2S3+4H2O→6AsO43−+28NO+9SO42−+8H+28NO_3^ - + 3A{s_2}{S_3} + 4{H_2}O \to 6AsO_4^{3 - } + 28NO + 9SO_4^{2 - } + 8{H^ + }
What will be the equivalent mass of As2S3A{s_2}{S_3} in above reaction?
a.) M.wt.2\dfrac{{M.wt.}}{2}
b.) M.wt.4\dfrac{{M.wt.}}{4}
c.) M.wt.24\dfrac{{M.wt.}}{{24}}
d.) M.wt.28\dfrac{{M.wt.}}{{28}}

Explanation

Solution

The equivalent mass of a substance is defined as the Molar mass of substance divided by total . The n-factor is the number of electrons gained by the substance.
Thus, Equivalent Mass= Molar massNumber of electrons gained\dfrac{{Molar{\text{ mass}}}}{{Number{\text{ of electrons gained}}}}
The total number of electrons gained can also be attributed to change in oxidation number.

Complete Answer:
We know the equivalent mass of a substance can be obtained using the formula as-
Equivalent mass= Molar massTotal change in oxidation number\dfrac{{Molar{\text{ mass}}}}{{Total{\text{ }}change{\text{ in oxidation number}}}}
From above reaction if we take only change in As, then the reaction is –
As2S3→AsO43−A{s_2}{S_3} \to AsO_4^{3 - }
The oxidation number of As in As2S3A{s_2}{S_3} is +3.
And the oxidation number of As in AsO43−AsO_4^{3 - } is +5.
So, the change in oxidation number per As atom can be written as-
5-3 = 2.
As we know two As atoms are involved. Thus, change in oxidation number of two As atoms is –
2*2 = 4.
Now, for the equivalent mass of As2S3A{s_2}{S_3}; we have the formula –
Equivalent mass of As2S3A{s_2}{S_3} = Molar mass of As2S3Total change in oxidation number of As\dfrac{{Molar{\text{ mass of A}}{{\text{s}}_2}{S_3}}}{{Total{\text{ change in oxidation number of As}}}}
Total change in oxidation state is 4.
Thus, Equivalent mass of As2S3A{s_2}{S_3} = M.wt.4\dfrac{{M.wt.}}{4}

So, option b is the correct answer.

Note : In this reaction change in oxidation number of As is +2 only. But in the reactant two atoms of As are taken. So, the value of change should be multiplied by 2 giving the total change in oxidation number.
We can also solve this question by taking the total number of electrons gained. The formula for which is the same just on the place of total change in oxidation number is replaced by total number of electrons gained. It can be written as –
Equivalent Mass= Molar massNumber of electrons gained\dfrac{{Molar{\text{ mass}}}}{{Number{\text{ of electrons gained}}}}