Question

The equivalent mass of sodium thiosulphate is (\[N{a_2}{S_2}{O_3}.5{H_2}O\]) in this reaction.
\[2N{a_2}{S_2}{O_3} + {I_2} \to 2NaI + N{a_2}{S_4}{O_6}\]
A) $248$
B) $124$
C) $596$
D) $62$

Answer 1

Hint: We have to remember that the calculation of equivalent mass should be different for acids and bases. For bases, equivalent mass can be calculated by dividing the molar mass with the hydroxyl group present in a particular base and for acids the molar mass is divided by the number of protons present in an acid.

Complete answer:
We will calculate the number of equivalents for one mole of sodium thiosulphate. The oxidation state for the elements leaving sodium and iodine is the same at both left hand side and right hand side. We need to calculate the oxidation states for other elements first to get equivalent mass.
Consider the oxidation state of sulphur in left hand side and right hand side be x and y
$\left( {2 \times 1} \right) + \left( {2x} \right) + \left( {3x - 2} \right) = 0$
$2x = + 4$
$ \Rightarrow x = 2$
Thus, oxidation state of sulphur in left hand side is $2$
For calculating y,
\[\left( {2 \times 1} \right) + \left( {4 \times y} \right) + 6x - 2 = 0\]
$4y = 10$
$y = \dfrac{5}{2}$
The change in oxidation state in sulphur is ½. Since sulphur is two in number in sodium thiosulphate thus \[2 \times \dfrac{1}{2} = 1\] therefore the number equivalent in one mole of sodium thiosulphate is $1$.
Equivalent weight of sodium thiosulphate= molecular weight of sodium thiosulpahte
Since molecular weight of sodium thiosulphate=\[2 \times \dfrac{1}{2} = 1\]
Molecular weight of sodium thiosulphate is:
\[2 \times 23 + 2 \times 32 + 3 \times 16 + 5 \times 18 = 248\]
Thus the equivalent mass of sodium thiosulphate is $248$.

Note:
We must have to remember that the electrons transferred should be taken by the total number of atoms, since there are two sulphur atoms therefore one mole of sodium thiosulphate contains two sulphur atoms. As we know that for acids and bases the calculation for equivalent mass will be different.