Question

In the reaction ${I_2} + 2N{a_2}{S_2}{O_3} \to 2NaI + N{a_2}{S_4}{O_6}$, what is the equivalent weight of sodium thiosulphate?

Answer 1

Hint: Equivalent weight is of an element is its gram atomic weight or molecular weight divided by its valence factor or n-factor. Also, it can be defined as the weight of a compound that contains one equivalent of a proton (for acid) or one equivalent of a hydroxide (for base). Equivalent weight can be determined by using the formula, $Equivalent\;Weight = \dfrac{{Molecular\;Weight}}{{n\;Factor}}$

Complete step by step answer:
In this question we have to determine the equivalent weight of sodium thiosulphate. We know that equivalent weight can be determined by using the formula, $Equivalent\;Weight = \dfrac{{Molecular\;Weight}}{{n\;Factor}}$
${I_2} + 2N{a_2}{S_2}{O_3} \to 2NaI + N{a_2}{S_4}{O_6}$is a redox reaction. In this reaction, sulphur and iodine is being oxidised and reduced respectively.
If we compare both sides of the reaction we can see that in ${I_2}$the oxidation number of iodine (I) is $0$, whereas in $NaI$the oxidation number of iodine (I) is $ - 1.$It shows that iodine is reduced in this reaction.
Now, we have to find the n-factor of $N{a_2}{S_2}{O_3}$.
First, we have to determine the oxidation number of sulphur (S) in $N{a_2}{S_2}{O_3}$
Let the oxidation number of S is x in $N{a_2}{S_2}{O_3}$. So, we can write,
$ \Rightarrow 2 \times ( + 1) + 2 \times x + 3 \times ( - 2) = 0$
$ \Rightarrow 2x = 4$
$ \Rightarrow x = 2$
Hence, in $N{a_2}{S_2}{O_3}$oxidation number of S is 2.
Similarly, we have to determine the oxidation number of sulphur (S) in $N{a_2}{S_4}{O_6}$
Let the oxidation number of S is y in $N{a_2}{S_4}{O_6}$. So, we can write,
$ \Rightarrow 2 \times ( + 1) + 4 \times y + 6 \times ( - 2) = 0$
$ \Rightarrow 4y = 10$
$ \Rightarrow y = 2.5$
Hence, in $N{a_2}{S_4}{O_6}$oxidation number of S in 2.5.
So, n-factor of $N{a_2}{S_2}{O_3}$ is $ = 2(2 - 2.5) = 1$
Therefore, now we can find the equivalent weight of sodium thiosulphate by using the formula,
$Equivalent\;Weight = \dfrac{{MolecularWeight}}{{nFactor}}$

$ \Rightarrow Equivalent\;Weight = \dfrac{{MolecularWeight}}{1}$

So, the correct answer is Option A.

Note: ‘n’ factor is also known as valence factor or conversion factor. In a redox reaction this factor is equal to the number of moles of lost or gained electrons per molecule and in a non-redox reaction it is equal to the product of displaced mole and its charge.