
In the reaction of sodium thiosulphate with iodine in aqueous medium, the equivalent weight of sodium thiosulphate is equal to?
(A) Molar mass of sodium thiosulphate
(B) The average molar masses of $N{{a}_{2}}S{{O}_{3}}\text{ and }{{\text{I}}_{2}}$
(C) Half the molar mass of sodium thiosulphate
(D) Molar mass of sodium thiosulphate $\times $ 2
Answer
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Hint: To solve this, firstly write the reaction that will take place between sodium thiosulphate and iodine. From this reaction, find the number of electrons lost or gained by ${{S}_{2}}{{O}_{3}}^{2-}$ ion and use it to find the equivalent weight of sodium thiosulphate.
Complete step by step answer:
Before solving this, we must know that the equivalent weight is the mass of one equivalent of a substance.
We can use different formulas to find out the equivalent weight of the required compound depending on the reaction.
If it is a redox reaction, the equivalent weight is the formula weight divided by the change in oxidation number, whereas for acid and base it is molecular weight divided by the basicity of acid and the acidity of the base respectively.
In the given question, we have sodium thiosulfate reacting with iodine to give us tetrathionate sodium and sodium iodide. We can write the reaction as:
\[2N{{a}_{2}}S{{O}_{3}}+{{I}_{2}}\to 2NaI+N{{a}_{2}}{{S}_{4}}{{O}_{6}}\]
Here, we can write the half reactions as:
\[\begin{align}
& {{I}_{2}}+2{{e}^{-}}\to 2{{I}^{-}} \\
& 2{{S}_{2}}{{O}_{3}}^{2-}\to {{S}_{4}}{{O}_{6}}^{2-}+2{{e}^{-}} \\
\end{align}\]
Here, we can also find the equivalent weight by n-factor.
Equivalent weight =$\dfrac{Molecular\text{ }weight}{~valency\text{ }factor}$
For two electrons lost by ${{S}_{2}}{{O}_{3}}^{2-}$ ion we get 2 molecules of NaI. So, for 1 ion we will get 1 molecule of NaI.
Therefore, n-factor i.e. the loss or gain in electrons is 1.
So, we can write that, equivalent weight = molecular weight.
We can understand from the above discussion that the equivalent weight of sodium thiosulphate is equal to the molecular weight of sodium thiosulphate.
Therefore, the correct answer is option (A) Molar mass of sodium thiosulphate.
Note: We should not be confused between equivalent weight and number of equivalents. Equivalent weight is the gram molecular weight of the substance divided by the valency factor of the substance. Valency factor is basically the charge it carries whereas the number of equivalents is the weight/ mass of the compound divided by its equivalent weight.
Complete step by step answer:
Before solving this, we must know that the equivalent weight is the mass of one equivalent of a substance.
We can use different formulas to find out the equivalent weight of the required compound depending on the reaction.
If it is a redox reaction, the equivalent weight is the formula weight divided by the change in oxidation number, whereas for acid and base it is molecular weight divided by the basicity of acid and the acidity of the base respectively.
In the given question, we have sodium thiosulfate reacting with iodine to give us tetrathionate sodium and sodium iodide. We can write the reaction as:
\[2N{{a}_{2}}S{{O}_{3}}+{{I}_{2}}\to 2NaI+N{{a}_{2}}{{S}_{4}}{{O}_{6}}\]
Here, we can write the half reactions as:
\[\begin{align}
& {{I}_{2}}+2{{e}^{-}}\to 2{{I}^{-}} \\
& 2{{S}_{2}}{{O}_{3}}^{2-}\to {{S}_{4}}{{O}_{6}}^{2-}+2{{e}^{-}} \\
\end{align}\]
Here, we can also find the equivalent weight by n-factor.
Equivalent weight =$\dfrac{Molecular\text{ }weight}{~valency\text{ }factor}$
For two electrons lost by ${{S}_{2}}{{O}_{3}}^{2-}$ ion we get 2 molecules of NaI. So, for 1 ion we will get 1 molecule of NaI.
Therefore, n-factor i.e. the loss or gain in electrons is 1.
So, we can write that, equivalent weight = molecular weight.
We can understand from the above discussion that the equivalent weight of sodium thiosulphate is equal to the molecular weight of sodium thiosulphate.
Therefore, the correct answer is option (A) Molar mass of sodium thiosulphate.
Note: We should not be confused between equivalent weight and number of equivalents. Equivalent weight is the gram molecular weight of the substance divided by the valency factor of the substance. Valency factor is basically the charge it carries whereas the number of equivalents is the weight/ mass of the compound divided by its equivalent weight.
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