Solveeit Logo
Login

Question

Question: What is the reaction between \( N{{a}_{2}}{{S}_{2}}{{O}_{3}}\cdot 5{{H}_{2}}O \) and \( {{I}_{2}} \)...

What is the reaction between Na2S2O3⋅5H2ON{{a}_{2}}{{S}_{2}}{{O}_{3}}\cdot 5{{H}_{2}}O and I2{{I}_{2}} ?

Explanation

Solution

We know that In order to solve the given problem and find the equivalent mass of sodium thiosulphate in the reaction. First we will understand the basic definition of equivalent and molar mass and the concept of how to find each of them. Further on the basis of the chemical reaction of sodium thiosulphate with iodine.

Complete answer:
First let us understand the basic concept behind molar mass and equivalent mass of any particular compound. Molar mass: Molar mass is the mass of any particular given compound is the mass of one mole of that molecule and equivalent weight is given on the basis of molecular weight of the compound as well as number of electrons that a molecule gains or losses in a particular chemical reaction. The equivalent weight of a molecule is calculated by dividing the molecular weight of the same compound with the number of electrons lost or gained. Now as we know the concept behind molar weight and equivalent weight, let us now proceed further to solve the problem.
We will need the reaction between the iodine and sodium thiosulphate to find the reaction between the iodine and sodium thiosulphate to find the equivalent weight. The chemical reaction between iodine and sodium thiosulphate is given as;
I2+2Na2S2O3→2NaI+Na2S4O6{{I}_{2}}+2N{{a}_{2}}{{S}_{2}}{{O}_{3}}\to 2NaI+N{{a}_{2}}{{S}_{4}}{{O}_{6}}
In order to find the number of electrons lost or gained in the reaction let us see the half reaction of iodine and thiosulphate ions and half of the reaction is given as;
I2+2e−→2I−{{I}_{2}}+2{{e}^{-}}\to 2{{I}^{-}}
2S2O32−→S4O62−+2e−2{{S}_{2}}O_{3}^{2-}\to {{S}_{4}}O_{6}^{2-}+2{{e}^{-}}
From the given reaction we can see that for 22 molecules of sodium thiosulphate, 22 electrons are lost by sodium thiosulphate.
Therefore, the final reaction is given by;
S2O2−3+4I2+5H2O→2SO2−4+10H++8I−{{S}_{2}}O_{2}^{-3}+4{{I}_{2}}+5{{H}_{2}}O\to 2SO_{2}^{-4}+10{{H}^{+}}+8{{I}^{-}}

Note:
Remember that in order to solve such types of problems students must be aware of general reactions between the compounds. In the problem we needed to find the relationship between different weights, so we did not find their numerical value. Students must remember the method to the molar weight of a molecule by the help of the molar weight of elements.