Question

A solution of$N{{a}_{2}}{{S}_{2}}{{O}_{3}}$ is standardised iodometrically by using ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$. The equivalent weight of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ in this method is:
A) mol.wt/2
B) mol.wt/6
C) mol.wt/3
D) equal to mol.wt.

Answer 1

Hint: The answer here is based on the concept of calculating equivalent weight based on the definition and the formula which is given by $Eq.Wt.=\dfrac{M}{n-factor}$ and by finding n- factor according to reaction gives you the required answer.

Complete step – by – step solution:
We have come across the chapters in physical chemistry that is based on several measurement techniques like conductometry, potentiometry, iodometry etc.
- Now, let us see what does iodometry means and how is equivalent weight calculated in this technique.
- Iodometry or which is also known as iodimetric titration is a method of volumetric analysis where the appearance or disappearance of iodine indicates the end point and it is the redox reaction.
The reaction is as follows,
$26{{H}^{+}}+3{{S}_{2}}Ol3+4C{{r}_{2}}{{O}_{7}}^{2-}\to 6S{{O}_{4}}^{2-}+8C{{r}^{3+}}+13{{H}_{2}}O$
Here, in ${{K}_{2}}C{{r}_{2}}{{O}_{7}}(C{{r}_{2}}{{O}_{7}}^{2-})$, chromium is in +6 oxidation state and in the product it changes to +3 state.
Thus, there is net change of +3 in oxidation state.
Now, the net change in oxidation per formula is from $4C{{r}_{2}}{{O}_{7}}^{2-}$ to $8C{{r}^{3+}}$ is 2
Thus, net change for oxidation = $2\times 3=+6$
Thus, n-factor = +6
Now according to the formula of equivalent weight, $Eq.Wt.=\dfrac{M}{n-factor}$ where M is the molecular weight.
Substituting values, we have$Eq.Wt.=\dfrac{M}{6}$

Therefore, the correct answer is option B) mol.wt./6

Note: There are chances to be confused between iodometry and iodimetry. Here, iodometry involves indirect titration of the iodine that is liberated by the reaction with that of analyte and iodimetry involves direct titration using iodine itself as the titrant.