Question

The molarity (M) of ${\text{KMn}}{{\text{O}}_4}$ solution in aqueous acidic medium is :
A ) 0.2 M
B ) 0.02 M
C ) 0.4 M
D ) 0.04 M

Answer 1

In a redox reaction, n factor gives the change in the oxidation number of the species undergoing oxidation or reduction. The number of milliequivalents is equal to the product of molarity, volume and n factor. The number of milliequivalents of oxidant are equal to the number of milliequivalents of reductants.

Complete step by step answer:
Consider the titration reaction between ${\text{100 mL KMn}}{{\text{O}}_4}$ and ${\text{100 mL 0}}{\text{.1 M N}}{{\text{a}}_{\text{2}}}{{\text{S}}_2}{{\text{O}}_3}$ solution. The oxidation number of manganese decreases from +7 to +2. The n factor for manganese will be 5.
Let M be the molarity of ${\text{KMn}}{{\text{O}}_4}$
Volume of ${\text{KMn}}{{\text{O}}_4}$ is $100{\text{ }}mL$
Calculate the milliequivalents of ${\text{KMn}}{{\text{O}}_4}$:

{\text{meq of KMn}}{{\text{O}}_4}{\text{ }} = {\text{ M }} \times {\text{ 100 }} \times {\text{ 5 }} \\\ {\text{meq of KMn}}{{\text{O}}_4}{\text{ }} = {\text{ 500M}} \\\ {\text{ }} \\\\$$ For $${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_2}{{\text{O}}_3}$$ consider the following half reaction $${{\text{S}}_2}{\text{O}}_3^{2 - }{\text{ }} \to {\text{ }}\dfrac{1}{2}{\text{ }}{{\text{S}}_4}{\text{O}}_6^{2 - }{\text{ + }}{{\text{e}}^ - }{\text{ }}$$ The n factor for $${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_2}{{\text{O}}_3}$$ is 1. The molarity of $${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_2}{{\text{O}}_3}$$ is 0.1 M. Volume of $${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_2}{{\text{O}}_3}$$ is 100 mL Calculate the milliequivalents of $${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_2}{{\text{O}}_3}$$: $${\text{meq of N}}{{\text{a}}_{\text{2}}}{{\text{S}}_2}{{\text{O}}_3}{\text{ }} = {\text{ molarity }} \times {\text{ Volume }} \times {\text{ n factor}} \\\ {\text{meq of N}}{{\text{a}}_{\text{2}}}{{\text{S}}_2}{{\text{O}}_3}{\text{ }} = {\text{ 0}}{\text{.1 }} \times {\text{ 100 }} \times {\text{ 1 }} \\\ {\text{meq of N}}{{\text{a}}_{\text{2}}}{{\text{S}}_2}{{\text{O}}_3}{\text{ }} = {\text{ 10}} \\\\$$ But the number of milliequivalents of $${\text{KMn}}{{\text{O}}_4}$$ is equal to the number of milliequivalents of $${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_2}{{\text{O}}_3}$$. $${\text{meq of KMn}}{{\text{O}}_4}{\text{ }} = {\text{ meq of N}}{{\text{a}}_{\text{2}}}{{\text{S}}_2}{{\text{O}}_3} \\\ {\text{500M }} = {\text{ }}10{\text{ }} \\\ {\text{M }} = {\text{ }}\dfrac{{10}}{{500}} \\\ {\text{M}} = 0.02{\text{M}} \\\\$$ The molarity of $${\text{KMn}}{{\text{O}}_4}$$ solution is $${\text{0}}{\text{.02 M}}$$ . _**Hence, the correct option is B ) $${\text{0}}{\text{.02 M}}$$.**_ **Note:** Take proper care while calculating the n factor of $${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_2}{{\text{O}}_3}$$. Do not write the half reaction as $${\text{2 }}{{\text{S}}_2}{\text{O}}_3^{2 - }{\text{ }} \to {\text{ }}{{\text{S}}_4}{\text{O}}_6^{2 - }{\text{ + 2 }}{{\text{e}}^ - }$$and say that n factor is 2. The half reaction is $${{\text{S}}_2}{\text{O}}_3^{2 - }{\text{ }} \to {\text{ }}\dfrac{1}{2}{\text{ }}{{\text{S}}_4}{\text{O}}_6^{2 - }{\text{ + }}{{\text{e}}^ - }{\text{ }}$$and the n factor is 1.