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

Hint: 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 \[{\text{KMn}}{{\text{O}}_4}\]:
\[{\text{meq of KMn}}{{\text{O}}_4}{\text{ }} = {\text{ molarity }} \times {\text{ Volume }} \times {\text{ n factor}} \\
{\text{meq of KMn}}{{\text{O}}_4}{\text{ }} = {\text{ M }} \times {\text{ 100 }} \times {\text{ 5 }} \\
{\text{meq of KMn}}{{\text{O}}_4}{\text{ }} = {\text{ 500M}} \\
{\text{ }} \\\]
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.