Question

If the molecular weight of $Ba{\left( {Mn{O_4}} \right)_2}$ is M, then the equivalent weight of $Ba{\left( {Mn{O_4}} \right)_2}$in acidic medium is
A) $M/5$
B) $M/10$
C) $M/3$
D) $M$

Answer 1

Hint: While doing these questions, we must first figure out whether the compound mentioned in the question will get oxidized or reduced in the given conditions. After that, we must know the electron change that the compound undergoes in the reaction. This will help us in the $n - factor$calculations. Then, we can easily relate the molecular weight and the equivalent weight.

Complete step-by-step answer:
We know that the highest oxidation state of barium is $ + 2$. Hence, the compound can’t undergo oxidation as there is absence of any species which can undergo an increase in the oxidation number.
Now, it is quite clear and obvious that the manganate species will undergo reduction in the acidic medium.

Writing reaction to know the change in the oxidation number:
We know that the given compound will ionize as follows and will give two manganate ions in the solution.
$Ba{\left( {Mn{O_4}} \right)_2}\underset {} \leftrightarrows B{a^{2 + }} + 2Mn{O_4}^ - $
Now, writing reduction reaction of manganate ion,
$Mn{O_4}^ - \to M{n^{2 + }}$(in acidic medium)
There is a change in oxidation number of manganese from $ + 7$to $ + 2$ in acidic medium.
Change $ = 7 - 2 = 5$
But since two moles of manganese are formed from one mole of the compound.
Hence, total change in oxidation number $ = 2 \times 5 = 10$
Now, we are left with calculating the equivalent weight
Equivalent weight $ = \dfrac{{Molar\,weight}}{{n - factor\,of\,the\,salt}}$
We are provided with molar weight of M. Hence, equivalent weight $ = \dfrac{M}{{10}}$

Hence, Option ‘B’ i.e, $M/10$ is the correct answer.

Note: We must be quite careful in calculations involving n-factor. n- factor is defined as molecular weight divided by equivalent weight. n-factor calculations are also done for acids and bases. In calculation of n-factor for oxidizing agents like $KMn{O_4}$, you have to pay special attention to the conditions mentioned in the question because you should remember that manganate ion undergo different changes in oxidation numbers in different mediums- acidic, basic or neutral.