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Question: A solution of 10mL \(FeS{{O}_{4}}\) was titrated with \(KMn{{O}_{4}}\) solution in acidic medium. Th...

A solution of 10mL FeSO4FeS{{O}_{4}} was titrated with KMnO4KMn{{O}_{4}} solution in acidic medium. The total amount of KMnO4KMn{{O}_{4}} used will be:
A. 10mL of 0.02M
B. 10mL of 0.5M
C. 10mL of 0.5M
D. 5mL of 0.1M

Explanation

Solution

The concept of n factor is to be used in this case. The number of moles of KMnO4KMn{{O}_{4}} can be found out using the volume of the solution and its molarity. After that, the number of moles from the options need to be matched.

Complete step by step answer:
- In order to answer our question, let us learn about potassium permanganate. Potassium permanganate can be prepared using two steps. Powdered pyrolusite is fused with KOH in the presence of air or an oxidising agent like KNO3KN{{O}_{3}}, or KClO3KCl{{O}_{3}} to give green coloured potassium manganate. The green mass is extracted with water and oxidised to KMnO4KMn{{O}_{4}} by passing chlorine or ozone into solution, or using electrolysis.. As a result of oxidation, the green colour of solution changes into purple. The purple solution upon concentration and then evaporation gives crystals of KMnO4KMn{{O}_{4}}. Let us study about some properties of the compound:
i. Potassium permanganate forms dark purple needles like crystals with a metallic luster. It's m.p. 523 K. It is moderately soluble in water giving a purple solution.
ii. Potassium permanganate changes into manganate and oxygen gas is evolved on heating.
iii. On heating with alkalies, potassium permanganate changes into manganate and oxygen gas is evolved.
iv. KMnO4KMn{{O}_{4}} is a powerful oxidising agent in acidic, neutral and alkaline medium.
Now the strength or concentration of FeSO4FeS{{O}_{4}} is M10\dfrac{M}{10}. The reaction can be represented as:
MnO4−+Fe2+→Mn2++Fe3+Mn{{O}_{4}}^{-}+F{{e}^{2+}}\to M{{n}^{2+}}+F{{e}^{3+}}
- Now, let us take out the n factor of KMnO4KMn{{O}_{4}}. The n- factor is 5 as valency of Mn changes from +7 to +2. The change in oxidation of Fe is 3 - 2 = 1. That means 1 mole FeSO4→15moles KMnO41\,mole\,FeS{{O}_{4}}\to \dfrac{1}{5}moles\,KMn{{O}_{4}}. As we know that
M=no of molevolumeM=\dfrac{no\,of\,mole}{volume}, so
n=110×10×(10−3l)n=\dfrac{1}{10}\times 10\times ({{10}^{-3}}l)
=10−3 mole={{10}^{-3}}\,mole.
- Now, by using unitary method, we can write that:
10−3 moles FeSO4→15×10−3 mol KMnO4  \begin{aligned} & {{10}^{-3}}\,moles\,FeS{{O}_{4}}\to \dfrac{1}{5}\times {{10}^{-3}}\,mol\,KMn{{O}_{4}} \\\ & \\\ \end{aligned}
Which gives 2×10−4 mol KMnO42\times {{10}^{-4}}\,mol\,KMn{{O}_{4}}. Now, let us see the options. As, number of moles is product of molarity and volume, so for option A, the number of moles is: 2×10−2×10−2=2×10−4mole2\times {{10}^{-2}}\times {{10}^{-2}}=2\times {{10}^{-4}}mole, which matches our question. So the correct answer is “A”:

Note: The uses of KMnO4KMn{{O}_{4}} are:
i. As an oxidising agent in the laboratory and industry.
ii. For volumetric estimation of ferrous salts, oxalates and other reducing agents.
iii. As a disinfectant for water.
iv. In dry cells.