\(25 \mathrm{~mL}\) of \(0.017 \mathrm{H}_{2} \mathrm{SO}_{3}\) in strongly acidic medium required \(16.9 \mathrm{~mL}\) of \(0.01 \mathrm{M} \mathrm{KMnO}_{4}\) and in neutral medium required \(28.6 \mathrm{~mL}\) of \(0.01 M\) \(\mathrm{KMnO}_{4}\) for complete conversion of \(\mathrm{SO}_{3}{ }^{2-}\) to \(\mathrm{SO}_{4}^{2-}\). Assign the oxidation no. of Mn in the product formed in each case.

Short Answer

Expert verified
Oxidation number of Mn in a strong acidic medium is +2, and in a neutral medium is +4.

Step by step solution

01

Write the balanced reactions

In a strongly acidic medium, the balanced reaction of permanganate ion and sulfite ion is: \[\mathrm{2KMnO_4 + 5H_2SO_3 + 3H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5H_2SO_4}\] In a neutral medium, the balanced reaction is: \[\mathrm{3KMnO_4 + 3H_2O + 5H_2SO_3 \rightarrow 3KOH + 2MnO_2 + 5H_2SO_4 + 2H_2O}\]
02

Determine the molarity of KMnO4 in a strongly acidic medium

Using the volume of KMnO4 solution and the molarity of H2SO3, calculate the molarity of KMnO4 in a strongly acidic medium: \[\frac{0.017\,M \times 25\,mL}{16.9\,mL} = 0.025 \, M\]
03

Determine the molarity of KMnO4 in neutral medium

Similarly, calculate the molarity of KMnO4 in a neutral medium: \[\frac{0.017\,M \times 25\,mL}{28.6\,mL} = 0.015 \, M\]
04

Calculate the oxidation number of Mn in each reaction

The oxidation number of Mn in KMnO4 is +7. For the product formed in a strongly acidic medium (MnSO4), the oxidation number of Mn is +2. In a neutral medium (MnO2), the oxidation number of Mn is +4.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Permanganate Titrations
Permanganate titrations are a type of redox titration where potassium permanganate (\( \text{KMnO}_4 \)) is commonly used as the titrant due to its strong oxidizing properties. These titrations rely on the change in oxidation states of the substances involved and are characterized by a distinct color change, typically from purple to colorless or brown, which indicates the end point of the titration.

In the original problem, permanganate titrations are used in different pH conditions to determine how much \( \text{KMnO}_4 \) is required to oxidize another substance, in this case, sulfite (\( \text{SO}_3^{2-} \)) to sulfate (\( \text{SO}_4^{2-} \)). The titration outcomes are different in strongly acidic versus neutral media because the oxidation potential of \( \text{KMnO}_4 \) varies with pH, and this can also affect which products are formed. It's crucial to consider the reaction conditions when setting up the titration and in interpreting the results.
Chemical Stoichiometry
Chemical stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It's indispensable for solving titration problems since it stipulates the ratios in which the reagents react and the products are formed. By writing out balanced chemical equations, one can deduce the stoichiometry of the involved substances.

In the exercise given, stoichiometry is pivotal for understanding the molarity calculations in steps 2 and 3 of the solution. It indicates that a specific volume of the sulfite solution reacts with a different volume of \( \text{KMnO}_4 \) solution depending on the medium's pH. The balanced chemical equations from step 1 show the molar ratios of sulfite to permanganate, which are crucial to find the correct molarity of \( \text{KMnO}_4 \) in each case. Understanding stoichiometry allows you to accurately calculate the concentration of one reactant based on the known concentration and volume of another.
Oxidation Numbers
Oxidation numbers are a concept used in chemistry to describe the degree of oxidation or reduction of an atom within a molecule or ion. They are fundamental in analyzing redox reactions, which include the transfer of electrons between species.

Oxidation states can change in a reaction, signifying the movement of electrons. In redox titrations like those involving \( \text{KMnO}_4 \), determining the oxidation numbers is essential to identify how many electrons are transferred in each reaction. For example, the exercise requires assigning oxidation numbers to manganese in the products formed in different media. In strongly acidic medium, Mn goes from +7 in \( \text{KMnO}_4 \) to +2 in \( \text{MnSO}_4 \), indicating a reduction. In a neutral medium, it goes to +4 in \( \text{MnO}_2 \), also reduction but to a lesser extent than in the acidic medium. This change helps us comprehend both the electron transfer and the stoichiometry of the reactions.

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Most popular questions from this chapter

\(\mathrm{Mn}^{2+}(\) aq \()\) can be determined by titration with \(\mathrm{MnO}_{4}^{-}(\) aq \()\) $$ 3 \mathrm{Mn}^{2+}+2 \mathrm{MnO}_{4}^{-} \longrightarrow 6 \mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{O} $$ A \(25.00 \mathrm{~mL}\) sample of \(\mathrm{Mn}^{2+}\) (aq) requires \(34.77 \mathrm{~mL}\) of \(0.05876 M\) \(\mathrm{KMnO}_{4}\) (aq) for its titration. What is the molarity of the \(\mathrm{Mn}^{2+}\) (aq)?

Reaction, \(2 \mathrm{Br}^{-}{ }_{(\mathrm{aq})}+\mathrm{Cl}_{2}(\mathrm{aq}) \longrightarrow 2 \mathrm{Cl}^{-}_{(\mathrm{aq} .)}+\mathrm{Br}_{2(\mathrm{aq} .)}\), is used for commercial preparation of bromine from its salts. Suppose we have \(50 \mathrm{~mL}\) of a \(0.060 \mathrm{M}\) solution of NaBr. What volume of a \(0.050 \mathrm{M}\) solution of \(\mathrm{Cl}_{2}\) is needed to react completely with the Br?

A \(1.100 \mathrm{~g}\) sample of copper ore is dissolved and the \(\mathrm{Cu}^{2+}(\mathrm{aq})\) is treated with excess \(\mathrm{KI}\). The liberated \(\mathrm{I}_{2}\) requires \(12.12 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solution for titration. What is \(\%\) copper by mass in the ore?

A particular acid-rain water has \(\mathrm{SO}_{3}{ }^{2-}\). If a \(25.00 \mathrm{~mL}\) sample of this water requires \(34.08 \mathrm{~mL}\) of \(0.01964 \mathrm{M} \mathrm{KMnO}_{4}\) for its titration, what is the molarity of \(\mathrm{SO}_{3}{ }^{2-}\) in acid-rain? \(2 \mathrm{MnO}_{4}^{-}+5 \mathrm{SO}_{3}^{2-}+6 \mathrm{H}^{+} \longrightarrow 5 \mathrm{SO}_{4}^{2-}+2 \mathrm{Mn}^{2+}+3 \mathrm{H}_{2} \mathrm{O}\)

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