
100 ml of given $KMn{{O}_{4}}$ titrates 50 ml of 0.1M Oxalic acid. Its normality against alkaline ${{H}_{2}}{{O}_{2}}$ is:
A. 0.1 N
B. 0.02 N
C. 0.06 N
D. None of the above
Answer
233.1k+ views
Hint: A certain amount of milliequivalents of $KMn{{O}_{4}}$ will titrate the same amount of milliequivalents of oxalic acid. Also, remember that the n-factor of $KMn{{O}_{4}}$ varies in acidic and basic medium.
Step-by-Step Solution:
Let us first go through the properties of Potassium Permanganate ($KMn{{O}_{4}}$) as a compound.
Potassium permanganate is an inorganic compound with the chemical formula $KMnO_4$ and composed of K+ and $MnO^{4-}$. It is a purplish-black crystalline solid that dissolves in water to give intensely pink or purple solutions.
Potassium permanganate is widely used in the chemical industry and laboratories as a strong oxidizing agent, and also as a medication for dermatitis, for cleaning wounds, and general disinfection.
Having established the properties and usage of $KMn{{O}_{4}}$, let us now move on to the particulars of this question.
Now, we know that Oxalic Acid’s chemical formula is $COOH-COOH$ and due to its presence in the initial titration, the medium would thus be acidic.
However, the later titration is done against alkaline ${{H}_{2}}{{O}_{2}}$ where the medium would be alkaline in nature.
With all this information now at our disposal along with the hint, let us now proceed to the step-by-step solution:
\[\begin{align}
& Milliequivalents\text{ }of\text{ }oxalic\text{ }acid~=50\times 0.1\times 2=10 \\
& \because Initial\text{ }medium\text{ }is\text{ }acidic~(\Rightarrow n=5~for~KMnO4). \\
& \therefore ~Molarity\text{ }of~KMn{{O}_{4}}=\dfrac{N}{5}=\dfrac{\dfrac{10}{100}}{5}=0.02M \\
& In\text{ }alkaline\text{ }medium,~\text{ }n=3~for~KMn{{O}_{4}}~ \\
& Hence,\text{ }normality~=0.02\times 3~=0.06N \\
\end{align}\]
Therefore, by our calculations, we can safely conclude that the required normality of $KMn{{O}_{4}}$ is 0.06N making the answer to this question c) 0.06N.
Note: Absolute caution must be taken during the titrations of any acidic or alkaline substance, regardless of its strength, so as to avoid any form of injury or damage. Also make sure that no mistakes are made in the n- factor calculation of Potassium Permanganate w.r.t change in mediums.
Step-by-Step Solution:
Let us first go through the properties of Potassium Permanganate ($KMn{{O}_{4}}$) as a compound.
Potassium permanganate is an inorganic compound with the chemical formula $KMnO_4$ and composed of K+ and $MnO^{4-}$. It is a purplish-black crystalline solid that dissolves in water to give intensely pink or purple solutions.
Potassium permanganate is widely used in the chemical industry and laboratories as a strong oxidizing agent, and also as a medication for dermatitis, for cleaning wounds, and general disinfection.
Having established the properties and usage of $KMn{{O}_{4}}$, let us now move on to the particulars of this question.
Now, we know that Oxalic Acid’s chemical formula is $COOH-COOH$ and due to its presence in the initial titration, the medium would thus be acidic.
However, the later titration is done against alkaline ${{H}_{2}}{{O}_{2}}$ where the medium would be alkaline in nature.
With all this information now at our disposal along with the hint, let us now proceed to the step-by-step solution:
\[\begin{align}
& Milliequivalents\text{ }of\text{ }oxalic\text{ }acid~=50\times 0.1\times 2=10 \\
& \because Initial\text{ }medium\text{ }is\text{ }acidic~(\Rightarrow n=5~for~KMnO4). \\
& \therefore ~Molarity\text{ }of~KMn{{O}_{4}}=\dfrac{N}{5}=\dfrac{\dfrac{10}{100}}{5}=0.02M \\
& In\text{ }alkaline\text{ }medium,~\text{ }n=3~for~KMn{{O}_{4}}~ \\
& Hence,\text{ }normality~=0.02\times 3~=0.06N \\
\end{align}\]
Therefore, by our calculations, we can safely conclude that the required normality of $KMn{{O}_{4}}$ is 0.06N making the answer to this question c) 0.06N.
Note: Absolute caution must be taken during the titrations of any acidic or alkaline substance, regardless of its strength, so as to avoid any form of injury or damage. Also make sure that no mistakes are made in the n- factor calculation of Potassium Permanganate w.r.t change in mediums.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Hydrocarbons Class 11 Chemistry Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Chemistry Chapter 5 CBSE Notes - 2025-26

Equilibrium Class 11 Chemistry Chapter 6 CBSE Notes - 2025-26

Organic Chemistry Some Basic Principles And Techniques Class 11 Chemistry Chapter 8 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Chemistry Chapter 7 Redox Reactions (2025-26)

