Normality Of KMnO₄ In Reaction With (NH₄)₃PO₄

Nov 7, 2025 by Admin 46 views

Hey guys! Let's dive into a fascinating chemistry problem today. We're going to explore how to determine the normality of a KMnO₄ solution when it reacts with (NH₄)₃PO₄ in an acidic medium. This is a classic redox reaction problem, and understanding the underlying principles is key to solving it. So, buckle up, and let's get started!

Understanding the Reaction

To really nail this, we need to break down the chemical reaction first. When ammonium phosphate ((NH₄)₃PO₄) reacts with potassium permanganate (KMnO₄) in an acidic environment, it undergoes a redox reaction. In this specific scenario, (NH₄)₃PO₄ is converted into nitrate ions (NO₃⁻) and phosphine gas (PH₃). KMnO₄, being a strong oxidizing agent, facilitates this conversion. The key here is recognizing that this is a redox reaction, meaning we've got oxidation and reduction happening simultaneously. Ammonium ions (NH₄⁺) are being oxidized, and permanganate ions (MnO₄⁻) are being reduced.

The balanced chemical equation, though not explicitly needed for this particular calculation, helps in visualizing the electron transfer. However, for normality calculations, focusing on the change in oxidation states is more direct. Oxidation involves the loss of electrons, and reduction involves the gain of electrons. Identifying these changes is crucial for determining the equivalence factor, which plays a vital role in normality calculations. This reaction isn't just some abstract chemical process; it's got real-world applications, like in wastewater treatment and chemical analysis. Understanding the electron transfer can give you insights into how these processes work on a fundamental level. Think about it – you're not just memorizing formulas; you're understanding how matter interacts at the electron level!

Calculating the n-factor

Now, let's get into the nitty-gritty of the calculation. The first thing we need to figure out is the n-factor for both (NH₄)₃PO₄ and KMnO₄. The n-factor, also known as the equivalence factor, represents the number of moles of electrons transferred per mole of the substance in a redox reaction. For (NH₄)₃PO₄, the nitrogen in the ammonium ion (NH₄⁺) goes from an oxidation state of -3 to +5 in nitrate (NO₃⁻), which is a change of 8 electrons per nitrogen atom. Since there are three ammonium ions in (NH₄)₃PO₄, the total change is 3 * 8 = 24 electrons. Additionally, the phosphorus in PO₄³⁻ goes from an oxidation state of +5 to -3 in PH₃, which is a change of 8 electrons. Therefore, the total change in electrons for one molecule of (NH₄)₃PO₄ is 24 + 8 = 32 electrons. So, the n-factor for (NH₄)₃PO₄ is 32. This might seem like a big number, but it reflects the significant change in oxidation states during the reaction. For KMnO₄, the manganese (Mn) goes from an oxidation state of +7 in MnO₄⁻ to +2 (we'll assume it's reduced to Mn²⁺ in acidic medium), a change of 5 electrons. So, the n-factor for KMnO₄ is 5.

Applying the Normality Equation

Alright, we've got our n-factors sorted. Now comes the exciting part – applying the normality equation! In a redox reaction, the key principle is that the number of equivalents of the oxidizing agent must equal the number of equivalents of the reducing agent. This is the cornerstone of our calculation. We can express this mathematically as:

N₁V₁ = N₂V₂

Where:

N₁ is the normality of (NH₄)₃PO₄
V₁ is the volume of (NH₄)₃PO₄ solution
N₂ is the normality of KMnO₄ solution (what we want to find)
V₂ is the volume of KMnO₄ solution

We're given that 50 ml of 0.2 M (NH₄)₃PO₄ solution reacts completely with 16 ml of KMnO₄ solution. But wait a minute! We have molarity for (NH₄)₃PO₄, not normality. No sweat! We know the relationship between normality (N) and molarity (M) is:

N = M * n-factor

So, the normality of the (NH₄)₃PO₄ solution is 0.2 M * 32 = 6.4 N. Now we have all the pieces of the puzzle. Let's plug the values into our equation:

4 N * 50 ml = N₂ * 16 ml

Solving for Normality

Time for some algebraic gymnastics! Let's isolate N₂ to find the normality of the KMnO₄ solution:

N₂ = (6.4 N * 50 ml) / 16 ml

N₂ = 320 / 16

N₂ = 20 N

Boom! We've got our answer. The normality of the KMnO₄ solution is 20 N. So, the correct answer is (d) 20. Isn't it satisfying when everything clicks into place? This problem highlights the importance of understanding the concepts of oxidation states, n-factors, and the relationship between normality and molarity. With these tools in your arsenal, you can tackle a wide range of redox reaction problems!

Key Concepts and Takeaways

Before we wrap up, let's quickly recap the key concepts we've covered. Understanding these concepts will make similar problems a breeze!

Redox Reactions: These reactions involve the transfer of electrons between species. Oxidation is the loss of electrons, and reduction is the gain of electrons.
Oxidation States: Knowing how to determine oxidation states is crucial for identifying the n-factor.
n-factor (Equivalence Factor): This represents the number of moles of electrons transferred per mole of substance.
Normality: Normality is the molarity multiplied by the n-factor (N = M * n-factor).
Equivalence Principle: In a redox reaction, the number of equivalents of the oxidizing agent equals the number of equivalents of the reducing agent (N₁V₁ = N₂V₂).

By mastering these concepts, you'll be well-equipped to handle a variety of stoichiometry and redox reaction problems. Remember, chemistry is all about understanding the underlying principles, not just memorizing formulas. Keep practicing, and you'll become a chemistry whiz in no time!

Practice Problems

Want to test your understanding? Here are a couple of practice problems similar to the one we just solved:

If 25 ml of 0.1 M K₂Cr₂O₇ solution reacts completely with 20 ml of a ferrous sulfate solution in acidic medium, what is the normality of the ferrous sulfate solution?
What is the molarity of a KMnO₄ solution if 30 ml of it is required to react completely with 40 ml of 0.15 N oxalic acid solution in acidic medium?

Try solving these problems, and feel free to share your solutions in the comments below! Let's learn and grow together. Keep up the great work, guys! Chemistry can be challenging, but with a solid understanding of the concepts and a bit of practice, you can conquer any problem. Remember to always break down complex problems into smaller, more manageable steps. And most importantly, have fun exploring the amazing world of chemistry!

This problem illustrates a very common type of redox titration calculation. Redox titrations are used extensively in analytical chemistry to determine the concentration of unknown solutions. By carefully controlling the reaction conditions and using appropriate indicators, we can precisely measure the amount of oxidant or reductant in a sample. Mastering these types of calculations is a valuable skill for anyone studying chemistry or related fields.

I hope this explanation has been helpful! If you have any questions or want to discuss other chemistry topics, feel free to ask. Happy studying!