Equivalent Weight Of K₂Cr₂O₇ In Acidic Medium

by Wholesomestory Johnson 46 views

Hello there! I understand you're curious about the equivalent weight of potassium dichromate (K₂Cr₂O₇) in an acidic medium. Don't worry; I'll break it down for you step-by-step, ensuring you grasp the concept clearly and confidently. We'll explore what equivalent weight means, why it's important, and how to calculate it specifically for potassium dichromate in the presence of an acid.

Correct Answer

The equivalent weight of potassium dichromate (K₂Cr₂O₇) in an acidic medium is 49 g/mol.

Detailed Explanation

Let's dive deep into this topic. To understand the equivalent weight, we need to first understand the concept of a redox reaction, the role of potassium dichromate, and the calculation involved.

Key Concepts

Before we begin, let's define some critical terms:

  • Potassium Dichromate (K₂Cr₂O₇): A bright orange crystalline salt. It's a powerful oxidizing agent commonly used in various chemical reactions.
  • Acidic Medium: A solution with a high concentration of hydrogen ions (H⁺), i.e., a low pH. Common acids used include hydrochloric acid (HCl) and sulfuric acid (H₂SO₄).
  • Oxidation: The process where a substance loses electrons.
  • Reduction: The process where a substance gains electrons.
  • Redox Reaction: A chemical reaction involving both oxidation and reduction.
  • Equivalent Weight: The mass of a substance that combines with or replaces 1 gram of hydrogen or 8 grams of oxygen, or 35.5 grams of chlorine. In redox reactions, it's the mass of a substance that gains or loses one mole of electrons.
  • Molar Mass: The mass of one mole of a substance, typically expressed in grams per mole (g/mol).

The Role of Potassium Dichromate in Redox Reactions

Potassium dichromate is a strong oxidizing agent. This means it readily accepts electrons from other substances, causing them to be oxidized. In an acidic medium, the dichromate ion (Cr₂O₇²⁻) is reduced to chromium(III) ions (Cr³⁺). This redox reaction is the core of understanding the equivalent weight.

Step-by-Step Calculation

Let's break down how to calculate the equivalent weight:

  1. Write the Balanced Redox Reaction:

    In an acidic medium, the half-reaction for the reduction of dichromate is:

    Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

    This equation tells us that one dichromate ion (Cr₂O₇²⁻) gains six electrons (6e⁻) in the process.

  2. Determine the Molar Mass of Potassium Dichromate:

    The molar mass is calculated using the atomic masses of each element:

    • Potassium (K): 2 × 39.1 g/mol = 78.2 g/mol
    • Chromium (Cr): 2 × 52.0 g/mol = 104.0 g/mol
    • Oxygen (O): 7 × 16.0 g/mol = 112.0 g/mol

    Total molar mass = 78.2 g/mol + 104.0 g/mol + 112.0 g/mol = 294.2 g/mol

  3. Determine the n-factor:

    The n-factor represents the number of moles of electrons gained or lost per mole of the substance. In this case, from the balanced equation, we see that 6 moles of electrons are gained when 1 mole of dichromate is reduced. Therefore, the n-factor is 6.

  4. Calculate the Equivalent Weight:

    The formula for equivalent weight is:

    Equivalent Weight = (Molar Mass) / (n-factor)

    So, for potassium dichromate:

    Equivalent Weight = 294.2 g/mol / 6 = 49.03 g/mol

    Therefore, the equivalent weight of potassium dichromate in an acidic medium is approximately 49 g/mol.

Why Is the Acidic Medium Important?

The acidic medium provides the necessary hydrogen ions (H⁺) for the reaction to occur. It also affects the products formed. In a neutral or basic medium, the reaction and the final products are different, which would lead to a different equivalent weight calculation.

Real-World Examples and Applications

Potassium dichromate is used in many applications. Let's look at a few:

  1. Titration: It is used in redox titrations to determine the concentration of unknown solutions because of its strong oxidizing properties. The reaction's color change (from orange to green) helps to indicate the endpoint.
  2. Leather Tanning: It is used to tan leather by cross-linking the collagen fibers.
  3. Photography: It is a component in some photographic processes, such as dichromated gelatin.
  4. Laboratory Reagent: It is a common oxidizing agent in laboratories for various chemical reactions.

Practical Tips and Common Mistakes

  1. Balancing the Equation: Ensuring the redox reaction is correctly balanced is crucial. Always check that both mass and charge are balanced on both sides of the equation.

  2. Identifying the n-factor: Accurately determine the number of electrons transferred. Look at the change in oxidation state of the element undergoing reduction or oxidation.

  3. Units: Always include units (g/mol) when expressing equivalent weight to avoid confusion.

  4. Common Mistakes to Avoid:

    • Using the wrong equation: Make sure you are using the correct balanced chemical equation for the reaction occurring in the specified medium (acidic, basic, or neutral).
    • Incorrect n-factor: Be very careful when determining the n-factor. This is the most common source of error.
    • Forgetting the medium: Remember that the acidic environment dictates the reaction products and the equivalent weight. Without the acid, the equivalent weight will be different.

Further Exploration and Practice

To solidify your understanding, let's try some additional practice questions:

  1. Calculate the equivalent weight of potassium permanganate (KMnO₄) in an acidic medium.
  2. What would be the equivalent weight of potassium dichromate if the reaction occurred in a basic medium?
  3. Explain how the n-factor changes depending on the medium of the reaction.

Key Takeaways

  • The equivalent weight of potassium dichromate in an acidic medium is 49 g/mol.
  • The equivalent weight is calculated using the formula: Equivalent Weight = (Molar Mass) / (n-factor).
  • The n-factor is the number of electrons gained or lost in the balanced redox reaction.
  • The acidic medium provides hydrogen ions (H⁺), which are essential for the reaction.
  • Potassium dichromate is a strong oxidizing agent used in various applications, including titration, leather tanning, and photography.

I hope this detailed explanation helps you. If you have any further questions, feel free to ask! Understanding equivalent weight is crucial in many areas of chemistry, so take your time, practice, and you'll master it!