Ammonia Water Reaction Calculating Percentage Of Reacted Molecules

by ADMIN 67 views

Hey guys! Let's dive into a fascinating topic in chemistry: the reaction between ammonia (NH₃) and water (H₂O). This reaction is super important in understanding the behavior of ammonia in aqueous solutions, and today, we're going to break down how to calculate the percentage of ammonia molecules that actually react in a 0.1 M (mol/dm³) solution at 298 K (which is about 25°C – room temperature!). We will assume that less than 5% of the ammonia molecules react. So, grab your thinking caps, and let's get started!

The Ammonia-Water Reaction: A Closer Look

At its core, the reaction between ammonia and water is a classic example of a Brønsted-Lowry acid-base reaction. In this scenario, ammonia acts as a base, accepting a proton (H⁺) from water, which acts as an acid. This results in the formation of the ammonium ion (NH₄⁺) and the hydroxide ion (OH⁻). Let's write out the balanced chemical equation to make it crystal clear:

NH₃ (aq) + H₂O (l) ⇌ NH₄⁺ (aq) + OH⁻ (aq)

This equation tells us a lot. First, it's an equilibrium reaction, denoted by the double arrow (⇌). This means the reaction can proceed in both directions: ammonia and water can react to form ammonium and hydroxide, and ammonium and hydroxide can react to reform ammonia and water. This dynamic equilibrium is crucial because it means that not all ammonia molecules will react completely with water. Some will remain as NH₃, while others will transform into NH₄⁺. The extent to which this reaction proceeds is governed by the equilibrium constant, which we'll touch on later.

Now, why is this reaction important? Well, it's fundamental to understanding the properties of aqueous ammonia solutions. The presence of hydroxide ions (OH⁻) makes the solution basic (alkaline), which has implications for various applications, from cleaning products to industrial processes. Understanding the equilibrium helps us predict and control the pH of the solution and the concentration of different species (NH₃, NH₄⁺, and OH⁻) present.

Calculating the Percentage of Ammonia Reacted

Now, let's get to the heart of the matter: how do we calculate the percentage of ammonia molecules that react in a 0.1 M solution? This is where the concept of the base ionization constant, Kb, comes into play. The Kb value is a measure of the strength of a base; the higher the Kb, the stronger the base and the more it will react with water. For ammonia at 298 K, the Kb value is approximately 1.8 x 10⁻⁵. This value is essential for our calculations.

To calculate the percentage of ammonia reacted, we'll follow these steps:

  1. Set up an ICE table: ICE stands for Initial, Change, and Equilibrium. This table helps us organize the concentrations of the reactants and products at different stages of the reaction.

    NH₃ H₂O NH₄⁺ OH⁻
    Initial (I) 0.1 Excess 0 0
    Change (C) -x - +x +x
    Equilibrium (E) 0.1 - x Excess x x
    • Initial: We start with 0.1 M of NH₃ and assume water is in excess (since it's the solvent). The initial concentrations of NH₄⁺ and OH⁻ are 0.
    • Change: Let 'x' represent the change in concentration as the reaction proceeds. Since ammonia reacts, its concentration decreases by 'x', while the concentrations of NH₄⁺ and OH⁻ increase by 'x'.
    • Equilibrium: The equilibrium concentrations are the initial concentrations plus the change.

  2. Write the Kb expression: The Kb expression is the equilibrium constant for the base ionization reaction. For the ammonia-water reaction, it is:

    Kb = [NH₄⁺][OH⁻] / [NH₃]

    Remember, we don't include water in the expression because it's a liquid and its concentration remains practically constant.

  3. Substitute equilibrium concentrations into the Kb expression: Using the equilibrium concentrations from the ICE table, we get:

    1. 8 x 10⁻⁵ = (x)(x) / (0.1 - x)

  4. Solve for x: This is where things get a little tricky, but we can make a simplifying assumption. Since the Kb value is small, we can assume that 'x' is much smaller than 0.1, meaning that the change in ammonia concentration is negligible. This allows us to simplify the denominator:

    1. 8 x 10⁻⁵ ≈ x² / 0.1

    Now, we can easily solve for x:

    x² ≈ 1.8 x 10⁻⁶ x ≈ √(1.8 x 10⁻⁶) ≈ 0.00134 M

    So, x represents the equilibrium concentration of both NH₄⁺ and OH⁻, as well as the amount of ammonia that reacted.

  5. Calculate the percentage of ammonia reacted: To find the percentage, we divide the concentration of ammonia reacted (x) by the initial concentration of ammonia (0.1 M) and multiply by 100%:

    Percentage reacted = (x / 0.1) * 100% Percentage reacted = (0.00134 / 0.1) * 100% Percentage reacted ≈ 1.34%

    Therefore, approximately 1.34% of the ammonia molecules react with water in a 0.1 M solution at 298 K. This confirms our initial assumption that less than 5% of the ammonia molecules react, justifying our simplification in step 4.

Why is this Percentage Important?

You might be wondering,