Calculating PH Of Ammonium Chloride Solution A Step By Step Guide

Hey guys, let's dive into a fascinating chemistry problem! We're going to figure out the pH of a solution made by dissolving ammonium chloride (NH₄Cl) in water. This is a classic acid-base equilibrium problem, and we'll break it down step by step so it's super clear.

Understanding the Problem

First, let's get our bearings. We have 21.4 grams of ammonium chloride (NH₄Cl) dissolved in 250 ml of water. We also know the molar mass (Mr) of NH₄Cl is 53.5 g/mol, and the base dissociation constant (Kb) for ammonia (NH₃) is 10⁻⁵. Our mission is to calculate the pH of the resulting solution.

Ammonium chloride (NH₄Cl) is a salt formed from a weak base (ammonia, NH₃) and a strong acid (hydrochloric acid, HCl). When this salt dissolves in water, it undergoes hydrolysis, meaning it reacts with water. The ammonium ion (NH₄⁺) acts as a weak acid, donating a proton (H⁺) to water and forming ammonia (NH₃) and hydronium ions (H₃O⁺). It's these hydronium ions that make the solution acidic, and we need to figure out their concentration to find the pH.So, to calculate the pH, we'll first need to determine the concentration of the ammonium chloride solution. Then, we'll set up an equilibrium reaction to see how much hydronium ion (H₃O⁺) is produced. Finally, we'll use the hydronium ion concentration to calculate the pH. We'll also use the Kb value to find the Ka (acid dissociation constant) for the ammonium ion, which will help us in our equilibrium calculations. Let's get started!

Step 1: Calculate the Molarity of NH₄Cl

Molarity (M) is defined as the number of moles of solute per liter of solution. So, first, we need to find out how many moles of NH₄Cl we have. We can do this using the formula:

Moles = Mass / Molar Mass

In our case:

Moles of NH₄Cl = 21.4 g / 53.5 g/mol ≈ 0.4 moles

Now, we need to convert the volume of the solution from milliliters to liters:

250 ml = 250 / 1000 = 0.25 liters

Finally, we can calculate the molarity:

Molarity (M) = Moles / Volume (in liters) Molarity (M) = 0.4 moles / 0.25 liters = 1.6 M

So, the concentration of our NH₄Cl solution is 1.6 M. This is a crucial piece of information because it tells us how much NH₄Cl we have to start with, which will then determine how much H₃O⁺ is formed.

Step 2: Determine the Ka of NH₄⁺

Since NH₄⁺ is the conjugate acid of a weak base (NH₃), we need to find its acid dissociation constant (Ka). We can use the relationship between Ka, Kb, and the ion product of water (Kw):

Kw = Ka * Kb

We know Kw is 1.0 x 10⁻¹⁴ at 25°C, and Kb for NH₃ is given as 10⁻⁵. So, we can rearrange the equation to solve for Ka:

Ka = Kw / Kb Ka = (1.0 x 10⁻¹⁴) / (10⁻⁵) = 1.0 x 10⁻⁹

This Ka value is essential for setting up our equilibrium expression. It tells us how much the ammonium ion will dissociate in water – a smaller Ka means weaker acid, so it will dissociate less. This value will directly impact the concentration of H₃O⁺ ions in the solution, and thus the pH.

Step 3: Set Up the Equilibrium Reaction

When NH₄Cl dissolves in water, the NH₄⁺ ion reacts with water according to the following equilibrium:

NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq)

To determine the equilibrium concentrations, we'll use an ICE table (Initial, Change, Equilibrium). This is a handy way to keep track of the concentrations of each species in the reaction:

Here's what each row represents:

• Initial: The initial concentrations before the reaction occurs. We start with 1.6 M of NH₄⁺ (calculated in step 1) and 0 M of both NH₃ and H₃O⁺. Water is a liquid and its concentration doesn't change significantly, so we ignore it in the equilibrium expression.
• Change: The change in concentration as the reaction reaches equilibrium. We use 'x' to represent the change. Since NH₄⁺ reacts with water, its concentration decreases by 'x', while the concentrations of NH₃ and H₃O⁺ increase by 'x'.
• Equilibrium: The concentrations at equilibrium. These are the initial concentrations plus the changes.

Step 4: Write the Ka Expression and Solve for x

The acid dissociation constant (Ka) expression for this reaction is:

Ka = [NH₃][H₃O⁺] / [NH₄⁺]

Plugging in the equilibrium concentrations from the ICE table, we get:

1. 0 x 10⁻⁹ = (x)(x) / (1.6 - x)

Since Ka is very small, we can assume that 'x' is much smaller than 1.6, so we can simplify the equation:

1. 0 x 10⁻⁹ ≈ x² / 1.6

Now, we can solve for x:

x² ≈ (1.0 x 10⁻⁹) * 1.6 x² ≈ 1.6 x 10⁻⁹ x ≈ √(1.6 x 10⁻⁹) x ≈ 4.0 x 10⁻⁵ M

So, x represents the equilibrium concentration of H₃O⁺, which is 4.0 x 10⁻⁵ M. This is a crucial value because it directly tells us how acidic the solution is. The smaller the concentration of H₃O⁺, the less acidic the solution will be.

Step 5: Calculate the pH

Finally, we can calculate the pH using the formula:

pH = -log[H₃O⁺]

Plugging in the concentration of H₃O⁺ we found:

pH = -log(4.0 x 10⁻⁵) pH ≈ 4.4

Therefore, the pH of the solution is approximately 4.4. This indicates that the solution is acidic, which makes sense since we dissolved the salt of a weak base and a strong acid. The pH value tells us quantitatively how acidic the solution is. A pH of 7 is neutral, values below 7 are acidic, and values above 7 are basic.

Conclusion

So, there you have it! We successfully calculated the pH of the ammonium chloride solution. We started by finding the molarity, then determined the Ka value, set up an ICE table, solved for the hydronium ion concentration, and finally, calculated the pH. This problem demonstrates how acid-base equilibrium works and how we can use equilibrium constants to find the pH of solutions. Understanding these principles is super important for chemistry, so great job following along!

Key Takeaways:

• The pH of a solution containing the salt of a weak base and a strong acid will be acidic.
• Using an ICE table is a systematic way to solve equilibrium problems.
• Approximations can simplify calculations when equilibrium constants are very small.

Hope this breakdown was helpful, guys! Keep exploring the fascinating world of chemistry!