Calculate the pH of a 0.10M ammonia solution. Calculate the pH after 50.0 mL of this solution is treated with 25.0 mL of 0.10M HCl. The dissociation constant of ammonia, Kb = 1.77 × 10 ?

To calculate the pH of a 0.10 M ammonia solution and then determine the pH after mixing it with hydrochloric acid (HCl), we need to follow a few steps involving equilibrium calculations and the concept of neutralization. Let's break this down step by step.

Step 1: Calculate the pH of the Ammonia Solution

Ammonia (NH₃) is a weak base that partially dissociates in water according to the following equilibrium reaction:

Nitrogen + Water ⇌ Ammonium Ion + Hydroxide Ion

In terms of the equilibrium expression, we have:

Kb = [NH₄⁺][OH⁻] / [NH₃]

Given that Kb for ammonia is 1.77 × 10⁻⁵, we can set up the equation. For a 0.10 M ammonia solution, let x be the concentration of OH⁻ produced at equilibrium:

• [NH₃] = 0.10 - x
• [NH₄⁺] = x
• [OH⁻] = x

Substituting these into the Kb expression gives:

1.77 × 10⁻⁵ = (x)(x) / (0.10 - x)

Assuming x is small compared to 0.10, we can simplify this to:

1.77 × 10⁻⁵ = x² / 0.10

Now, solving for x:

x² = 1.77 × 10⁻⁵ × 0.10

x² = 1.77 × 10⁻⁶

x = √(1.77 × 10⁻⁶) ≈ 0.00133 M

This x value represents the concentration of OH⁻ ions. To find the pOH:

pOH = -log[OH⁻] = -log(0.00133) ≈ 2.87

Now, we can find the pH using the relationship:

pH + pOH = 14

pH = 14 - 2.87 ≈ 11.13

Step 2: Mixing Ammonia with HCl

Next, we need to determine the pH after mixing 50.0 mL of the 0.10 M ammonia solution with 25.0 mL of 0.10 M HCl. First, we calculate the moles of each component:

• Moles of NH₃ = 0.10 M × 0.050 L = 0.005 moles
• Moles of HCl = 0.10 M × 0.025 L = 0.0025 moles

When HCl is added to the ammonia solution, it reacts with ammonia to form ammonium ions:

NH₃ + HCl → NH₄⁺ + Cl⁻

Since we have 0.005 moles of NH₃ and 0.0025 moles of HCl, the reaction will consume all of the HCl:

• Remaining NH₃ = 0.005 - 0.0025 = 0.0025 moles
• Produced NH₄⁺ = 0.0025 moles

After the reaction, we have a total volume of:

Total Volume = 50.0 mL + 25.0 mL = 75.0 mL = 0.075 L

Now, we can find the concentrations of NH₃ and NH₄⁺:

• [NH₃] = 0.0025 moles / 0.075 L = 0.0333 M
• [NH₄⁺] = 0.0025 moles / 0.075 L = 0.0333 M

Step 3: Calculate the pH of the Resulting Solution

In this case, we have a buffer solution consisting of NH₃ and NH₄⁺. We can use the Henderson-Hasselbalch equation to find the pH:

pH = pKa + log([base]/[acid])

First, we need to find pKa. Since we know Kb for ammonia, we can find Ka for NH₄⁺ using the relationship:

Kw = Ka × Kb

Where Kw = 1.0 × 10⁻¹⁴. Thus:

Ka = Kw / Kb = 1.0 × 10⁻¹⁴ / 1.77 × 10⁻⁵ ≈ 5.64 × 10⁻¹⁰

Now, we can find pKa:

pKa = -log(5.64 × 10⁻¹⁰) ≈ 9.25

Now substituting into the Henderson-Hasselbalch equation:

pH = 9.25 + log(0.0333 / 0.0333) = 9.25 + log(1) = 9.25

Thus, the final pH of the solution after mixing is approximately 9.25.