in a solution obtained bymixing 100ml of 0.25M triethylamine (k_b= 6.4 ×10^-5) and 400ml of 1/18 M NH₄OH (k=1.8 ×10^-5) : find ph of solutiion and

[NH₄⁺] and [C₆Ndiv6⁺]

To determine the pH of the solution formed by mixing triethylamine and ammonium hydroxide, we need to analyze the contributions of both components to the overall equilibrium. Let's break down the problem step by step.

Step 1: Calculate Moles of Each Component

First, we need to find the number of moles of triethylamine and ammonium hydroxide in the solution.

• Triethylamine: The concentration is 0.25 M and the volume is 100 mL (0.1 L).
• Moles of triethylamine = Concentration × Volume = 0.25 mol/L × 0.1 L = 0.025 moles.
• Ammonium Hydroxide: The concentration is 1/18 M and the volume is 400 mL (0.4 L).
• Moles of ammonium hydroxide = (1/18) mol/L × 0.4 L = 0.0222 moles.

Step 2: Determine the Total Volume of the Mixture

The total volume after mixing the two solutions is:

• Total Volume = 100 mL + 400 mL = 500 mL = 0.5 L.

Step 3: Calculate Concentrations After Mixing

Next, we calculate the concentrations of triethylamine and ammonium hydroxide in the final solution.

• [Triethylamine]:
  • Concentration = Moles/Total Volume = 0.025 moles / 0.5 L = 0.05 M.
• [Ammonium Hydroxide]:
  • Concentration = Moles/Total Volume = 0.0222 moles / 0.5 L = 0.0444 M.

Step 4: Establishing the Equilibrium

Triethylamine (a weak base) will react with ammonium hydroxide (which can act as a weak acid). The equilibrium can be represented as:

Triethylamine + NH4OH ⇌ C6NH16+ + OH-

We can use the Kb values to find the pH. The Kb for triethylamine is 6.4 × 10^-5, and for ammonium hydroxide, it is 1.8 × 10^-5.

Step 5: Finding the pH

To find the pH, we need to calculate the pOH first. We can use the Kb expression for triethylamine:

Kb = [C6NH16+][OH-] / [Triethylamine]

Assuming x is the change in concentration due to the reaction, we can set up the equation:

6.4 × 10^-5 = (x)(x) / (0.05 - x)

For simplicity, we can assume x is small compared to 0.05, so:

6.4 × 10^-5 ≈ (x²) / 0.05

Solving for x gives:

x² = 6.4 × 10^-5 × 0.05 = 3.2 × 10^-6

x = √(3.2 × 10^-6) ≈ 0.00179 M (this is [OH-]).

Now, we can find the pOH:

pOH = -log[OH-] = -log(0.00179) ≈ 2.75.

Finally, we can find the pH:

pH = 14 - pOH = 14 - 2.75 = 11.25.

Step 6: Concentrations of Ions

Now, let's find the concentrations of [NH4⁺] and [C6NH16⁺].

• [NH4⁺]: This is primarily from the ammonium hydroxide, which dissociates slightly:
• [NH4⁺] ≈ 0.0444 M (since it doesn't significantly change).
• [C6NH16⁺]: This is equal to x, which we found to be approximately 0.00179 M.

Summary of Results

In summary, after mixing the solutions, we find:

• pH of the solution: 11.25
• [NH4⁺]: 0.0444 M
• [C6NH16⁺]: 0.00179 M

This analysis shows how the interaction of weak acids and bases can influence the overall pH of a solution. If you have any further questions or need clarification on any part, feel free to ask!