What is the pH at the equivalence point when titrating a 35.0 mL sample of 0.450 M NH3 with 0.450 M HCl, given that the Kb of NH3 is 1.8 x 10^-5?

To find the pH at the equivalence point for the titration of NH3 with HCl, we need to consider what happens at that point. At the equivalence point, all the NH3 has been converted into its conjugate acid, NH4⁺, since HCl is a strong acid and fully ionizes in solution.

First, let’s calculate the concentration of NH4⁺ formed at the equivalence point. Since both solutions are 0.450 M and we have equal volumes of titrant and titrand (35.0 mL), the concentration of NH4⁺ will also be 0.450 M at the equivalence point.

Next, we can find the pKa of the conjugate acid NH4⁺ using the relationship between Ka and Kb:

1. Calculate K_a:
K_a = K_w/K_b, where K_w = 1.0 x 10^-14 at 25°C.

K_a = (1.0 x 10^-14) / (1.8 x 10^-5) ≈ 5.56 x 10^-10

2. Calculate pK_a:
pK_a = -log(5.56 x 10^-10) ≈ 9.25

3. Now we can calculate the concentration of H⁺ ions from the dissociation of NH4⁺:
NH4⁺ ⇌ NH3 + H⁺

Using the formula: K_a = [NH3][H⁺]/[NH4⁺], we let [H⁺] = x, so:

K_a = (x)(x)/(0.450 – x) ≈ (x²/0.450) (since x is small compared to 0.450).

Setting up the equation:

5.56 x 10^-10 = x²/0.450

Now, rearranging gives:

x² = 5.56 x 10^-10 * 0.450

x² ≈ 2.502 x 10^-10

x ≈ √(2.502 x 10^-10) ≈ 1.58 x 10^-5 M

4. Finally, calculate pH:
pH = -log([H⁺])

pH ≈ -log(1.58 x 10^-5) ≈ 4.80

Therefore, the pH at the equivalence point is approximately 4.80.