Answer :
The pH values at different titrant volumes for 30.0 mL of 0.050 M NH₃ titrated with 0.025 M HCl are calculated and provided, starting with pH 10.98 for 0 mL and decreasing to pH 2.49 for 73.4 mL.
Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added: 0 mL, 20 mL, 59.1 mL, 60.0 mL, 71.4 mL, 73.4 mL.
Let's solve this step-by-step:
0 mL of HCl added: NH₃ is a weak base. Using the formula for the base dissociation constant (Kb) and its initially given concentration (0.050 M), calculate the initial pH.
NH₃ + H₂O ⇌ NH4+ + OH⁻
Kb = 1.8 × 10⁻⁵
[OH-] ≈ √(Kb × [NH₃]) = √(1.8 × 10⁻⁵ × 0.050) ≈ 9.5 × 10⁻⁴
pOH = -log(9.5 × 10⁻⁴) ≈ 3.02
pH = 14 - pOH = 14 - 3.02 = 10.98
20 mL of HCl added:
Moles NH₃ initially = 0.030 L × 0.050 M = 0.0015 mol
Moles HCl added = 0.020 L × 0.025 M = 0.0005 mol
After reaction: Moles NH₃ left = 0.0015 mol - 0.0005 mol = 0.001 mol
Moles NH⁴⁺ formed = 0.0005 mol
Use the Henderson-Hasselbalch equation:
pH = pKa + log([base]/[acid])
pKa = 14 - pKb = 14 - 4.74 = 9.26
pH = 9.26 + log(0.001/0.0005) = 9.26 + 0.301 = 9.56
59.1 mL of HCl added:
Moles HCl added = 0.0591 L × 0.025 M = 0.0014775 mol
Moles NH₃ left = 0.0015 mol - 0.0014775 mol = 0.0000225 mol
Moles NH⁴⁺ formed = 0.0014775 mol
pH = 9.26 + log(0.0000225/0.0014775) ≈ 6.06 (using the ratio in the HH equation)
60.0 mL of HCl added:
Moles HCl added = 0.060 L × 0.025 M = 0.0015 mol
This is the equivalence point where all NH₃ has been converted to NH⁴⁺.
pH at equivalence point is determined by NH⁴⁺ hydrolysis:
NH⁴⁺ + H₂O ⇌ NH₃ + H₃O⁺
Ka = Kw/Kb = 1 × 10⁻¹⁴ / 1.8 × 10⁻⁵ ≈ 5.56 × 10⁻¹⁰
pH ≈ 4.75 (since pKa of NH⁴⁺ is 9.26)
71.4 mL of HCl added:
Moles HCl added = 0.0714 L × 0.025 M = 0.001785 mol
Excess HCl = 0.001785 mol - 0.0015 mol = 0.000285 mol
Concentration of H⁺ in the total volume ( 30 mL + 71.4 mL = 101.4 mL) = 0.000285 mol / 0.1014 L ≈ 0.00281 M
pH = -log(0.00281) ≈ 2.55
73.4 mL of HCl added:
Moles HCl added = 0.0734 L × 0.025 M = 0.001835 mol
Excess HCl = 0.001835 mol - 0.0015 mol = 0.000335 mol
Concentration of H⁺ in the total volume (30 mL + 73.4 mL = 103.4 mL) = 0.000335 mol / 0.1034 L ≈ 0.00324 M
pH = -log(0.00324) ≈ 2.49
