To determine the pH of the reaction mixture during the titration of 100 mL of 0.1 M aqueous ammonia (NH₃) with 0.1 M hydrochloric acid (HCl) at [tex]\(25^\circ C\)[/tex], we need to analyze the situation at two specific points:
(i) When 20 mL of HCl was added:
1. Initial moles of NH₃:
- Volume of NH₃ solution = 100 mL = 0.100 L
- Molarity of NH₃ = 0.1 M
- Moles of NH₃ = Volume (L) × Molarity (M) = 0.100 L × 0.1 M = 0.010 moles
2. Moles of HCl added:
- Volume of HCl added = 20 mL = 0.020 L
- Molarity of HCl = 0.1 M
- Moles of HCl = Volume (L) × Molarity (M) = 0.020 L × 0.1 M = 0.002 moles
3. Reaction between NH₃ and HCl:
- NH₃ + HCl → NH₄⁺ + Cl⁻
- Initial moles of NH₃ = 0.010 moles
- Initial moles of HCl = 0.002 moles
- Moles of NH₃ after reaction = 0.010 moles - 0.002 moles = 0.008 moles
- Moles of NH₄⁺ formed = 0.002 moles
4. Total volume of the solution:
- Volume of NH₃ solution + Volume of HCl solution = 100 mL + 20 mL = 120 mL = 0.120 L
5. Concentration of NH₃ and NH₄⁺ in the mixture:
- Concentration of NH₃ = Moles / Volume = 0.008 moles / 0.120 L = 0.0667 M
- Concentration of NH₄⁺ = Moles / Volume = 0.002 moles / 0.120 L = 0.0167 M
6. Use the Henderson-Hasselbalch equation for the buffer solution:
- [tex]\( \text{pH} = \text{pK}_b + \log \left(\frac{[\text{NH}_4^+]}{[\text{NH}_3]}\right) \)[/tex]
- pKₐ of NH₄⁺ = [tex]\( - \log K_b = - \log (1.85 \times 10^{-5}) \approx 4.73 \)[/tex]
- Converting this to pKₐ, since [tex]\(K_w = K_a K_b\)[/tex]:
[tex]\( K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.85 \times 10^{-5}} \approx 5.41 \times 10^{-10} \)[/tex]
- [tex]\( \text{pK}_a = - \log (5.41 \times 10^{-10}) \approx 9.27 \)[/tex]
7. Plugging in values:
- [tex]\( \text{pOH} = \text{pK}_b + \log \left(\frac{[\text{NH}_4^+]}{[\text{NH}_3]}\right) = 4.73 + \log \left(\frac{0.0167}{0.0667}\right) \)[/tex]
- [tex]\( \text{pOH} = 4.73 + \log (0.25) \)[/tex]
- [tex]\( \text{pOH} = 4.73 + (-0.60) = 4.13 \)[/tex]
- [tex]\( \text{pH} = 14 - \text{pOH} = 14 - 4.13 = 9.87 \)[/tex]
So, the pH of the reaction mixture when 20 mL of HCl is added is approximately 9.87.
(ii) At the equivalence point:
1. Moles of HCl needed to reach equivalence point:
- At equivalence point, moles of HCl added = moles of NH₃ initially present
- Moles of HCl = 0.010 moles (same as initial moles of NH₃)
2. Volume of HCl needed:
- Volume of HCl = Moles / Molarity = 0.010 moles / 0.1 M = 0.100 L = 100 mL
3. Total volume of the solution at equivalence point:
- Volume of NH₃ solution + Volume of HCl solution = 100 mL + 100 mL = 200 mL = 0.200 L
4. Concentration of NH₄⁺ at equivalence point:
- At equivalence point, all NH₃ is converted to NH₄⁺
- Moles of NH₄⁺ = Initial moles of NH₃ = 0.010 moles
- Concentration of NH₄⁺ = Moles / Volume = 0.010 moles / 0.200 L = 0.050 M
5. NH₄⁺ hydrolysis in water:
- NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺
6. Ka of NH₄⁺:
- [tex]\( K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.85 \times 10^{-5}} = 5.41 \times 10^{-10} \)[/tex]
7. Calculate [H₃O⁺]:
- For NH₄⁺ hydrolysis, we use [tex]\( K_a = \frac{[H_3O^+][NH_3]}{[NH_4^+]} \)[/tex]
- Assuming [H₃O⁺] = x, and hydrolysis results in equal concentrations of H₃O⁺ and NH₃:
[tex]\( K_a = x^2 / [NH₄^+] \)[/tex]
[tex]\( 5.41 \times 10^{-10} = \frac{x^2}{0.050} \)[/tex]
[tex]\( x^2 = (5.41 \times 10^{-10}) \times 0.050 \)[/tex]
[tex]\( x^2 = 2.705 \times 10^{-11} \)[/tex]
[tex]\( x = \sqrt{2.705 \times 10^{-11}} \approx 5.2 \times 10^{-6} \)[/tex]
8. Finding pH:
- [tex]\([H_3O^+] = 5.2 \times 10^{-6} \)[/tex]
- [tex]\( \text{pH} = - \log (5.2 \times 10^{-6}) \approx 5.28 \)[/tex]
So, the pH of the solution at the equivalence point is approximately 5.28.
Given the complexity of steps and calculations, it is crucial to follow through methodically to avoid mistakes and ensure accuracy.
