Question

1. Find the pH at each of the following points in the titration of 50 mL of 0.3 M HF with 0.3 M NaOH. The ka value is 6.6×10-4.

1. The initial pH

2. After adding 10 mL of 0.3 M NaOH

3. After adding 25 mL of 0.3 M NaOH

4. After adding 50 mL of 0.3 M NaOH

5. After adding 50 mL of 0.4 M NaOH

Answer 1

[AI-generated answer — Bot GPT 3.5]

To solve this problem, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

where [A-] is the concentration of the conjugate base, [HA] is the concentration of the acid, and pKa is the negative logarithm of the acid dissociation constant.

1. The initial pH:
Before any NaOH is added, the HF is just in water, so [HF] = 0.3 M
pH = -log(6.6×10^(-4)) + log([0]/0.3) = -log(6.6×10^(-4)) = 3.18

2. After adding 10 mL of 0.3 M NaOH:
Total volume = 50 mL + 10 mL = 60 mL
[HF] = 0.3 * (50/60) = 0.25 M
[NaF] = 0.3 * (10/60) = 0.05 M
pH = -log(6.6×10^(-4)) + log(0.05/0.25) = -log(6.6×10^(-4)) - 0.698 = 4.38

3. After adding 25 mL of 0.3 M NaOH:
Total volume = 50 mL + 25 mL = 75 mL
[HF] = 0.3 * (50/75) = 0.2 M
[NaF] = 0.3 * (25/75) = 0.1 M
pH = -log(6.6×10^(-4)) + log(0.1/0.2) = -log(6.6×10^(-4)) - 0.301 = 4.98

4. After adding 50 mL of 0.3 M NaOH:
Total volume = 50 mL + 50 mL = 100 mL
[HF] = 0.3 * (50/100) = 0.15 M
[NaF] = 0.3 * (50/100) = 0.15 M
pH = -log(6.6×10^(-4)) + log(0.15/0.15) = -log(6.6×10^(-4)) = 3.18

5. After adding 50 mL of 0.4 M NaOH:
Total volume = 50 mL + 50 mL = 100 mL
[HF] = 0.3 * (50/100) = 0.15 M
[NaF] = 0.4 * (50/100) = 0.2 M
pH = -log(6.6×10^(-4)) + log(0.2/0.15) = -log(6.6×10^(-4)) + 0.176 = 3.36

Therefore, the pH at each point in the titration is:
1. 3.18
2. 4.38
3. 4.98
4. 3.18
5. 3.36