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

1 answer

1. The initial pH can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])

First, calculate the initial concentration of HF and NaF after mixing:
- Initial moles HF: (0.3 mol/L) x 0.05 L = 0.015 mol
- Initial moles NaOH: (0 mol/L) x 0 L = 0 mol

Next, calculate the moles of NaF and HF after reacting with NaOH:
- Moles NaOH added: (0.3 mol/L) x 0 L = 0 mol
- Moles NaOH reacted: (0 mol + 0.3 mol/L x 0.01 L) = 0.003 mol
- Moles NaOH remaining: 0.015 mol - 0.003 mol = 0.012 mol
- Moles NaF: 0.003 mol

Now, calculate the concentrations of HF and NaF after reacting with NaOH:
- HF concentration: (0.015 mol - 0.003 mol) / (0.05 L + 0.01 L) = 0.24 mol/L
- NaF concentration: (0.003 mol) / (0.05 L + 0.01 L) = 0.048 mol/L

Lastly, calculate the pH using the Henderson-Hasselbalch equation:
pH = -log(6.6x10^-4) + log(0.048/0.24) = 3.07

2. After adding 10 mL of 0.3 M NaOH:
- Moles NaOH added: (0.3 mol/L) x 0.01 L = 0.003 mol
- Moles NaOH reacted: (0.003 mol) = 0.003 mol
- Moles NaOH remaining: 0.012 mol - 0.003 mol = 0.009 mol
- Moles NaF: 0.003 mol
- HF concentration: (0.015 mol - 0.003 mol) / (0.05 L + 0.01 L) = 0.24 mol/L
- NaF concentration: (0.003 mol) / (0.05 L + 0.01 L) = 0.048 mol/L
- pH = -log(6.6x10^-4) + log(0.048/0.24) = 3.07

3. After adding 25 mL of 0.3 M NaOH:
- Proceed with the same calculations as above to determine the pH

4. After adding 50 mL of 0.3 M NaOH:
- Proceed with the same calculations as above to determine the pH

5. After adding 50 mL of 0.4 M NaOH:
- Proceed with the same calculations as above, but utilize the new NaOH concentration to determine the pH.