Titrations and Buffers

Question

What is the pH of a 1.0 L buffer made with 0.300 mol of HF (Ka = 6.8 × 10⁻⁴) and 0.200 mol of NaF to which 0.100 mol of HCl were added?

Answer 1

For these types of problems, we can use the Henderson Hasselbalch equation:

pH = pKa + log [conj. base] / [weak acid]

The pKa is the negative log of the Ka. pKa = -log 6.8x10^-4 = 3.17

Now, we have to find the concentration of the conjugate base [F-] and the weak acid [HF].

Initial [HF] = 0.300 mol / L = 0.300 M

Initial [F-] = 0.200 mol / L = 0.200 M

After adding 0.100 mols HCl, we have the following reaction:

F^- + HCl ==> HF + Cl^-, so [F-] will decrease by 0.100 mols and [HF] will increase by 0.100 mols.

Final [HF] = 0.300 + 0.100 = 0.400 M

Final [F-] = 0.200 - 0.100 = 0.100 M

Use these values in the Henderson Hasselbalch equation and solve for pH:

pH = pKa + log [F-] / [HF]

pH = 3.17 + log (0.100 / 0.400)

pH = 3.17 + log 0.25 = 3.17 -0.60

pH = 2.57