Exercise 8 | Buffers and Acid-Base Titrations

The titration equation is:

Reaction: F^- + H₃O⁺ $\rightleftarrows$ HF + H₂O

After adding 0.20 mol of HCl:

n_F-,0 = 0.40 mol

n_H+,0 = 0.20 mol

H₃O⁺ is the limiting reagent.

n_F- = n_F-,0 - n_H+,0 = 0.20 mol

n_HF = n_HF,0 + n_H+,0 = 0.50 mol

Henderson-Hasselbalch equation can be used:

pH = pKa + log $\frac{{\left[\mathrm{base}\right]}_0}{{\left[\mathrm{acid}\right]}_0}$

pH = 3.14 + log 

pH = 2.74