When a 24.7 mL sample of a 0.500 M aqueous hydrofluoric acid solution is titrated with a 0.477 M aqueous potassium hydroxide solution, what is the pH at the midpoint in the titration?

Answer 1

Let’s look at the reaction taking place....

HF + KOH ==> H₂O + KF

Since HF is a weak acid, this forms a buffer system with a weak acid and the salt of that acid, KF.

At the mid point, 1/2 of the HF is converted to KF and thus the [HF] = [KF].

Original moles HF = 0.0247 L x 0.500 mol/L = 0.01235 moles

How many mls of 0.477 M KOH is needed to react with HALF of this, i.e. 0.006175 moles?

(xL)(0.477 mol/L) = 0.006175 moles and x = 0.01295 L

Thus, total final volume at the midpoint = 0.0247 L + 0.0130 L = 0.0377 L

pH = pKa + log [salt]/[acid]

pKa HF = 3.1 (look it up in a table)

[salt] = 0.006175 moles/0.0377 L = 0.2638 M

[acid] = [salt] = 0.2638 M

pH = 3.1 + log 1

pH = 3.1