When a 18.0 mL sample of a 0.308 M aqueous hydrofluoric acid solution is titrated with a 0.361 M aqueous sodium hydroxide solution, what is the pH after 23.0 mL of sodium hydroxide have been added?

pH = 12.8

Explanation:

HF + NaOH → F⁻ + Na⁺ + H₂O

1 mole of HF reacts with 1 mole of NaOH

Initial moles of HF and NaOH are:

HF = 0.018L × (0.308mol / L) = 5.544x10⁻³mol HF

NaOH = 0.023L × (0.361mol / L) = 8.303x10⁻³mol NaOH

That means moles of NaOH remains after reaction are:

8.303x10⁻³mol - 5.544x10⁻³mol = 2.759x10⁻³moles NaOH

Total volume is 18.0mL + 23.0mL = 41.0mL = 0.0410L

Molar concentration of NaOH is

2.759x10⁻³moles NaOH / 0.0410L = 0.0673M = [OH⁻]

pOH = - log [OH⁻] = 1.17

As pH = 14 - pOH

pH = 12.8

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