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DerKrebs [107]
2 years ago
7

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 soluti

on, what is the pH after 23.0 mL of sodium hydroxide have been added?
Chemistry
1 answer:
sleet_krkn [62]2 years ago
4 0

Answer:

pH = 12.8

Explanation:

HF + NaOH → F⁻ + Na⁺ + H₂O

<em>1 mole of HF reacts with 1 mole of NaOH</em>

<em />

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 = <em>2.759x10⁻³moles NaOH</em>

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

<em>pH = 12.8</em>

<em></em>

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