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25 August, 22:33

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?

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  1. 25 August, 22:51
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    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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