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The pH of a 0.25 M aqueous solution of hydrofluoric acid, HF, at 25.0 °C is 2.03. What is the value of Ka for HF?

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Anonymous

  1. pH = 2.03 = -log(H^+). That will be about 0.01 but you need a more exact answer.
    .................HF --> H^+ + F^-
    I................0.01.....0.........0
    C................-x........x..........x
    E..............0.01-x....x..........x

    Ka = (H^+)(F^-)/(HF). You know x, evaluate 0.01-x, plug into Ka expression and solve for Ka.
    Post your work if you get stuck.

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    DrBob222

  2. Can you solve it fully and explain it

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    Kulwinder

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