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Pertanyaan

natrium askobat 0.001 M degan pH=12. ditanya harga Ka asam askobart adalah?

1 Jawaban

  • pH = -log [H+]
    12 = -log [H+]
    [H+] = 10^-12

    [H+] = √Ka.Ma
    10^-12 = √Ka.10^-3
    (10^-12)^2 = Ka.10^-3
    10^-24 = Ka.10^-3
    Ka = 10^-24 / 10^-3
    Ka = 10^-21

    Maaf kalau salah

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