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Pertanyaan

pH HF 0.1 M (Ka=2x10^-4)

1 Jawaban

  • [H^+] =√Ka.Ma
    [H^+] = √2.10^-4*10^-1
    [H^+] = √2.10^-5 = √20.10^-6
    [H^+] = 4,47.10^-3

    pH = -log[H^+]
    pH = -log[4,47.10^-3]
    pH = 3-log4,47

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