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Pertanyaan

Jika pH larutan asam 2+ Log 5 berapa konsentrasi H+ dari larutan asam tersebut

1 Jawaban

  • pH = 2 + log5

    pH = -log[H+]
    2 + log5 = -log[H+]
    - 2 - log5 = log[H+]
    10^(-2 - log5) = [H+]
    [H+] = 1 / (10^(2 + log5))
    [H+] = 1 / (10^2 . 10^log5)
    [H+] = 1 / (10^2 . 5)
    [H+] = 1/ 500
    [H+] = 0,002 M

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