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Pertanyaan

hitung pH larutan 0,10 M asam asetat apabila diketahui harga Ka asam asetat 1,75×10^-5

1 Jawaban

  • Diketahui:
    M CH3COOH = 0,1 M
    Ka CH3COOH = 1,75x10^-5
    Ditanya : pH?
    Jawab:
    [H+] = √Ka x Ma
    [H+] = √1,75x10^-5 . 10^-1
    [H+] = √1,75x10^-6
    [H+] = 1,32x10^-3
    pH = -log[H+]
    pH = -log 1,32x10^-3
    pH = 3-log1,32

    Semoga membantu.

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