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Pertanyaan

berapakah pH larutan asam formiat 0,01 M. Ka= 1.96*0.0004

1 Jawaban

  • [H+]=√0.01×1.96×10^-4
    [H+]= 1.4 × 10^-3
    pH = -log[H+]
    = -log 1.4×10^-4
    = 4-log 1.4
    = 3.398

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