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Pertanyaan

hitunglah ph larutan 100ml NH4CL 0,1 M, jika kb NH3 = 1 x 10^-5 !

1 Jawaban

  • pKb = -logKb
    pKb = -log10^-5
    pKb = 5

    pH = 1/2 (14 - pKb - log[G])
    pH = 1/2 (14 - 5 - log10^-1)
    pH = 1/2 (9 + 1)
    pH = 1/2 (10)
    pH = 10

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