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Pertanyaan

tentukan ph 10ml larutan nh4oh 0.001M kb=1,8x10*-5

1 Jawaban

  • OH- = akar Kb Mb
    = akar 18.10^-6 x 10^-3
    = akar 1,8.10^-8
    = 1,34.10^-4 M

    pOH = 4 - log 1,34
    pH = 10 + log 1,34

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