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Pertanyaan

hitunglah hasil kali kelarutan suatu senyawa M(OH)2 mempunyai pH=10

1 Jawaban

  • pH = 14-10 = 2
    pOH = -log 10^-2
    [OH-] = 10^-2

    [OH-] = b × Mb
    10^-2 = 2 × Mb
    Mb = 5×10^-3

    s (kelarutan) = Mb
    s = 5×10^-3

    M(OH)2 -> M2+ + 2OH-
    s s 2s
    Ksp = [M2+] [2OH-]
    Ksp = (s) (2s)²
    Ksp = 4s³
    Ksp = 4 (5×10^-3)³
    Ksp = 4 (125×10^-9)
    Ksp = 500×10^-9
    Ksp = 5×10^-7

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