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Pertanyaan

1.berapa ML larutan H2SO4 0.1 M diperlukan untuk menetralkan 200 ML larutan NaOH yg PHnya 12 + log 5 ??
2.hitunglah [h+] dan [OH-} yang terdapat dalam NH4OH 0,4 m (KB = 1 X 10 -5)

1 Jawaban

  • No.1
    pH NaOH = 12 + log 5
    pOH = 2 - log 5; [OH-] = 5 x 10^-2 M
    M NaOH = [OH-] = 5 x 10^-2
    Masam . Vasam . jml H = Mbasa . Vbasa . jml OH
    (0,1) Vasam (2) = (0,05) (200) (1)
    Vasam = 50 mL

    No.2
    [OH-] = √(Ka x [NH4OH])
    = √(10^-5 x 0,4)
    [OH-] = 2 x 10^-3 M
    [H+] = (10^-14/2x10^-3) = 5 x 10^-12 M

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