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pH 150 mL larutan CH3COONH4 0,1 Mdengan Ka CH3COOH = 4 × 10-5 dan KbNH4OH = 10-5 adalah ....

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  • [H+] = √Kw . Ka / Kb
    [H+] = √10^-14 . 4 x 10^-5 / 10^-5,
    [H+] = √4 x 10^-14
    [H+] = 2 x 10^-7

    pH = -log[H+]
    pH = -log 2 x 10^-7
    pH = 7 - log2

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