Pertanyaan
pH 150 mL larutan CH3COONH4 0,1 Mdengan Ka CH3COOH = 4 × 10-5 dan KbNH4OH = 10-5 adalah ....
Ditanyakan oleh: USER9452
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92 Jawaban
Jawaban (92)
[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
[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