pH HF 0.1 M (Ka=2x10^-4)
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Pertanyaan
pH HF 0.1 M (Ka=2x10^-4)
1 Jawaban
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1. Jawaban rizkysmaga
[H^+] =√Ka.Ma
[H^+] = √2.10^-4*10^-1
[H^+] = √2.10^-5 = √20.10^-6
[H^+] = 4,47.10^-3
pH = -log[H^+]
pH = -log[4,47.10^-3]
pH = 3-log4,47