Jika pH larutan asam 2+ Log 5 berapa konsentrasi H+ dari larutan asam tersebut
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Pertanyaan
Jika pH larutan asam 2+ Log 5 berapa konsentrasi H+ dari larutan asam tersebut
1 Jawaban
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1. Jawaban ayynanu
pH = 2 + log5
pH = -log[H+]
2 + log5 = -log[H+]
- 2 - log5 = log[H+]
10^(-2 - log5) = [H+]
[H+] = 1 / (10^(2 + log5))
[H+] = 1 / (10^2 . 10^log5)
[H+] = 1 / (10^2 . 5)
[H+] = 1/ 500
[H+] = 0,002 M