suatu larutan memiliki harga pH=12. Tentukan lah [OH] larutan tersebut!
Jawaban:
[OH]= 10^(-2)
Penjelasan:
pH + pOH= 14
12 + pOH= 14
pOH= 2
[tex]poh = - log(oh) \\ 2 = - log(oh) \\ - 2 = log(oh) \\ {10}^{ - 2} = {10}^{ log(oh) } \\ oh = {10}^{ - 2} [/tex]
[OH]= 10^(-2)
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