If 3.3 Is -log[H+]
Then -log[OH-] =14 -3.3 = 10.7
[OH-] =10E-10.7
2. If 3.3 Is n(H+)
Then
C(H+) = 3.3 mol/dm3
-log[OH-]= 14 + log(C(H+))
-log[OH-]= 14.519
[OH-]= 3.0303E-15
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