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![pH=-log[H_3O^+]\\ \\ {[H_3O^+]}=10^{-pH}\\ \\ {[H_3O^+]}=10^{-8.3}=0.00000000501](/tpl/images/0225/2000/ffe17.png)
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What is the classification of a solution of naoh with a ph of 8.3?...
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