pH=−log[H3O+]=7.3log[H3O+]=−7.3[H3O+]=10−7.3or[H3O+]=antilog of −7.3[H3O+]=5×10−8MpH=−log[H3O+]=7.3 log[H3O+]=−7.3 [H3O+]=10−7.3or[H3O+]=antilog\;of\;−7.3 [H3O+]=5×10−8MpH=−log[H3O+]=7.3log[H3O+]=−7.3[H3O+]=10−7.3or[H3O+]=antilogof−7.3[H3O+]=5×10−8M
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