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Chemistry 14 Online
OpenStudy (anonymous):

Pleaseeeee help! fan and medal In a solution with a pH of 4, the [OH-] is: 1 x 10^-10 4 1 x 10^-4 10 1 x 10^-8

OpenStudy (aaronq):

use: pH+pOH=14 to find pOH, then \(pOH=\log[OH^-]\)

OpenStudy (aaronq):

sorry it's \(pOH=-log[OH^-]\)

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