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Oh no!
When the [OH
-
] is given, the other values are readily determined by the series of steps shown below.
[H
+
]=10
-14
/[OH
-
]
=>
10
-14
/5.89e-5=1.70e-10
pH=-log[H
+
]
=>
-log1.70e-10=9.77
pOH=-log[OH
-
]
=>
-log5.89e-5=4.23
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