What is the OH concentration of a pH of 12.33

[From: ] [author: ] [Date: 11-12-15] [Hit: ]

00-12.33 = 1.[OH-] = 10^-1.[OH-] = 0.......

1.67 was not an option (mult choice) so I did the inverse log of -12.33.

Is that wrong?

If pH = 12.33, then pOH = 14.00-12.33 = 1.67
Then [OH-] = 10^-pOH
[OH-] = 10^-1.67
[OH-] = 0.021M

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