Calculate the [OH–] if the pH is 11.6.?
Answer:
[H]=10^-11.6
[H]=2.51e-12
[OH]=kw/[H]
[OH]=1e-14/2.51e-12
[OH]= .0040 M
OK, so think of how pH and pOH are related. pH+pOH=14, so 14-pH=pOH.
So your pOH is 2.4
Take the negative inverse log of 2.4 (10^-2.4)
OH concentration is 3.98x10^-3 (.00398), this is in moles per liter
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