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Chemistry 17 Online
OpenStudy (anonymous):

Please help, I am confused! Which of the following equations can be used to calculate pOH? a) pOH=-logKw b)pOH=pKw+pH c)pOh=pKw-pH d)pOH=-log[H+] I am confused because I always thought that the equation we used for pOH was pOH=-log[OH-] But non of my answers give me that option! >.<

OpenStudy (frostbite):

Hehe I can see the problem. It sure is :) but there is a another answer that go along with your answer. How good are you with logarithms?

OpenStudy (anonymous):

waiitt! I GET IT! it's pOH=pKw-pH because pKw= pOh+pH!

OpenStudy (anonymous):

right? 0.o haha

OpenStudy (frostbite):

Exactly. :)

OpenStudy (anonymous):

pKa+pKb = pKw which is same thing as =14 pOH+pH = 14 Goob job!

OpenStudy (anonymous):

@Frostbite Oxidation half ?

OpenStudy (anonymous):

Thank you guys! I didn't really think things through! I appreciate your help(:

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