[H+] and pH of 100 mM solution of KOH

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Homework Statement


What is [H+] and pH for a 100 mM solution of KOH


The Attempt at a Solution



This seems very simple..its only 2 marks but I don't know what I'm doing wrong.

I understand KOH will dissociate into K+ and OH-

pOH=-log(OH) ?
ph + pOH = 14
p[OH]=-log[100x10^-6?]
where and what am i doing wrong?

Thanks for the help!
 
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