"[KOH]=[OH-]=0.00028 mol\/L"
"[H3O+][OH-]=Kw"
"[H3O+]=Kw\/[OH-]=1* 10^{-14}\/0.00028=3.57*10^{-11}"
"pH=-log[H3O+]=-log(3.57*10^{-11})=-10.45"
It is basic solution
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