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