How many moles of KOH are present in 2 L, if pH of solution is 12. 

Answer

0.02

Explanation

0.02
`pH = 12`
`H^(+) = 10^(-12) M`
`[OH^(-)] = 10^(-2)M`
`n_(KOH) = n_(OH^(-)) = 10^(-2) xx 2 = 0.02`