Discussion
The pH of \(10^{-2}\) M KOH will be :
Answer: D
The pH is 12.
KOH is a strong base and dissociates completely, so \([OH^{-}] = 10^{-2}\) M.
\[ \begin{aligned} pOH &= -\log(10^{-2}) = 2 \\ pH &= 14 - pOH = 14 - 2 = 12 \end{aligned} \]
2 is the pOH, or the pH of \(10^{-2}\) M acid, which comes from forgetting that KOH is a base. \(-2\) and \(-12\) come from dropping the minus sign in the logarithm; a dilute solution like this cannot have a negative pH.
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