determine the ph of 0.010 m koh solution. koh is a strong base.

Explanation:

Step1: Find $[OH^-]$ concentration

Since KOH is a strong base and dissociates completely as $KOH\rightarrow K^+ + OH^-$, for a 0.010 M KOH solution, $[OH^-]=0.010\ M = 1\times10^{- 2}\ M$.

Step2: Calculate pOH

The formula for pOH is $pOH=-\log[OH^-]$. Substituting $[OH^-]=1\times10^{-2}\ M$ into the formula, we get $pOH = -\log(1\times10^{-2})=2$.

Step3: Calculate pH

We know the relationship $pH + pOH=14$. So, $pH=14 - pOH$. Substituting $pOH = 2$ into the formula, we have $pH=14 - 2=12$.

Answer:

12