# What is pH of a solution 1.2xx10^-2*mol*L^-1 with respect to KOH?

Jun 16, 2017

$p H = 12.08$

#### Explanation:

Well, first off we calculates $p O H$

And $p O H = {\log}_{10} \left[H {O}^{-}\right] = - {\log}_{10} 1.2 \times {10}^{-} 2 = 1.92$.

Now it is a given that $p O H + p H = 14$.

Why? Because ${K}_{w} = \left[H {O}^{-}\right] \left[{H}_{3} {O}^{+}\right] = {10}^{- 14}$, and when we take ${\log}_{10}$ of both sides, we come up with that relationship.

And so pH=14-pOH=14-11.92=....??