# How do you find the pH, the pOH, [H3O+], and [OH-] in equations?

May 16, 2016

$2 {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$

${K}_{a} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{- 14}$

#### Explanation:

${K}_{a} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{- 14}$

This ion product, this equilibrium, has been established by experiment at $298$ $K$. At higher temperatures, given that this is a bond-breaking reaction, how would you expect this equilibrium to evolve?

Now, given the equation, we can manipulate it, as long as we do it to both sides of the equation. So we take ${\log}_{10}$ of BOTH sides to give:

${\log}_{10} {K}_{a}$ $= {\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left[H {O}^{-}\right]$. And on rearrangement:

$- {\log}_{10} \left[{H}_{3} {O}^{+}\right] - {\log}_{10} \left[H {O}^{-}\right] = - {\log}_{10} {K}_{a}$

But by definitions, $- {\log}_{10} \left[{H}_{3} {O}^{+}\right] = p H$, $- {\log}_{10} \left[H {O}^{-}\right] = p O H$, and $- {\log}_{10} {10}^{-} 14 = 14$.

And thus $p H + p O H = 14.$

Now if you are at A level, you only have to know this last result. You do have to be able to use it to solve problems. There should be many examples of problems on these boards so get cracking.