# What is pOH and how does it relate to pH?

Oct 15, 2016

In aqueous solution under standard conditions $p H + p O H = 14$

#### Explanation:

But what is $p H$? Simply, $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, and $p O H = - {\log}_{10} \left[H {O}^{-}\right]$.

We know that water undergoes autoprotolysis, which we could represent in the following fashion:

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

As with any equilibrium, we can define and measure an equilibrium constant, i.e.

$K {'}_{w}$ $=$ $\frac{\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right]}{{\left[{H}_{2} O\right]}^{2}}$

But because ${\left[{H}_{2} O\right]}^{2}$ is so large, it does not change much effectively, and can be incorporated on the left hand side of the equation to give:

${K}_{w}$ $=$ $\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right]$ $=$ ${10}^{-} 14$ under standard conditions of $1 \cdot a t m$, and at $298 \cdot K$. ${K}_{w}$ is low because the reaction involes the breaking of a strong $O - H$ bonds, and thus the equilibrium lies strongly to the left. Now we can manipulate this expression, provide we do it to BOTH sides of the equation. One thing that we can do is to take ${\log}_{10}$ of BOTH SIDES.

And thus ${\log}_{10} {K}_{w}$ $=$ ${\log}_{10} \left[{H}_{3} {O}^{+}\right] + \log \left[H {O}^{-}\right]$.

And, mulitiplying both sides by $- 1$, we get:

$- {\log}_{10} {K}_{w} = - {\log}_{10} \left({10}^{-} 14\right) = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] - \log \left[H {O}^{-}\right]$

But $- {\log}_{10} \left({10}^{-} 14\right) = 14$ by definition, i.e. ${\log}_{a} \left({a}^{x}\right) = x$, the power to which you raise the base $a$ to get ${a}^{x}$ is clearly $x$.

And, by definition, $p O H = - \log \left[H {O}^{-}\right]$, and $p H = - \log \left[{H}_{3} {O}^{+}\right]$

So $p {K}_{w} = 14 = p H + p O H$.