# How do you get pH and pOH given pka or pkb?

Oct 11, 2016

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

#### Explanation:

In water, under standard conditions, we know that the following equilibrium occurs:

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

As for any equilibrium, we can measure the equilibrium constant, ${K}_{w}$, of this reaction:

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

Given that ${K}_{w} = {10}^{-} 14$, and taking $- {\log}_{10}$ of both sides, we get the defining relationship:

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

Thus $14 = p H + p O H$, and this is something that you have to commit to memory.

And thus when quoted $p {K}_{a}$ for weak acids etc. you have to solve the equilibrium expresssion:

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

There should be many model answers on these boards. Here is a start, and here is another attempt.