# What is pH?

Sep 23, 2015

To expand on the previous answer, $p H$, pouvoir hydrogene , is a logarithmic measure of hydrogen ion concentration, ${H}^{+}$.

That is $p H$ $=$ $- {\log}_{10} \left[{H}^{+}\right]$.

#### Explanation:

Water undergoes self-ionization:

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

Alternatively,

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

Both (i) and (ii) are equivalent representations; (ii) tends to be more common, and characterizes the cation of water as ${H}_{3} {O}^{+}$.

This is an equilibrium reaction, whose extent may be very accurately measured.

Under standard conditions, $\left[{H}_{3} {O}^{+}\right] \left[O {H}^{-}\right]$ $=$ ${10}^{- 14}$. Taking logarithms (base 10) of both sides, and multiplying by $- 1$, we get:

$p H + p O H$ $=$ $14.$ That is $p H$ $=$ $- {\log}_{10} \left[{H}^{+}\right]$, and $p O H$ $=$ $- {\log}_{10} \left[O {H}^{-}\right]$

This relationship is always obeyed. At high concentrations of ${H}_{3} {O}^{+}$, there are low concentrations of $O {H}^{-}$, at low concentrations of ${H}^{+}$, there are high concentrations of $O {H}^{-}$. But the sum of $p H + p O H$ always equals $14.$

Students tend to have problems with the logarithmic function. When I write ${\log}_{a} b = c$, I am asking to what power I raise the base $a$ to get $b$; in other words, ${a}^{c} = b$. So ${\log}_{10} 100 = 2$, ${\log}_{10} 10 = 1$, and ${\log}_{10} 0.1 = - 1$.

So now, assuming complete ionization of $H C l$ (reasonable!), can you tell me the $p H$ of $0.1$, $1.0$, and $10$ $m o l$ ${L}^{-} 1$ solutions?

$H C l \left(g\right) + {H}_{2} O \left(l\right) \rightarrow {H}_{3} {O}^{+} + C {l}^{-}$.