# What is the concentration of hydronium ions in a solution with a pH = 3.9?

Jun 25, 2016

If $p H$ $=$ $3.9$, then $\left[{H}_{3} {O}^{+}\right]$ $=$ ${10}^{- 3.9} \cdot m o l \cdot {L}^{-} 1$.

#### Explanation:

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

In water, under standard conditions $p H + p O H = 14$.

Both $p H$ and $p O H$ measure the extent of the equilibrium:

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

At $298 \cdot K$, the dissociation constant of water,

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

The $p$ means take the logarithm to the base 10, and mulitply this by $- 1$.

And thus $p {K}_{w}$ $=$ $- {\log}_{10} {K}_{w}$ $=$ $- {\log}_{10} {10}^{- 14}$ $=$ $14$ $=$ $p H + p O H$.