# What can you say about the proportion of hydrogen ions and hydroxide ions in a solution that has a pH of 2?

Dec 7, 2015

You can say that it contains ${10}^{10}$ more hydronium ions that hydroxide ions.

#### Explanation:

In order for an aqueous solution to be neutral, you need it to contain equal concentrations of hydronium ions, ${\text{H"_3"O}}^{+}$, and hydroxide ions, ${\text{OH}}^{-}$.

Now, a solution's pH is calculated by taking the negative common logarithm (this is simply a base 10 log) of the concentration of hydronium ions.

"pH" = - log( ["H"_3"O"^(+)])

Likewise, a solution's pOH is calculated by taking the negative common logarithm of the concentration of hydroxide ions

"pOH" = - log( ["OH"^(-)])

For aqueous solutions, you can say that

$\textcolor{b l u e}{\text{pH" + "pOH} = 14}$

Here $14$ is actually equal to $- \log \left({K}_{W}\right)$, ${K}_{W}$ being the ion product constant for water's self-ionization reaction.

So, take a look at your solution. You know that its pH is equal to $2$. Automatically, you can say that its pOH will be

$\text{pOH" = 14 - "pH}$

$\text{pOH} = 14 - 2 = 12$

Now, a lower pH is equivalent to a higher concentration of hydronium ions, and implicitly a lower concentration of hydroxide ions.

Use the log definitions of the pH and pOH to get

["H"_3"O"^(+)] = 10^(-"pH") = 10^(-2)"M"

and

["OH"^(-)] = 10^(-"pOH") = 10^(-12)"M"

This means that a solution that has a pH equal to $2$ will have ${10}^{10}$ more hydronium ions that hydroxide ions, since

(["H"_3"O"^(+)])/(["OH"^(-)]) = (10^(-2)color(red)(cancel(color(black)("M"))))/(10^(-12)color(red)(cancel(color(black)("M")))) = 10^10