Question

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

Answer 1

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, `"H"_3"O"^(+)`, and hydroxide ions, `"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

`color(blue)("pH" + "pOH" = 14)`

Here `14` is actually equal to `- log(K_W)`, `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

`"pOH" = 14 - "pH"`

`"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`