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

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#