The value of #K_w# is #5.13 * 10^-13#, at an elevate temperature. What is the #H_3O^+# if the #OH^-# concentration is #2.5 * 10^-2# M?

As you know, water undergoes a self-ionization reaction that leads to the formation of hydronium cations, #"H"_3"O"^(+)#, and hydroxide anions, #"OH"^(-)#.

In this reaction, water exhibits its amphoteric character by acting both as an acid, and as a base. The balanced chemical equation for this equilibrium reaction looks like this

#"H"_2"O"_text((l]) + "H"_2"O"_text((l]) rightleftharpoons "H"_3"O"_text((aq])^(+) + "OH"_text((aq])^(-)#

Now, the equilibrium constant for this reaction looks like this

#K_a = ( ["H"_3"O"^(+)] * ["OH"^(-)])/(["H"_2"O"]^2)#

In aqueous solution, the concentration of water can be assumed to be constant. This means that you can write

#overbrace(K_a * ["H"_2"O"]^2)^(color(blue)(K_W)) = ["H"_3"O"^(+)] * ["OH"^(-)]#

Here #K_W# is called the self-ionization constant, or Ion product constant, of water.

Now, you know that a certain temperature, #K_w = 5.13 * 10^(-13)#. Moreover, you know that this solution contains

#["OH"^(-)] = 2.5 * 10^(-2)"M"#

Notice how large the concentration of hydroxide ions is compared with #K_W#. This tells you that you're dealing with a very basic solution for which you can expect a very high pH value.

Rearrange the above equation and solve for #"H"_3"O"^(+)# to get

#["H"_3"O"^(+)] = K_W/(["OH"^(-)])#

#["H"_3"O"^(+)] = (5.13 * 10^(-13))/(2.5 * 10^(-2)) = color(green)(2.1 * 10^(-11)"M")#

As a side note, the pH of the solution will be

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

#"pH" = - log(2.052 * 10^(-11)) = 10.69#

Another interesting thing to do here is figure out what the pH of pure water would be at this temperature.

#"pH" = - log (sqrt(K_W))#

#"pH" = - log( sqrt(5.13 * 10^(-13))) = 6.14#

At this temperature, pure water is neutral at pH equal to #6.14#.