Question

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?

Answer 1

`2.1 * 10^(-11)"M"`

Explanation:

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`.