# 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?

Dec 4, 2015

$2.1 \cdot {10}^{- 11} \text{M}$

#### Explanation:

As you know, water undergoes a self-ionization reaction that leads to the formation of hydronium cations, ${\text{H"_3"O}}^{+}$, and hydroxide anions, ${\text{OH}}^{-}$.

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

${\text{H"_2"O"_text((l]) + "H"_2"O"_text((l]) rightleftharpoons "H"_3"O"_text((aq])^(+) + "OH}}_{\textrm{\left(a q\right]}}^{-}$

Now, the equilibrium constant for this reaction looks like this

${K}_{a} = \left({\left[\text{H"_3"O"^(+)] * ["OH"^(-)])/(["H"_2"O}\right]}^{2}\right)$

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, ${K}_{w} = 5.13 \cdot {10}^{- 13}$. 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 ${\text{H"_3"O}}^{+}$ to get

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

$\left[\text{H"_3"O"^(+)] = (5.13 * 10^(-13))/(2.5 * 10^(-2)) = color(green)(2.1 * 10^(-11)"M}\right)$

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

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

$\text{pH} = - \log \left(2.052 \cdot {10}^{- 11}\right) = 10.69$

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

$\text{pH} = - \log \left(\sqrt{{K}_{W}}\right)$

$\text{pH} = - \log \left(\sqrt{5.13 \cdot {10}^{- 13}}\right) = 6.14$

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