Question

Question #c0369

Answer 1

`["H"_3"O"^(+)] = ["OH"^(-)] = 2.35 * 10^(-7)"M"`

Explanation:

The balanced chemical equation for water's self-ionization reaction looks like this

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

The equilibrium constant for this reaction would look like this

`K_(eq) = ( ["H"_3"O"^(+)] * ["OH"^(-)])/(["H"_2"O"]^2)`

Now, when dealing with dilute aqueous solutions, the concentration of water can be thought of as being constant. This means that you can write

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

This expression, `K_(eq) * ["H"_2"O"]^2`, is called water's self-ionization constant, or water's ion product constant.

Now, you know that this constant is equal to `5.5 * 10^(-14)` at `50^@"C"`. Since the self-ionization reaction produces equal number of moles of hydronium and hydroxide ions, you can say that `x` represents the molarity of these species in aqueous solution.

This means that you can write

`K_W = x * x = x^2`

In your case, you would have

`5.5 * 10^(-14) = x^2 implies x= sqrt(5.5 * 10^(-14)) = 2.35 * 10^(-7)`

Therefore,

`["H"_3"O"^(+)] = ["OH"^(-)] = color(green)(2.35 * 10^(-7)"M")`

As an interesting comment, the pH of water at this temperature would be

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

`"pH" = - log( 2.35 * 10^(-7)) = 6.63`

The important thing to realize here is that the water would still be neutral because you have equal concentrations of hydronium and hydroxide ions.