Question #c0369
1 Answer
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,
Now, you know that this constant is equal to
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.