When we look at water what we see is that water can actually undergo
auto-ionization. Two water molecules
can react and produce an H_30+ ion and hydroxide ion.
Notice that this process is at equilibrium and the equilibrium lies far
to the left.
We only get a very small amount to be H_3O+ ion and and the hydroxide ion.
We can also show this in a little bit about shorthand notation by showing a
single water molecule
going to H+ a quiz and OH- aqueous.
Remember that we cannot actually have H+ ions free in solution.
If we write a H+, what we really mean is that we have
H_3O+ present. If we look at some of the properties of water
one thing we notice is that it is amphoteric. This means that it can act as
an acid or a base.
Just as when we talked about the Bronsted-Lowry definition of acids and bases
we said whether water acts as an acid or base depends on the identity and the
other reactant.
K_w is in equilibrium constant specifically we call it the ionic product constant
for water.
We can also call it a dissociation constant, but regardless of what we call
it, it is still an equilibrium constant.
The W in the subscript let us know that this is for water.
We can still write ab equilibrium expression for this just as we would
for any other reaction.
So how do we write the equilibrium expression?
Well we have a reaction in here we are using our shorthand notation but either
one will show us the same thing.
We have H_2O liquid in equilibrium with H+ an OH- ions.
When I'm writing an equilibrium constant expression or a law of mass action,
I am writing the concentration that the products raised to their powers
which we have coefficients have one for both H+ and OH-.
So we don't have any powers in the expression. On the bottom in the
denominator we put the concentration of our reactants.
However, we remember that we exclude both are solids and are pure liquids.
So we do not include water in our denominator.
So the equilibrium expression or the ion product expression in this case
is K_w equals H+ times
OH-. Another thing that's important to remember is that the value of K_w.
Notice that this is only true at 25 degrees Celsius that K_2 equals
1.00 time 10 ^ -14.
As we change temperatures we change the value of k
that equilibrium constant just as the values of equilibrium constants for
other reactions also change when we have a change
in temperature.
Let's look and see what we can find about the values at the H+
and OH- concentration for a sample pure water
given that we know the value of K_w. So we are at 25 degrees Celsius
are K_w value is 1.0 x 10 ^-14.
That equals the H+ concentration times the OH- concentration.
If we are looking at a sample a pure water are H+ concentration must be
equal to our OH- concentration because the balanced reaction
and so we represent them both as X therefore we end up with 1.0 x 10 ^-14
equals X^2. When I solve for X, I get 1.0 x 10 ^-7.
That is going to be equal to the concentration of H+
as well as the concentration ever OH-
in a sample of pure water. We are also going to look at later about what happens
when we are not at 25 degrees Celsius when we have a different value
K_w and what that means for the concentrations of H+ and OH-
both in pure water
as well as in either an acidic or basic solution.
First, let's look at example of how we find OH-
if we know the H+ concentration. So for example if we have a solution that has an
H+ concentration equal to 3.5 x 10^ -5
we know we are at 25 degrees so we know that out K_w value is going to be
equal to 1.0 x 10 ^-14.
We can set up our expression 1.0 x 10 ^-14
equals 3.5 x 10 ^ -5
times are hydroxide concentration thats are unknown that we're trying to find.
When I divide both sides by 3.5 x 10 ^ -5
I end up with 2.9 x 10 ^ -10 equals the OH- concentration.
I could do a similar calculation if I knew the OH- concentration
and needed to find the H+ concentration, but at 25 degrees Celsius
this expression will always be true.
1.00 x 10 ^-14 equals
H+ times the OH- concentration.
