A chemistry course to cover selected topics covered in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. Prerequisites: Students should have a background in basic chemistry including nomenclature, reactions, stoichiometry, molarity and thermochemistry.
3.02 pH and Kw
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来自 University of Kentucky 的课程
高级化学
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从本节课中
Acid-Base Equilibria
The concept of equilibrium is applied to acid and base solutions. To begin, the idea of weak acids and bases is explored along with the equilibrium constants associated with their ionization in water and how the value of the equilibrium constant is associated with the strength of the acid or base. The autoionization of water is discussed and how temperature affects this process. A variety of problem types are covered including calculations of pH, pOH, [OH-], and [H+] for both strong and weak acids and bases.
Aqueous salt solutions are classified as acids and bases and the multi-step ionization of polyprotic acids is discussed. Finally, the concept of Lewis acids and bases is discussed and demonstrated through examples.
与讲师见面
Dr. Allison SoultLecturer
Chemistry
Dr. Kim WoodrumSr. Lecturer
Chemistry
In this module we're going to look at the values of pH
and K_w the equilibrium constant associated with the ionization of
water. Our objective for this unit is to understand the auto ionization of water
and to calculate the pH values based on the hydronium ion concentrations.
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.
Most the time are when we looking at solutions we're looking at them at 25 degrees Celsius
and so we most commonly use this K_w value 1.0 x 10 ^-14
However, if we decrease the temperature
we see that the value of K_w changes or if we increase the temperature
we see that the value K_w changes as well.
You should know the value at 25 degrees Celsius but if you doing calculations at
a different temperature
then it's okay to be able to look at these values. Now note that some of these values
are not written in the correct format
for scientific notation because we wanted to be able to compare the numbers
and have the powers be all the same.
So when we are at 25 degrees Celsius if we have a sample a pure water
we know that it has an H+ concentration at 10 x 10 ^ -7.
And what we want to be able to do is find the pH from that
and so we see the pH is equal to the negative log at the H+ concentration.
So we can find that the pH of water
when the concentration of H+ is 1 x 10 ^ -7.
The pH is equal to 7.
Now when the pH is equal to 7
we know that the pOH is also equal to 7 and we can do this two different ways.
We could take the pOH equals minus log of OH-.
Remember if the H+ concentration is equal to 1 x 10 ^ -7
that the OH- concentration is also equal to the 1 x 10 ^ -7.
So I can take the negative log of 1 x 10 ^ -7 and
find the pOH value
is seven as well, or we can take advantage of the fact that pH plus pOH
equals 14.
Now note this only applies at 25 degrees Celsius because our K_w value is equal to
1 x 10 ^ -14.
If we had a different K_w value because we read a different temperature
then this expression would no longer be true. We would still be able to say pH
plus pOH equals
plus something, but we have to have a new value instead of 14.
Using a pH meter we can easily measure the pH of a solution.
However that doesn't directly give us the concentration at the H+ ions.
But we can use the pH and be able to calculate that H+ ion concentration.
This is our expression for finding the pH from our H+ concentration. This
is the relationship that we know:
pH equals minus log H+ concentration.
Now I'm going to rearrange this to solve
for H+ so I have -pH, I divided both sides by -1.
So -pH equals the log of H+. Now to get rid of the log I am going to take
10 to both sides.
Remember that 10 and log are
opposite functions. In the same way that if I divide by six or multiplied by
6 they do opposite things to the number.
What I can also see is that X equals
10 to the log of X.
The log and the 10 cancel each other out
and we are left with X. We can take advantage if this by saying 10^-pH
equals 10 to the log of H+ and I can simplify that
because those two terms, the effect will cancel each other out
so we end up with 10 ^-pH equals the H+ ion concentration.
So if I were able to measure the pH of a solution
I can then find the H+ concentration.
I could also do the same thing with my pOH if I had
pOH value, I could find the OH- concentration.
Now we want to look at some examples where we are not at 25 degrees Celsius.
Where we are at a different temperature in this case 40 degree Celsius
and out K_w value is 2.92 x 10 ^ -14.
Now note that this first part is just about looking at the conversion
of H+ from the value of the pH
that was measured.
You should have found that the H+ concentration was 3.98 x 10 ^-5
molar. We can do this by saying 10 ^-pH equals
the H+ concentration. This is the expression we derive from the
formula for finding the pH from pH equals negative log of
H+. Here I put in 10 ^ -4.40
equals my H+ concentration
and I find that equals to 3.98 x 10 ^-5
molar. Now
we are going to take this one step further and look at what the OH-
concentration of this same solution.
But we have to remember that we are not, in fact, at
25 degrees Celsius we have a different value of K_w.
So what is the OH- concentration of a solution with a pH of 4.40
at 40 degrees noted that the K_w value
at that temperature is given.
Now we can solve for that OH- concentration. We know that K_w
is equal to H+ times the OH- concentration.
We know that K_w is equal to 2.92 x 10 ^ -14
and we know from our previous problem that our H+ concentration is 3.98 x 10 ^-5
molar. So now we can solve for our OH- concentration
and what we find is that it's 7.5 x 10 ^ -10 molar.
We can go back and check our work by multiplying 3.98 x 10 ^-5
times 7.5 x 10 ^ -10 and we should get a value somewhere close to
the 2.92 x 10 ^ -14.
However we have to ask ourselves why can't we use pH plus pOH equals 14?
Remember this is only true at 25
degrees celsius. This is only true because K_w at 25 is equal to 1.0 x 10 ^-14.
A different value K_w we cannot say pH plus pOH equals
14. Let's look at an example where we look at the OH- concentration given
the H+ concentration.
We are 25 degrees Celsius so we can remember that K_w equals 1.0 x 10 ^ -14.
Once we find those two values we can then find pH
and pOH by knowing the H+ concentration pH equal -log H+.
We could find the pH value for pOH we can do the same thing
9.54 and what we do notice is that 4.46
plus 9.54
is equal to 14. This works out because we're at 25 degrees Celsius.
When we're looking at calculations involving logs we have to be careful
about our significant figures.
Note that this value the 3.5
has two significant figures that will be the number of digits in our mantissa
or the number of decimal places in our answer.
So we have two sig figs here.
We have two decimal places here, but as a result we end up with 3 sig figs in our
answer. So the number of sig figs equals the number of decimal places.
In the next module we will talk about acid strength and compare strong and weak acids.
