- [Instructor] Before we get into the topic
of weak acid equilibria, let's look at a strong acid first.
So let's say that HA is a strong acid and reacts with water
to produce the hydronium ion and A-minus,
the conjugate base to HA.
Technically, the reaction comes to an equilibrium.
However, the equilibrium favors the products so much
that we don't draw an equilibrium arrow.
We simply draw an arrow going to the right
indicating at equilibrium, we have essentially
all H30-plus and A-minus, and no HA.
So we assume that a strong acid ionizes 100% in solution,
and we can see that in our particulate diagrams.
And keep in mind that these particulate diagrams
are not meant to represent the entire solution,
just a small portion of it so we can get an idea
about what's happening in the entire solution.
We can think about the particulate diagram on the left
representing the initial concentration of our acid, HA.
So there are five HA particles in solution
and also five water particles.
Since the reaction goes to completion,
the HA particles are going to react with the water particles
to produce five hydronium, H3O-plus, and five A-minus.
We can see that in our second particulate diagram,
which represents the equilibrium mixture
and there are five hydroniums present
and also five A-minuses.
We can write the equilibrium constant expression
for this reaction from the balanced equation.
we look at our products and say it'd be the concentration
of H3O-plus, and since there's a coefficient of one,
it'd be the concentration of H30-plus
raised to the first power, times the concentration
of A-minus also raised to the first power, divided by
the concentration of HA raised to the first power,
and pure water is left out
of the equilibrium constant expression.
So that's equal to the equilibrium constant K,
and since this shows the ionization of an acid,
we could call this Ka value,
the acid ionization constant,
or the acid dissociation constant.
Since the equilibrium lies so far to the right
for a strong acid, at equilibrium, there's going to be
essentially all products and very, very little reactants.
That means the Ka value for a strong acid
will be much greater than one.
Now let's look at a weak acid.
We're gonna use the same general equation here.
HA plus H2O yields H3O-plus plus A-minus.
However, if HA is a generic weak acid,
we're going to include an equilibrium arrow here.
The equilibrium arrow needs to be included
because weak acids only partially ionize,
and we can see that in the particulate diagrams.
In the particulate diagram on the left,
this is supposed to represent the initial concentration
of our weak acid HA and there are five particles of HA
and five particles of water.
Since weak acids only partially ionize, let's say
that only one HA particle reacts with one water molecule
to produce one hydronium ion and one A-minus anion.
So if one HA particle reacts with one H2O particle,
that gives us one hydronium, H30-plus and one A-minus.
So out of the original five HA particles,
only one of them ionized, leaving four HA particles
still in solution at equilibrium, therefore HA
only partially ionized and is a weak acid,
the equilibrium constant expression for a weak acid
looks the same as it did for our strong acid,
because we use the same general equation.
However, since a weak acid only partially ionizes
at equilibrium, there's gonna be a small amount
of products and a large amount of reactants.
Therefore, the Ka value for a weak acid
will be less than one.
Let's look at some Ka values
for some weak acids at 25 degrees Celsius.
Hydrofluoric acid has a Ka value
of 6.3 times 10 to the negative fourth,
and acetic acid has a Ka value
of 1.8 times 10 to the negative fifth.
Since both of these have Ka values less than one,
they're both considered to be weak acids.
However, the higher the Ka value, the stronger the acid.
So out of these two weak acids, since hydrofluoric acid
has the higher Ka value, hydrofluoric acid
is the stronger acid of the two.
Now let's do a calculation using a weak acid.
Let's say we have a 0.10 molar solution of benzoic acid,
which is a weak acid, and the pH of the solution
is 2.60 at 25 degrees Celsius.
Our goal is to calculate the Ka value
for benzoic acid at 25 degrees Celsius.
Since we're given the pH in the problem,
we can go ahead and plug the pH of 2.60 into this equation.
So 2.60 is equal to the negative log of the concentration
of hydronium ions, and to solve for the concentration
of hydronium ions, first moved the negative sign
to the left, which gives us negative 2.60,
is equal to log of the concentration of hydronium ions.
And to get rid of the log, we take 10 to both sides.
This gives us the concentration of hydronium ions
is equal to 2.5 times 10 to the negative third molar.
It's also important to recognize this is
the equilibrium concentration of hydronium ions.
The next step is to write out the equation
showing benzoic acid reacting with water.
So here's the chemical formula for benzoic acid
reacting with water to form the hydronium ion
and the conjugate base to benzoic acid,
which is the benzoate anion.
To help us find the Ka value for benzoic acid,
we're gonna use an ICE table,
where I stands for the initial concentration,
C is the change in concentration,
and E is the equilibrium concentration.
In our problem, we were told the initial concentration
of benzoic acid was 0.10 molar, and if we assume
that benzoic acid hasn't reacted yet,
the initial concentration of hydronium ion would be zero,
and the initial concentration of benzoate
would also be zero.
We just calculated the equilibrium concentration
of hydronium ions, which we found to be
2.5 times 10 to the negative third molar.
So we're going to go ahead and put that under
our hydronium ion in for the equilibrium concentration.
So if the initial concentration of hydronium ion was zero
and the equilibrium concentration
is 2.5 times 10 to the negative third, there was
an increase in 2.5 times 10 to the negative third,
so we're going to write that under hydronium
for the change in the concentration.
And looking at the mole ratio of hydronium ion
to benzoate, it's a one-to-one mole ratio.
So if hydronium ion increased
by 2.5 times 10 to the negative third,
so did the benzoate anions, I can go ahead and write here
plus 2.5 times 10 to the negative third,
which means the equilibrium concentration
of benzoate anion would also be
2.5 times 10 to the negative third.
Since we saw an increase in the concentration
of hydronium ion and benzoate, that increase
in the concentration of the products
came from the ionization of our weak acids.
In the balanced equation, there's a one
in front of benzoic acid, so if we write
plus 2.5 times 10 to the negative third for hydronium,
and it's a one-to-one mole ratio from hydronium
to benzoic acid, we would have to write
minus 2.5 times 10 to the negative third,
since we lost some of the benzoic acid when it ionized,
therefore the equilibrium concentration of benzoic acid
would be 0.10 minus 2.5 times 10 to the negative third.
Since our goal is to calculate the Ka value
for benzoic acid, the next step is to write out
the equilibrium constant expression,
which we get from the balanced equation.
It would be the concentration of hydronium ions
times the concentration of benzoate anions
divided by the concentration of benzoic acid.
Since these are equilibrium concentrations,
we can get the equilibrium concentrations
from the ICE table and plug them in
to our equilibrium constant expression.
So we can plug in the equilibrium concentration
for hydronium ions, we can plug in
the equilibrium concentration for benzoate anions,
and we can plug in the equilibrium concentration
for benzoic acid.
Here we have the equilibrium concentrations
plugged in to the equilibrium constant expression,
and when we solve for Ka, we find that Ka
for benzoic acid at 25 degrees Celsius
is equal to 6.4 times 10 to the negative fifth.
Notice that Ka is less than one
because benzoic acid is a weak acid.
Finally, let's talk about the idea of percent ionization.
And to help us think about percent ionization.
Let's go back to our example with a strong acid.
We started off with five HA particles
and all five of those HA particles ionized
to produce five hydronium ions in solution
and five A-minus anions in solution.
Therefore 100% of our acid ionized.
So we can write 100% ionization
for this hypothetical strong acid HA.
Now let's go back to our weak acid.
For our weak acid, we started off with five HA particles,
but only one of them ionized to produce
one H3O-plus and one A-minus.
So four remain unionized at equilibrium,
therefore one out of five ionized, so the percent ionization
for this hypothetical weak acid would be 20%.
Notice that the one particle of HA that ionized
made one hydronium ion, so we can also use
the concentration of hydronium ion at equilibrium
to calculate percent ionization.
So the equation for percent Ionization
is equal to the equilibrium concentration
of hydronium ions divided by the initial concentration
of the acid HA times 100%.
So looking at our ICE table for our benzoic acid problem,
the equilibrium concentration of hydronium ions
was 2.5 times 10 to the negative third molar,
and the initial concentration of our acid was 0.10 molar,
so we can plug those into our equation.
So we plug in our concentrations and molar cancels,
and we find for the final answer, the percent ionization
of benzoic acid is equal to 2.5%.