Sal: We now know all about strong acids.
For example, hydrochloric acid in an aqueous solution.
It's strong because it disassociate completely into
hydrogen ions or hydronium molecules, depending on how
you want to view it.
And chloride anions.
And it's an acid in the Arrhenius definition, because
once again, this is an aqueous solution.
In the Arrhenius definition because it is increasing the
hydrogen concentration in the water.
It's an acid in the Bronsted-Lowry definition
because this thing is donating these hydrogen protons.
And it's a Lewis acid because you can kind of view it as
hydrochloric acid.
By giving away this, it's taking these-- the chloride,
as you can kind of say, it's taking the
electrons for itself.
So it's an electron acceptor.
So that's another reason why it's a Lewis acid.
But in every case, it completely disassociates.
So it's a strong acid.
Now, by implication, when I call this a strong acid, that
must mean that there must be things called weak acids.
And you could probably guess what they do.
They don't disassociate completely.
So let's say you have hydrofluoric acid in an
aqueous solution.
It does disassociate somewhat, so you are going to have some
hydrogen break-off of the fluorine molecules.
So a hydrogen ion, and that's of course
in an aqueous solution.
I was doing aqueous in blue.
And then that's going to disassociate with the
fluoride, the fluorine cation.
But this is a weak acid, so it's
not just in one direction.
It actually goes in both.
This is an equilibrium reaction.
We know all about these now.
Which means that some of this goes in that direction, some
of that goes in that direction, and both of these
species exist in the molecule.
And just as aside, you might wonder hey, aren't fluorine
and chlorine-- they're both halogens.
Why is this a weak acid while this is a strong one?
And this is just an aside.
This is something I always used to wonder about.
So in general, hydrogen likes to form bonds.
Remember, alkali metals we included
everything below hydrogen.
Right?
We included all of these guys.
All of these guys, they have this 1 extra electron and
they're on a valence shell, so they just want to give it up.
And they tend to be very willing to give
it up to these guys.
So if you have any of these period 2 and below with a
halogen, you normally form this ionic salt.
Where this guy gives away an electron, and one of these
guys takes it, and you have an ionic compound.
Hydrogen's a little different.
Hydrogen does have exactly 1 electron in
it's outermost shell.
It just has one shell.
But the first shell is interesting because it's
complete when it only has 2 electrons.
These guys wouldn't have been happy.
If you gave 1 electron to these guys-- they
need to get to eight.
So they're not happy.
But if you give 1 electron to hydrogen,
hydrogen is pretty happy.
So when hydrogen forms bonds with these guys here, it
doesn't just completely give up the electron.
When hydrogen forms bonds with our halogens over here, it
doesn't completely give up the electron.
Because hydrogen says hey, if you give me some of the
electron, then I get to 2.
And then if you share the electron, you get to 8.
So it does definitely like the bond with these guys, but
doesn't give it completely up.
Now.
Hydrochloric acid, hydrobromic acid, and hydroiodic-- I think
I think I'm seeing it right-- these are all strong acids.
When hydrogen bonds with any of these guys.
Which means it disassociates completely in water.
But when hydrogen bonds with flourine, it doesn't
disassociate completely in water.
And the reason why is flourine is even more electronegative.
So this bond between hydrogen and fluorine-- so if you
compare hydrogen fluorine or hydrofluoric acid to hydrogen
chloride, hydrogen fluoride to hydrogen chloride, this bond
is much stronger.
Not much stronger.
I mean, you can measure in the energy of the bond, and we'll
do that eventually.
But this is a stronger bond than this right here.
And hydrobromic is even weaker.
Let me draw it.
So this one is a little bit weaker than that one and then
hydrobromic-- I'll draw it as one line right there.
It's like that.
So these are easier to break, so when you put this in water,
hydrogen sees these water molecules that have these
extra electron pairs floating around.
And and it hogged electrons from here.
So there's a partial negative charge, it's tempting for it
to disassociate with the water molecules.
Well, in this case, yeah, it's tempting, but it still like to
be hanging around.
Oxygen is very
electronegative, but so is fluorine.
So it likes to hang around with both.
So it kind of doesn't completely disassociate.
So that's why this one is a hydrofluoric acid, is a weak
acid, while all of the other halogenic acids are strong.
Well, with that said, the point of this isn't to just
look at the periodic table.
Let's figure out what the pH is when I dissolve-- let's say
I have 1.5 molar of hydrofluoric acid.
And obviously, by definition, since I've given this as
molarity, I have it in some solutions.
In this case, I have it in an aqueous solution.
Now.
You could look it up.
Well actually, I'll just write it.
So every equilibrium reaction has some equilibrium constant.
So the equilibrium constant for this reaction-- I'll do it
in green-- and since this is an acidic reaction, we'll say
it's the equilibrium constant, or an acid constant, however
you want to say it, for hydrogen fluorine.
Hydrofluoric acid.
Hydrofluoric acid is equal to the products.
So it's the concentration of your hydrogen ions times your
concentration of your fluorine ions divided by your
concentration of your hydrogen fluoride, or
hydrofluoric acid.
So HFl.
And if you look this up at 25 degrees Celsius at roughly
room temperature-- I believe that is, but let's say it is--
you get a constant value here of 7 times 10 to the minus 4.
So if I have 1.5 molar of hydrofluoric acid, can I
figure out what the pH is?
Remember, esssentially, I want to solve for this right here.
If I solve for this, then I know the hydrogen
concentration.
And then if I take the negative log of that, then I
know its pH.
So let's see if we can do that.
So.
I'm throwing 1.5 molar onto this side of
the equation, right?
But some of it's going to go and form that stuff, right?
So let's say x molar gets disassociated.
So on this side-- let me see if I can do this.
So I'm throwing 1.5 molar on this side.
Let's say x molar goes to that side, right?
And so let's say I have 1.5 molar on this side.
Once I'm in equilibrium, x goes to
that side of the equation.
Or x reacts to form this stuff, right here.
Disassociates.
So once I'm in equilibrium, I'll have my concentration of
hydrogen fluoride will be 1.5 molar minus x.
And what's my concentration of hydrogen?
Well, if x molar, or x moles of this, the way you can view,
disassociate, then we have x moles or x molar of that.
And we'll also have x molar of that.
Right?
If one of the disassociates, then you get one of that and
one of that.
So these will be the equilibrium concentrations.
And I think you can smell some algebra coming our way.
So these are the equilibrium concentrations.
Let's see if we can solve for x.
So we know what the equilibrium constant is.
It's 7 times 10 to the minus fourth.
We know that the final equilibrium concentration of
our hydrofluoric acid is the original 1.5 molar we put in,
minus whatever disassociates into the stuff up here.
And since x disassociated this stuff here, I have x molar of
hydrogen, and I have molar of fluoride.
This is just pure algebra now.
Let's see if we can solve this.
So this is equal to-- I'm going to
need space to do this.
So we get 7 times 10 to the minus 4.
Let's multiply both sides times 1.5 minus x.
Times 1.5 minus x.
Is equal to
x times x, is equal to x squared.
And then if we multiply-- 7 times 10 to the minus fourth
times-- 7 times 1.5, that's 10.5, right?
So we have 10.5 times 10 to the minus fourth.
That's that times that.
Minus 7 times 10 to the minus fourth x
is equal to x squared.
This looks complicated, but it's just a quadratic
equation, and there's many videos on that.
So let's just throw all of this stuff onto to the
right-hand side because I like having a
positive coefficient here.
So if I subtract this from both sides and add this to
both sides, I get 0 is equal to x squared, plus this, so
plus 7 times 10 to the minus fourth x, right?
I just add this to both sides, so it shows up on this side of
the equation, which is right here.
Minus this.
I'm subtracting this from both sides of the equation.
So I get minus 10.5 times 10 to the minus 4.
Now, this is a quadratic equation here.
And actually, I'll throw you a little--
We can solve this quadratic equation.
We'll get the answer.
In fact, I'm about to solve it.
But there is an approximation you can do.
Because if you look at this quadratic equation, this term
here, x squared, dominates this term right here.
Even when you have relatively small values of x, it
dominates that term.
So if chemists look at this and they say
hey, you know what?
If this term is so much smaller than what's going on
here, in a lot of the cases, you might be able
to just ignore it.
And then the problem becomes very simple.
Because if you ignore this term, it's still a quadratic,
but you don't have to use a quadratic equation.
Actually I'll do it on the side right here.
I'll just do it in a box.
Because this isn't the exact answer, but it's going to get
it's pretty close to the exact answer.
So in this-- I'll make a nice little box here.
I'll make a box.
So if we just ignored this term, then we'd have x squared
is equal to this.
Is equal to minus-- you would add this to the other side, so
you would have plus 10.5 times 10 to the minus 4.
Or x would be equal do the square root of this.
x would be equal to-- what?
See, 10.5-- I see you have to put the exponent thing here.
So exponent for minus all right.
And I want to take the square root, so to the 0.5 power is
equal to 0.0324.
So this is 0.0324.
So if we make that simplifying assumption, we've essentially
solved four our hydrogen concentration, right?
That was x.
This is x.
So this is 0.0324.
And if we want to figure out the pH, we just take the
negative log of this.
So let's do that.
So you take the log of it.
If you just press log, that's base 10.
So you take the log.
And we want the negative of that, so you just say minus
1.489, so about 1.49.
So it's a pH of 1.49.
But remember, I did that using my little simplifying
assumption.
It does not solve the quadratic equation.
Now let's solve it, just to kind of confirm in our heads
that we get the right answer.
Because that's actually the right way to
get the right answer.
So quadratic equation.
Minus b.
So x is going to be equal to-- let me do it here.
x is equal to minus b.
So minus 7 times 10 to the minus 4 plus or minus the
square root of b squared.
So b squared is 49 times 10 to the minus 8, right?
7 squared.
And then 10 to the minus 4 squared.
Is minus 4 times 2 is minus 8.
Minus 4 times a, which is 1, times c.
Well, c is a minus sign there-- we have this minus
sign, so it becomes plus.
So 4 times 10.5 times 10 to the negative 4.
That's what?
That's 42 times 10 to the minus fourth power.
All of that over 2a.
a is just 1.
All of that over 2.
So let's see what we can get in this
square root right here.
And notice, this is a very small number compared to this.
This is really all we should care about with the square
root, but let's just do it.
So 49 e 8 negative plus, so that's that small number--
plus 42 times 10 to the negative-- or to the 4
negative, is equal to that.
So that's this value right here.
And now we want to take the square root.
So to the 0.5 power is equal to 0.06.
So so far, we get x is equal to-- I will arbitrarily switch
colors-- minus 7 times 10 to the minus 4
plus or minus 0.0648.
All of that over 2.
And it should be clear.
If we subtract it right here, we get a negative number.
This is already a negative, so if we subrtact it, we get a
negative numbers.
So let's just worry about the plus sign because we can't
have a negative concentration.
So if we add that to this right there, so we have 0.06,
and then we have minus--
Actually let me make sure that the minus is there.
So it's minus 7 times 10.
Right.
I just want to make sure I'm not making a mistake by
putting a minus there.
No.
It looks right.
So let's see.
So we just do our math.
So we take this and we subtract that.
So it's minus-- I want to make sure you can
see what I'm doing.
So we have minus 7 e 4 negative is equal to that and
then we have to divide by 2.
Divided by 2 is equal to 0.032.
So 0.032.
And remember, this is our hydrogen concentration.
It's also going to be our concentration of, of course,
our fluorine ions.
We don't care about that.
We just want to figure out the pH.
So this is 0.032, which is very close to our
approximation before.
But now we have the exact number.
So let's take the log of it, and then take
the minus of that.
So the log of that.
And then you want to put a minus in front of the log.
So the exact pH is right there.
So our pH of hydrofluoric acid is equal to 1.49.
So once again, pretty low pH.
But notice, this pH-- this is 1.5 molar of
hydrofluoric acid.
While 1 molar of hydrochloric acid was even more acidic.
It had a pH of 0.
So even though we put more acid here, we didn't get our
pH as low as when we used only 1 molar of hydrochloric acid.
Anyway, see you in the next video.