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JOCELYN: Hi.
Jocelyn here.
And today we're going to go over fall 2009's final exam,
problem 10.
As always we're going to start by reading the question.
You have 333 milliliters of alkaline solution at a pH
equal to 9.9.
You wish to neutralize this by reacting it with 222
milliliters of acid.
What must be the value of the pH of the acid?
So first off let's write down the information that was given
and the information that he's asking for.
So we have 333 milliliters of a solution
with pH equal to 9.9.
And he's asking you what must the pH be if you want to put
in 222 milliliters in order to completely neutralize that
basic solution.
And the key here is to recognize
what neutralize means.
And a neutralization is when you take acid, or protons, and
base and it reacts to form water, thus
creating a neutral solution.
So the next step would probably be to figure out how
much excess hydroxide we have in this solution.
And then we can figure out how much hydrogen we need to react
with that hydroxide and to form water in order to
completely neutralize this solution.
So first step is to find out what our
concentration of OH minus is.
And we do that by remembering a few things
about acid/base solutions.
Right?
We know that if you have the pH plus the pOH it equals 14.
And this is from the equilibrium constant of this
neutralization reaction.
And then we can figure out what the pOH is because we're
given the pH of our initial solution--
9.9--
and that equals 4.01.
Now using the definition of pOH, which is that the pOH
equals the negative log of the concentration of the OH minus,
we can rearrange that and get that our OH minus
concentration equals 10 to the negative pOH.
Right?
Just using our log rules there.
So this gives us that our current concentration of the
hydroxide ion is 10 to the negative 4.1.
That gives us the concentration of the
hydroxide, but what we really want to know, if we're going
to use our neutralization reaction here, is we want to
know how many moles of hydroxide we
have in that solution.
And we can just do that with stoichiometry.
Right?
The concentration is in moles per liter.
And we know--
so our moles hydroxide equals our concentration, which is 10
to the negative 4.1 moles per liter.
Times--
if we want to get rid of that liter unit we multiply it by
our volume, which is 333 milliliters, converting it
from milliliters to liters.
And this gives us 2.65 times 10 to the negative 5 moles of
OH minus in solution.
So that's just the first step of this problem.
Now that we have the number of moles of OH minus we need to
find the number of moles of H plus necessary to react with
this excess hydroxide, and then we will be able to find
the concentration of hydrogen, or protons,
in that second solution.
So first off we're going to look at our neutralization
reaction over here and realize that it's a one-to-one
stoichiometric ratio.
That means in order to react with 2.65 times 10 to the
negative 5 moles of hydroxide, we need the same number of
moles of hydrogen.
So over here I'm going to say that this also equals the
moles of H plus needed.
All right.
Now that we have our moles H plus needed, we need to do the
same process here but in reverse.
So first we need to find the concentration of protons that
we have. So given that we want to only have 222 milliliters
of this solution, right?
So we have--
and we just divide that by our volume, 0.222 liters.
Right?
Because we divide by 1,000 to convert from milliliters.
And that equals 1.19 times 10 to the
negative 4 moles per liter.
So this is another problem where you could get the answer
but not actually be answering the question.
Right?
Because this is the concentration of protons that
we need, but now we need to convert that to pH.
So moving over here we'll just write the final steps.
We know the definition of pH is the negative log of the
concentration.
And so because we know the concentration of protons we
know that the--
we just need to plug this into our calculator and we get our
answer, which is a pH of 3.92.
Put a box around that one because it's our correct final
answer, and you'll be good to go for
this part of the problem.
Now moving on to part b we are asked, name the conjugate base
of each of the following.
So we'll just start part b right here.
And the first thing to recognize is what does a
conjugate base mean.
And in an acid/base pair, or when you have a kind of
dissociation of an acid, in the most simple terms we can
think about it just losing a proton.
In the leftover molecule, that's the conjugate base
right there.
So in this problem we're asked if the molecule loses a
proton, what is the resulting molecule?
Ion, basically.
So part i we have HPO4, and that will dissociate to
hydrogen phosphate ion.
Simple enough.
This is your conjugate base.
So that's your answer for part i.
Moving to part ii.
We have CH3NH3 plus.
So it's kind of like an ammonia ion but not quite--
sorry, ammonium.
And we know that this dissociates by losing a
hydrogen off of the nitrogen, because that's where the
positive charge is.
It makes more sense there.
And we get--
sorry, keeping the same order as above--
a proton plus.
And this would be your conjugate base.
If you were a little bit confused on this one it might
help to draw the Lewis structure, and you would see
that the formal charge does actually reside on nitrogen
and, therefore, it makes sense that the nitrogen
would lose a proton.
In addition, if you think about the ammonia/ammonium ion
relationship you know that you have NH3 going to NH4.
And in this case we just have one of the hydrogens replaced
with a methyl group, and so you can make the analogy there
that, oh, the hydrogen would come from the
nitrogen in that case.
All right.
So hopefully this was pretty straightforward.
As long as you're comfortable with what a conjugate base is
and what the dissociation reactions look like for acids.
So now on part c we are asked to classify each of the
following as a Lewis acid or a Lewis base.
Now first we need to remember what a Lewis acid and
a Lewis base are.
So if you remember from lecture or in your reading, we
know that a Lewis acid is an electron-pair acceptor and a
Lewis base is an electron-pair donor.
These definitions are just a little more general than the
classic proton acceptor or proton donor.
The first part is asking us about the cyanide ion.
And to kind of clarify it for yourself and because we're
asked about Lewis acids and bases, which are to deal with
electron pairs, it might be helpful to
draw the Lewis structure.
So for this molecule a good Lewis structure looks like
this, and it has a net negative charge.
And you can see that we have a couple of lone electron pairs
there that aren't involved in bonding.
And because it has a net negative charge you can
imagine that these electron pairs would want to kind of
grab onto something else that's a little bit more
electron positive.
And because of that we know that this is
actually a Lewis base.
It acts as a base and would grab something like a proton
and thus become more, I don't know, stable in some cases.
All right.
Moving to part 2.
We are asked about water, which is a molecule that we
deal a lot with in acid/base chemistry.
Again, you might be able to answer this right off the bat,
but we're going to look at the Lewis structure here.
Water has two lone pairs.
So that means it could very well act as a Lewis base.
Hopefully that makes a little bit of sense.
Right?
Because these electrons can go and grab something like a
proton and form the hydronium ion with a positive charge,
thus acting as a Lewis base.
It's important to remember that even though we're
creating an acid in the hydronium ion, this water
molecule in converting to the acid acts as a Lewis base.
However, we also know that water can lose a hydrogen and
accepts those electrons to form the hydroxide ion, and
thus acts as a Lewis acid.
So in the case of water either Lewis acid or Lewis base was a
correct answer in this case.
Now moving on to part d.
We will go over here, I think.
I'm sorry.
d, not b.
Again, it's going to be about acids and bases.
Consider the effect each of the following substances has
on the ionization of the weak base ammonia.
For each state whether the substance, one, suppresses
ionization, two, enhances ionization, or, three, has no
effect on the ionization of ammonia.
In each instance give a reason for your choice.
Going back to the first part of this question we see that
it's asking us about the ionization of ammonia.
A good place to start from here would be to write down
the reaction, just so we have that on our paper, in our
mind, ready to think about it.
So we have ammonia--
and aqueous means that it's in water, right?
And that is reacting with water to form
the ammonium ion--
sorry--
by adding a proton and a hydroxide ion.
OK.
So this is the reaction we're asked about.
Now we need to look at each of the three substances listed
and decide how it affects this ionization.
So number i, we have KOH.
Again, a good place to start is to write what happens to
potassium hydroxide when it goes into water.
And we know that KOH is a strong base and will
dissociate into potassium plus hydroxide.
So we see that it produces hydroxide.
And this should indicate to you that, well, it has the
same product as one of the products of the
ionization of ammonia.
And if we remember back a little bit before this
section, we know that if you have the same ion in solution
it will definitely affect the solubility, or ionization.
So that's one way to think about it, is that I now have a
common ion effect that will, because there's more hydroxide
in solution now, it will push this reaction, this
equilibrium, back to the left.
Another way to think about that is
Le Chatelier's Principle.
Kind of the same thing, although it's a little bit
more versatile, I think, so pretty easy to think about.
Right?
If you add a product your reaction will
shift to the left.
If you add a reactant your reaction will
shift to the right.
So here we're adding a product, our reaction will
shift to the left.
Same as a common ion effect, these two combine to decrease,
or both suggest that we will decrease, ionization.
Moving to the second part of this problem.
We're asked about hydrogen chloride.
Again, let's look at what that does in water.
We know that it's a strong acid, so it will dissociate
into hydrogen, or protons, and chloride.
Now this one's a little bit trickier because we don't have
either of these ions in our initial ionization reaction,
but if we think about the fact that ammonia is a weak base we
can write an additional ionization reaction.
So that additional would be ammonia reacting with a proton
to form ammonium.
Right?
So this is another way that the ammonia
molecule would be ionized.
And we can see that if we add protons from this acidic
molecule there we will increase the ionization,
because it will allow the ammonia to work more as a base
because of this reaction here.
So we are increasing ionization.
Another way to think about this is completely valid and
kind of the same thing, but if we look at this ionization
reaction here where the hydrogen chloride is
dissociating into protons and chlorides we can recognize
that these protons will then react with the hydroxide that
is produced in the initial ionization reaction.
This will result in a decrease of hydroxide
in solution, right?
And so by Le Chatelier's Principle we know that a
decrease in a product causes the reaction to shift towards
the product, to make more hydroxide.
So that's just another way of thinking about the same
process, and it results in the same conclusion that addition
of hydrogen chloride increases the ionization of ammonia.
OK.
Moving to part iii.
We have ammonium chloride, and this will dissociate into
ammonium and chloride.
And even if you're not sure how much it will dissociate,
it will still dissociate to some amount.
So it will have at least a little bit of an effect on our
initial ionization reaction.
So here we see the dissociation produces ammonium
ions, and just like the first case that is one of the
products in our ionization reaction.
And so Le Chatelier's Principle, common ion effect,
whatever you want to call it, tells us that it will shift
the equilibrium to the left and decrease ionization.
So with that we're done with problem 10 from the final.
It had a lot to do with acid/bases, so hopefully you
feel more comfortable with them after this.