Acid Base Titration


Uploaded by khanacademy on 15.10.2010

Transcript:

I've taken this problem from Chapter 4 of the Chemistry &
Chemical Reactivity book by Kotz, Treichel and Townsend,
and I've done it with their permission.
So let's do this example.
A 1.034 gram sample of impure oxalic acid is dissolved in
water and an acid-base indicator added.
The sample requires 34.47 milliliters of 0.485 molar
sodium hydroxide to reach the equivalence point.
What is the mass of oxalic acid, and what is its mass
percent in the sample?
So before we even break into the math of this, let's just
think about what's happening.
We have some oxalic acid, which looks like this.
It's really two carbolic acid groups joined together, if
that means anything to you.
Watch the organic chemistry play list if you want to learn
more about that.
So we have a double bond to one oxygen, and then another
bond to a hydroxide.
We have that on the other carbon as well.

This right here is what oxalic acid is.
And it's an interesting acid, because it can actually donate
two protons.
This proton can be nabbed off, and this proton can also be
contributed.
And it's actually resident stabalized.
If that doesn't mean anything to you, don't worry.
You'll learn more about that in organic chemistry.
But the important thing to realize here is that there's
two protons to nab off of it.

Now each molecule of sodium hydroxide-- remember when you
put it in the water it really just dissolves, and you can
really just think of it as hydroxide-- each molecule of
hydroxide can nab one of the hydrogen protons.
So for every one molecule of oxalic acid, you're going to
need two hydroxides-- one to nab this hydrogen proton, and
then another one to nab that hydrogen proton.
So let's write down the balanced equation that we're
dealing with here.
So we're going to start off with some oxalic acid.
So that has two hydrogens-- so it's H2-- two carbons, and
then four oxygens-- O4.
It's dissolved in water, so it's an aqueous solution.
And to that, we're going to add sodium hydroxide.

Now I just told you that you're going to need two of
the hydroxides to fully neutralize the oxalic acid.
So you're going to need two of them.
And this is also in our aqueous solution.
And once the reaction happens, this guy will have lost both
of the hydrogen protons, so let me draw that.
So it will look like this.
No more hydrogen, so it's C2O4.
It'll have a negative 2 charge.
And actually, you could imagine that it might be
attracted to these positively charged sodiums. And 2 sodiums
in particular.
So this has a negative 2 charge.
We could even write it there if you want-- 2 minus.
And then you could have the sodiums over here.
You have these two sodiums that have two plus.
And this entire molecules becomes neutral.
They are attracted to each other.
They are still in an aqueous solution.
And then, the hydroxide nabs the protons, and then you are
left with just water.
So plus 2 moles-- or 2 molecules depending on how
we're viewing this-- plus two waters.
I'll just use that same orange color.
Plus two H2Os.
One of the hydrogens in each of the water molecules are
coming from the oxalic acid, and so two of these hydrogens
in these two moles of the water are coming from one
entire molecule of oxalic acid.
Now let's actually do the math.
We have 34.47 milliliters of the solution that has the
sodium hydroxide.
And I'm just going to convert that to liters just so it's
easier to deal with the molarity right over there.
So we have 34.47 milliliters-- we could write it of the
solution, but we understand that, that's the case.
So let's just So this is times, we have one liter for
every 1,000 milliliters.
And then this will give us-- the milliliters cancel out--
34.47 divided by 1,000 is 0.03447 liters of this 0.485
molar sodium hydroxide solution.
So let's figure out how many actual molecules of sodium
hydroxide we have. This is the solution.
And we know its concentration, 0.485 molar-- so let me do
that in a different color-- 0.485 molar, this information
allows us to figure out the actual
molecules of sodium hydroxide.
So we want to multiply this by-- we have 0.485 moles of
sodium hydroxide for every 1 liter of this solution.
That's what the molarity tells us.
We have 0.485 moles per liter.
So the liters cancel out, and then now we're going to
actually have to get a calculator out.
And this'll tell us how many moles of sodium hydroxide we
have in this solution.
So let me get my calculator.
There we go.
All right, let me just multiply these two numbers.
So we have 0.03447 times 0.485 is equal to-- let me put this
down here-- 0.167.
And we only have three significant digits here, so
we're going to round to three significant digits.
So we'll just go with 0.0167.
So let me move that over off the screen.
So this is going to be equal to 0.0167, and all we have
left here are moles of sodium hydroxide.

Now we know that this many moles of sodium hydroxide are
going to completely react with however many moles of oxalic
acid we have. Now we know that we need two moles of this for
every mole of oxalic acid.
Or for every mole of oxalic acid that completely reacts,
we need two moles of this.
So let's write that down.

And then you color.
So we need two moles of sodium hydroxide, we got that from
our balanced equation right there, and it's obvious it
needs one mole, or one molecule will take this
proton, and then you need another molecule
to take that proton.
So we need two moles of sodium hydroxide for every one mole
of oxalic acid.
For every one mole of H2C2O4.

So essentially, we are just going to divide
this number by 2.
Let me get the calculator back.
So we're just going to divide 0.0167 divided by 2.
Once again, three significant digits 0.00835.
So this is going to 0.00835 moles of oxalic acid, H2C2O4.

So we have the number of moles, but we to
figure out the mass.
And we know the molar mass of oxalic acid.
Let me write these down.
We know that hydrogen has a molar mass-- let me write it
this way-- molar mass if you have a mole of hydrogen, it
has a molar mass of one gram.
If you have-- and this comes from its atomic weight-- if
you have carbon its molar mass is 12 grams. And if you have
oxygen, its molar mass is 16 grams.
So what's the molar mass of oxalic acid?
Well we have two hydrogens, so that's going to
be two grams, right?
2 times 1 gram.
That's the hydrogens there.
We have two carbons, so it's going to be plus 24 grams. 12
grams for each of these carbons.
And then we're going to have four oxygens that weigh in, if
we have a mole of them, at 16 grams. So that's
going to be plus 64.
So what does this come out to?
24 plus 64 is 88, right?
2 plus 6 is 88, right.
So it's 88 plus 2 more is 90 grams. So if you had a mole of
oxalic acid, it would be 90 grams. So we could say 90
grams per mole of H2C2O4.

So let's go back to the math here.
I'll rewrite it over here.
We know we're dealing with 0.00835 moles of
oxalic acid, H2C2O4.

And now we know its molar mass.
We know that there are 90 grams-- let me do this in a
different color, this color's getting motonous-- we know
that there are 90 grams of H2C2O4 for
every one mole of H2C2O4.
This is its molar mass.
And now we just multiply this number, and we'll figure out
the grams of oxalic acid.
That and that cancels out.
and.
Then we just take the number that we had and multiply it by
90, so times 90.
This just says the previous answer, which is the number of
moles of oxalic acid times its molar mass will tell us the
grams of oxalic acid.
So we get 0.75.
I'll just round it to 2 since we only have
three significant digits.

The 90 isn't exact, so it's a little bit-- but I'll just
round it to three significant digits.
So 7--.752.
This is equal to 0.752 grams of oxalic acid: H2C2O4.
And I think we've answered part of the question.
So the first question is, what is the mass of oxalic acid?
We've just answered it.
That answer right there is 0.752 grams.
Now the next part is, what is its mass
percent in the sample?
Well the sample of impure oxalic acid right over here,
was 1.034 grams. So we just have to say, what percentage
is 0.752 of 1.034?
So let's get the calculator back.
So we have the 0.752 divided by 1.034 and we get 72.7%.
So the answer to the second part right
over there is, 72.7%.
We were able to figure out that this impure oxalic acid
sample is 72.7% actual oxalic acid.