In this video, we will learn how to
use the results from a titration experiment to calculate unknown properties of a
Exactly how titration experiments
work is not really the purpose of this video. Instead, we’re going to look
specifically at performing calculations from the results of these titration
experiments. Remember that during a titration
experiment, you add small amounts of a solution with a known concentration to a
solution with an unknown concentration. The solution of known concentration
might be referred to as the titrant or a standard solution. The solution whose concentration is
unknown might be referred to as a titrand or an analyte.
During our titration experiment, we
continue adding our standard solution until our unknown concentration solution has
been neutralized or fully reacted. These are all important terms that
we’ll come across later when looking at example questions. And the key point of getting all
this data is that we can then use the amount of titrant added to calculate the
unknown concentration. So let’s see how we do this.
To carry out calculations with data
obtained from titration experiments, we’re going to need a couple of key
equations. The first key equation we’ll need
is 𝑛 equals 𝑐𝑣, where 𝑛 is the amount in moles, 𝑐 is the concentration in moles
per liter, and 𝑣 is the volume in liters. Let’s practice using this equation
before we move on.
How many moles of NaCl are there in
1.5 liters of 0.3-molar solution?
First, we need to remember that
molar means moles per liter. In the question, we’ve been given
the value of 𝑣 and 𝑐, and we’re asked to find the value of 𝑛. We start with our equation 𝑛
equals 𝑐𝑣. And we can substitute in the values
from the question. Next, we calculate 0.3 molar
multiplied by 1.5 liters, which gives us 0.45 moles.
We can double check our calculation
by looking at the units. By doing moles per liter multiplied
by liters, the liters cancel. And we should be left with
moles. Since moles are the appropriate
units for the value of 𝑛, it looks like we’ve done this correctly.
Now, we need to use this same
equation to perform our titration calculations where we have an unknown
concentration. So let’s look at an example.
A 0.05-liter solution of HCl was
titrated against a 0.1-molar solution of NaOH. The addition of 0.025 liters of
NaOH was found to neutralize the HCl. What is the concentration of the
Perhaps one of the trickiest things
about titration calculations is working out exactly what you have to do from quite a
wordy question. One way to approach this is to
simply strip out all of the really important information and lay it out in a logical
way. But we should also try to
understand exactly what’s going on here. So to do that, let’s try writing
out a really simple balanced equation.
The question tells us that we are
titrating HCl with NaOH. So these are our reactants, but
what do they form? This is actually an acid–base
titration. The question uses the word
“neutralize,” which gives us a clue. Now that we know this, we can work
out what our reactants form. Of course, we get a salt, in this
case sodium chloride, and water. The important thing now is to check
that this is balanced. You’ll see why this is important a
bit later on.
Now let’s get the important data
out of our question. Let’s try laying out all of our
information in a handy table. First of all, we know that we have
0.05 liters of HCl. So we can put this in the
appropriate volume box. Next, we’re told that we have a
0.1-molar solution of NaOH. Remember that molar is the same as
moles per liter. We’re told that we need 0.025
liters of our NaOH to neutralize the acid. So again, we can add this to our
volume box. The question is asking us for the
concentration of the acid. So we can label this on our table
as our goal.
Now we just need to fill in enough
gaps on our table until we manage to fill in the concentration of the acid box. Remembering our key equation 𝑛
equals 𝑐𝑣, we can work out the amount of moles of NaOH. To calculate the number of moles of
sodium hydroxide, we multiply the concentration, 0.1 molar, by the volume, 0.025
liters, which gives us 0.0025 moles.
Now comes the important part of why
titrations are really useful. From our balanced equation, we can
have a look at our molar ratio. We can see that we have a
one-to-one ratio of acid to base. This means that one mole of HCl
will react completely with exactly one mole of NaOH. However, in this experiment, we
don’t have a full mole of NaOH. Instead, we have 0.0025 moles. But because of our one-to-one
ratio, we know that to react with 0.0025 moles of our base, we need exactly 0.0025
moles of our acid. So we can simply copy the number of
moles of our NaOH into the box for the number of moles of our acid.
Now that we have 𝑛 and 𝑣 for our
acid, we can rearrange our key equation and work out the concentration. The concentration is equal to 𝑛
divided by 𝑣. We can double check this by looking
at the units, where the units for 𝑛 are moles, volume is liters. So moles divided by liters gives us
moles per liter, which are the right units for concentration. So doing the number of moles
divided by volume gives us a concentration of 0.05 molar.
Now let’s look at an example where
we need another key equation.
28.5 grams of KOH was dissolved to
make 0.2 liters of solution. 0.067 liters of this KOH solution
completely reacted with 0.12 liters of nitric acid of unknown concentration. What is the concentration of the
Again, let’s start by writing out a
balanced equation. We’re told that our reactants are
KOH and nitric acid with the formula for nitric acid being HNO3. Again, we have an acid–base
reaction. So we’re going to produce a salt
and water. In this case, we have KNO3 and
H2O. And this equation is balanced. Let’s again try drawing out a table
to help us.
This time, however, we’ll need to
be careful to pick out the right data. For example, we’re given two
different volumes of KOH. So which one is the one that we
need? Let’s take this first sentence in
isolation. We’re told that we have 28.5 grams
of KOH to dissolve into a solution of 0.2 liters. This sounds like it’s leading us
towards a concentration of the KOH. And if we read the rest of the
question, we’re not actually told the concentration of the KOH. And that’s important. So let’s remember another key
Here, we have 𝑛 equals 𝑚 over
capital 𝑀, where 𝑛 is the number of moles, lowercase 𝑚 is mass in grams, and
capital 𝑀 is molar mass in grams per mole. Let’s use this to work out the
concentration of our KOH solution. The question tells us that we use
28.5 grams of KOH. So we can substitute this in for
lowercase 𝑚. The molar mass we can get from our
periodic table. We simply add together the masses
of K, O, and H. This means that the number of moles
is 28.5 grams divided by 56.105 grams per mole, which gives us 0.50798 moles.
Now we need to return to our first
key equation to work out our concentration. By rearranging our equation, we get
concentration equals number of moles divided by volume, which works out as 0.50798
moles from where we just calculated it and 0.2 liters given in the question, leaving
us with a concentration of 2.5399 molar, remembering that molar is the same as moles
per liter. So let’s add this to our table.
Now, we need the amount or volume
of KOH solution used in the titration. The question tells us that this is
0.067 liters. It then tells us that we reacted
this with 0.12 liters of acid. So we can add that to our table as
well. And the question is asking us for
the concentration of nitric acid. So now, let’s fill in as many boxes
as we can until we can work out the concentration.
Let’s start by working out the
number of moles of KOH. We do this by multiplying the
concentration by the volume, which gives us 0.17017 moles. Now, let’s go to our molar
ratio. The coefficients for both KOH and
HNO3 are one, so it’s a one-to-one molar ratio. This means that if we have 0.17017
moles of KOH, we need exactly the same number of moles of acid. Now that we have the number of
moles, we can work out the concentration. We do 𝑛 divided by 𝑣, giving us a
concentration of 1.418 molar.
Now that we have a rough idea of
how to do these calculations, let’s have a quick look at unit conversions. In our key equation of 𝑛 equals
𝑐𝑣, we’ve mainly been using the units of moles, liters, and moles per liter. Likewise, in our other key equation
of 𝑛 equals 𝑚 over capital 𝑀, we’ve been using the units of moles, grams, and
grams per mole. But what if the question doesn’t
give our data in these specific units?
Remember that if you see decimeters
cubed, these are the same units as liters. And centimeters cubed are the same
as milliliters. We’re often given values in
milliliters and need to convert to liters. So let’s look at how to do
that. Since there are 1000 milliliters in
every liter, we can convert milliliters to liters by multiplying by one liter per
1000 milliliters. Remember that milli- means
thousandth. So if we were to convert 200
milliliters into liters, we would multiply by one liter per 1000 milliliters, giving
us 0.2 liters. Notice that by using this method,
the milliliters cancel, and we’re left with liters.
The same conversion technique works
for converting milligrams into grams or millimolar into molar. So let’s try combining unit
conversions with carrying out titration calculations.
A 30-millimeter solution of nitric
acid was titrated against a 0.1-molar solution of potassium hydroxide. The addition of 26.6 millimeters of
potassium hydroxide was found to neutralize the nitric acid. What is the concentration of the
nitric acid? Give your answer to two decimal
This question is asking us to
perform a titration calculation. This means we’re going to need the
equation 𝑛 equals 𝑐𝑣, where 𝑛 is the amount in moles, 𝑐 is the concentration in
moles per liter, and 𝑣 is volume in liters. Let’s start by working out exactly
what reaction is occurring here.
We’re told that we’re reacting
nitric acid with potassium hydroxide. So this is an acid–base
titration. Let’s finish writing out our
reaction equation. When writing out our equation, we
should check that it’s balanced. And luckily, it is. Now, let’s extract the important
information from the question and put it into a table.
We’re told that we need 30
milliliters of nitric acid and 26.6 milliliters of potassium hydroxide. So these are our values for 𝑣. However, currently, they’re not in
the units that we want. To convert milliliters to liters,
we multiply by one liter per 1000 milliliters. By converting these values for 𝑣,
we get 0.03 liters of acid and 0.0266 liters of potassium hydroxide. Next, we’re given the concentration
of the potassium hydroxide as 0.1 molar. Remember that molar means moles per
liter. And the question is asking us to
work out the concentration of the nitric acid. So let’s fill in the boxes of our
table until we can work this out.
First, we can work out the number
of moles of potassium hydroxide. We do this by multiplying the
concentration and the volume, giving us 0.00266 moles. But how do we know how many moles
of acid we need? To work this out, we need to look
at the molar ratio. We can see from our balanced
chemical equation that both the nitric acid and the potassium hydroxide have a
coefficient of one. This means we have a one-to-one
molar ratio. So to neutralize one mole of acid,
we need exactly one mole of potassium hydroxide.
But we have 0.00266 moles of
potassium hydroxide instead. So this means that we need exactly
the same number of moles of acid. Now that we have 𝑛 and 𝑣 for our
acid, we can work out the concentration. We do the number of moles divided
by the volume in liters, giving us 0.0887 molar. However, we’re asked for two
decimal places. So our answer is 0.09 molar or
moles per liter.
So let’s look at the key
points. Our key equations for titration
calculations are 𝑛 equals 𝑐𝑣 and 𝑛 equals 𝑚 over capital 𝑀. In these calculations, we use the
volume of a known concentration to calculate an unknown concentration. The trick to doing these
calculations is to use a balanced chemical equation and to carefully extract the key
information, remembering to be careful with your units.