>> narrator: In this video,
we'll take a closer look at
titrations and calculations
involving titrations.
A titration is an analytical
lab technique in which
a standard solution is used
to determine the concentration
of an unknown solution.
The standard solution
is used as the titrant
in the titration.
It is the substance that is
added to, and passed through,
the burette during the course
of the titration.
The equivalence point of
the titration is the point
at which stoichiometric
equivalent amounts
of the hydrogen ion
and the hydroxide ion
have been combined.
The endpoint of the titration
is the experimental point
where no more titrant
should be added.
And this is determined
by an indicator
that changes colors.
There is also a dramatic
change in pH at or near
the equivalence point.
The equation we use when
solving titration calculations
is the following.
The molarity of the acid
times the volume of the acid
times the number of
hydrogen ions present
in the formula of the acid
is equal to
the molarity of the base
times the volume of the base
times the number of hydroxide
ions that are present
in the formula of the base.
Here's an example of
a titration calculation.
41.3 milliliters of a 1.5 molar
sodium hydroxide solution
are required to neutralize
45.0 milliliters
of sulfuric acid.
Find the molarity of
the sulfuric acid.
We will be using
the titration equation
that we just went over.
We set up a data table
where we indicate
what's given to us
and what we are
trying to solve for.
For the hydrogen ion,
which relates to
the sulfuric acid,
what we know is that
the volume of the acid
was 45.0 milliliters.
The "n" is the number
of moles of the acid,
which actually refers to
the number of hydrogen ions
that are present in
the formula of the acid.
And that number is 2
because the formula is H2SO4.
We do not know
the molarity of the acid.
That is what we're
trying to solve
through the calculation.
For the base, or the hydroxide,
what we know is the molarity
of the base is 1.5 molar.
We also know that
41.3 milliliters of that
1.5 molar solution was required
to complete the titration.
For "n," the hydroxide,
we put a 1 because there is
1 OH in the formula
for sodium hydroxide.
When we plug in the given values
to the titration equation,
and solve for the molarity
of the acid, we find that
the sulfuric acid has
a molarity of 0.69 molar.
Here is another
titration problem.
26.2 milliliters of a 1.8 molar
sodium hydroxide solution
are required to neutralize
25.0 milliliters
of hydrochloric acid.
Find the molarity of
the hydrochloric acid.
So, as in the previous question,
we're going to use
the titration equation.
We'll tabulate the data
that we're given and what
we're going to solve for.
For the acid,
or the hydrogen ion,
we know that 25.0 milliliters
of the hydrochloric acid
was titrated.
For "n," we choose 1,
because there is 1 hydrogen
in the formula of HCl.
We don't know the
molarity of the acid.
That's what the question
is asking us to solve for.
For the base, the hydroxide,
we know that
the molarity is 1.8.
For the volume of the base,
we know that it took
26.2 milliliters to reach
the endpoint of the titration.
For "n," we would choose 1.
Again, because we are
using sodium hydroxide,
and there's 1 hydroxide
in the formula.
Taking all the given values
and plugging them
into the equation,
and solving for the
molarity of the acid,
we find that this
hydrochloric acid solution
is 1.9 molar.
Here's one for you to try.
35.0 milliliters of
a 2.0 molar calcium hydroxide
solution are required to
neutralize 20.0 milliliters
of sulfuric acid.
Find the molarity
of the sulfuric acid
When you're ready, press pause,
try the problem on your own,
and then press play
and check your work.
Did you get 3.5 molar
for the sulfuric acid?
If you did, good job.
Here's how we solved for this.
For the acid, the hydrogen ion,
what we knew was that
20.0 milliliters of the
sulfuric acid was titrated.
For H2SO4,
we chose a 2 for "n"
because there are 2 hydrogens
in the formula.
Again, we were trying to
solve for the molarity
of the sulfuric acid,
so that was our unknown.
For the base, or the hydroxide,
we knew the molarity
of the base was 2.0 molar.
We knew that 35.0 milliliters
was required, in terms of
the volume of the base.
For "n," we chose 2,
because in calcium hydroxide,
there are 2 hydroxides
in the formula.
So when we took
all of those values
and plugged them into
our titration equation,
that's how we found that
the molarity was 3.5 molar
for the sulfuric acid.
And that's a closer look
at titration calculations.
For more "Chemistry Matters,"
go to our home page,
where the entire series
is available to you.