Titration Lab and Lecture
Saturday, February 22, 2014
4:17 PM
Slide
Notes
Lesson Helps Menu
The interactive is at
http://www.virtualhomeschoolgroup.com/file.php/1/03
_Science/04_Chemistry/General_Chemistry/M10_Acid_Base/L3
_Titration_Lab_and_Lecture/Published/SWF_and_HTML5/multis
creen.html
Lab starter is at …
http://www.virtualhomeschoolgroup.com/file.php/1/03
_Science/04_Chemistry/General_Chemistry/M10_Acid_Base/L3
_Titration_Lab_and_Lecture/Lab_Starter/Moore_Timothy_M10
_Lab_10_2_Acid_Base_Titration.doc
Acid Base Titration
This lab walk-through will give you an overview of the lab you
will do at home and show you the steps and equipment that
would be used if you had access to more expensive lab gear.
Titration is a lab method used to find the concentration of an
unknown acid or base through acid/base neutralization.
The Equipment
Apologia Chemistry Page 1
The Equipment
When you get to college, you will be working with a burette to
do the titration. Here you see one attached to the ring stand. The
reason a burette is used is that it gives very good control of the
delivery of the known acid or base. You can think of it as a very
large syringe where gravity is used instead of a plunger. It has a
tap near the bottom where you can adjust the flow rate. This is
much like the faucet you have at home. It also makes very
accurate measurements because the long narrow tube with
marked measurements gives much more accuracy than a
graduated cylinder can.
Audio correction:
I said mL divided by # drops. The monitor shows the correct way.
The equipment is several hundreds of dollars which is far too
much to spend on one lab done at home, so you will be using
inexpensive items that you probably already have around the
house. Instead of the burette, you will use an eye dropper or a
more scientifically accepted name would be a bulb pipette. You
have a significant disadvantage though. Your pipette doesn't
have the fine scale markings of the burette. That means you will
have to do a bit of extra work. You will have to calibrate how
many mL is in an average drop from your pipette. To do this, you
will fill your pipette and count the number of drops it takes to
reach the first scale marking on your beaker or graduated
cylinder. Then you will divide the volume by the number of drops
to get the volume of an average drop. Once you calculate this, I
recommend writing it down for long term use because you will
need this again in other labs.
Record the data for your calculations that you will do later in the
lab.
While you are working with your pipette, you might as well rinse
it with distilled water. Every item that you use in this lab will
need to be thoroughly rinsed with distilled water. If you have any
soap residue, it will throw off your pH readings.
Indicator
A pH indicators will help you identify the end point of the
titration. There will be a color change that you will need to watch
for. Different indicators are used for different types of acids and
bases. A commonly used indicator in laboratory environments is
phenolphthalein. It will change to pink when the end point is
reached.
For your home lab, you will make your own indicator from red
cabbage. Merely collect the juice from the steeped leaves. You
can store it for 1 week in the refrigerator if you do not use it right
away.
Known: Vinegar (C2H4O2)
In a real lab setting, you would know the molarity of the stock
solution that would be your known. In a home lab with only
access to locally available acids and bases, you will need to
calculate the molarity of your known which is the vinegar.
To do this, you will measure out 0.0500 L (50.0 mL) of vinegar
and find its mass. Record the measures in your lab data. You will
Apologia Chemistry Page 2
In a real lab setting, you would know the molarity of the stock
solution that would be your known. In a home lab with only
access to locally available acids and bases, you will need to
calculate the molarity of your known which is the vinegar.
To do this, you will measure out 0.0500 L (50.0 mL) of vinegar
and find its mass. Record the measures in your lab data. You will
use it later in math calculations to find the number of moles of
your known that you have.
You would add the vinegar into the burette if you were using
that apparatus. If using a pipette, place the vinegar into a
container that has been well-rinsed with distilled water and
allowed to dry. You will use this solution in your pipette.
Unknown: Ammonia
Now you will prepare your ammonia. Take 10 mL of household
ammonia and 90 mL of distilled water.
You will add about 25 mL of your indicator to this solution. It will
turn blue.
The Titration
Now we are ready to perform the titration itself. I will use the
burette here, but at home you will be using your pipette to add
the vinegar. With the pipette method you will have to count the
drops. Take care not to lose count or you will have to begin
again. With a burette there is no need to count because it has
measured markings that you can use once you reach your end
point.
Begin to add the titrant, which is vinegar, into the unknown. You
will likely see a blush of a color change right as the drop lands in
the solution but then the color will return to normal. As the color
change begins to linger longer slow the rate to mere drops at a
time and stir or swirl the unknown. Watch closely to catch the
endpoint because it can suddenly go past what you want. You
want to stop as soon as when you stir the solution the blush
doesn't go away. If you have a deep pink, more than likely you
added the drops too fast.
Record the number of drops if you used a pipette and if you have
the burette, record the measurement of how much titrant you
used to reach the end point.
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Lab Calculations
This calculation method is the one you would follow if you used
the pipette. It has a different first step than you would do if you
used a burette. To see how the burette method would be
calculated pay particular attention to the word problems in the
following slides.
The numbers from the lab here are entirely made up and are
provided merely to help you understand the math. Do not worry
if your lab results differ considerably.
For our pipette method, we do not have the molarity of the
vinegar, so we will convert grams to moles instead. To get the
grams, our given is how many mL of the vinegar we needed. We
will use a method in which we convert the number of mL of the
vinegar to the mass of the vinegar. That is why we took the mass
of 50 mL of vinegar. Essentially we use the data to bring the
density of the vinegar in to help us convert from mL to grams. I
simplified mine down here to how many grams there are in 1
mole. This will get us to our usual starting point for calculations
of the given over 1 for grams.
Audio correction:
I said grams per mole when I should have said grams per mL for the 0.98/1mL
From here our calculations are quite typical. We will find the
number of moles of vinegar. We will convert moles to moles
using a balanced chemical equation. Then we will compute the
molarity of our unknown.
Now it is time to work on the math from a more typical titration
in which a burette would be used and there would be no need to
work grams to moles in the first step. You will go from the
molarity of the known to the molarity of the unknown.
OYO 10.10
This slide will have a more typical titration and it is the type of
problem you are likely to see in the assessment. 125 mL of nitric
acid (HNO3) with unknown concentration is titrated against 2.3M
Mg(OH)2. If 35.4 mL of base is required in order to reach the
endpoint, what was the concentration of the acid?
Let's get the balanced equation. We can go ahead and get our
reactants in place. The hydrogen ion will be combined with the
polyatomic ion portion of the base to make H2O.
The magnesium will combine with the nitrate. We need to make
sure our subscripts are placed. Magnesium is in the second
column on the periodic table, so it gives its partner a 2. Nitrate
has a charge of 1 to give its partner. Ones can be dropped since
they are assumed if you do not see a subscript.
Audio Correction:
At the very end I said 1.3 moles instead of 1.3 molar
Now we will balance it. We will use the even/odd technique. The
magnesium is odd on both sides. The nitrogen is one on the
reactant side and two on the product so we will make a note that
the reactant side must be even. The smallest even number is
two. The magnesium on the product side is already two, so the
coefficient will be one there. That will mean there is a one for
the coefficient of Mg(OH)2. Counting up our totals for the
hydrogen and oxygen will give use a 2 for the water. It is
balanced.
From the problem, we have 125 mL or 0.125 L. Remember that
you just need to move the decimal over three places to convert
between mL and L. We are solving for the molarity.
In the problem we are given that we have 35.4 mL or 0.00354 L.
The molarity is 2.3 molar.
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The molarity is 2.3 molar.
We will be going from the molarity of the magnesium hydroxide
to the molarity of the nitric acid. When you have two substances
like that you will need all three parts of the solution process with
molarity at both ends.
We will plug in our numbers of the given and solve for moles.
For our moles to moles conversion we have a 1:2 ratio. When we
cross multiply and divide we will have 0.16 moles of nitric acid.
Now we calculate the molarity of the nitric acid. We have 1.3
moles.
Practice 10.9
Practice 10.10
Contact: teacher.tammy.moore@gmail.com
If you have any questions or comments
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If you have any questions or comments
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