Chapter 6 Good Titrations
Problems 1, 2, 4, 6, 8, 10, 12, 14, 15, 17,19
In this analysis we determine the volume of a reagent needed to react with our analyte
6-1 Principles of Volumetric Analysis
Usually in the form of a Titration
In a titration increments of a reagent solution called the titrant are added to the
analyte until they have completely reacted. The titrant us usually added from a buret
so we can determine its exact volume
For a titration two things must occur
1. The K of the reaction must be very large
2. The reaction n must go quickly and completely.
3. There must be some way to determine when the reaction is complete
A. An indicator (or some reagent) can change color
B. A chemical potential may be monitored using a potentiometer (similar
to a pH meter)
C. The absorbance of the solution may be measured
In determining when the reaction is complete you need to be aware of two terms
1. Equivalence point This is the point at which you have added titrant that is
exactly equivalent to the analyte
2. End point - This is the point that you observe some change in solution.
Ideally the Equivalence point and the endpoint are exactly the same, but many
times they are a little off from each other. This lead to titration error. It is usually
possible to perform a blank titration where the procedure is repeated without an
analyte to see how large the titration error is.
In preparing solution for a titration you need to have a reagent that is of
sufficient purity (>99.9% pure) and stability that it can be quantified directly by
weighing. Such material are called Primary standards .
Often the titrant we use in a titration is not a primary standard, but is a second
reagent which we have made to approximately the correct concentration, and have then
accurately determined the concentration by comparing it to our primary standard. The
second reagent is called a secondary standard, and the process of calibrating the
secondary standard against our primary standard is called a standardization .
There are two general types of titrations; direct titrations, where the titrant is
added do the analyte until a reaction goes to completion, and a back titration, where
excess reagent is added to the analyte, and then the excess of this added reagent is
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determined.
6-2 Titration Calculations
The point here is that at the equivalence point you have equivalent amounts of titrant
and analyte. And trying to transofrm volume of titrant into amount of analyte.
There are three major steps
1. From volume of titrant determine moles of titrant used
2. From moles of titrant used determine moles of analyte consumed
(need a conversion factor here)
3. From moles of analyte consumed determine chosen unit of analyte to finish
problem
Summary Volume of titrant 6mole of titrant6Mole of analyte6 chosen units of analyte
In many reactions with 1:1 stoichiometery the above process is simple. Let’s try
an example like the text
RXN Ag+(aq) + Cl- (aq) W AgCl(s)
Say we have 12 ml of a Cl- solution that requires 22.50 ml of a .05269M Ag+
solution to form a complete precipitation reaction. What was the concentration of the
Cl- solution?
Step 1 Volume of titrant to moles of titrant
.02250 mole X .05269 mole/l = 1.186x10-3 moles
Step 2 moles of titrant to moles of analyte
1.186x10-3 moles of Ag+ X 1 mole Cl-/1 mole Ag+ = 1.186x10-3 moles Cl(Using conversion factor from equation)
Step 3 moles of analyte to final unit analyte
1.186x10-3/ .012l = .09879M Clthis step can be tricky depending on what I want the answer to be. Maybe
O wanted g of Cl- or maybe % Cl- in a sample of known weight. You have to do a little
thinking here.
In many reaction, like acid-base and complexometric, the stoichiometry of the
reaction is simple and one mole of analyte reacts with one mole of titrant. However in
other cases, oxidation reduction reactions in particular the stoichiometry can be messy
and you have to pay attention.
3
Take the Book example of Oxalic acid and permanganate
5C2O4H2
Oxalate
+ 2MnO4-+
6H+ 6 10CO2- + 2Mn2+ + 8 H20
Permanganate
If we have 1.063 x10-4 moles of oxalate, how many mole do permanganate does that
react with?
The only thing you have to change is in step 2 the conversion of moles of titrant
to moles of analyte. Lets assume the permanganate is the titrant, then in step 2 we
have
Mole MnO4- X (5 mole C2O4H2 / 2 moles MnO4-2)
So it is really just another unit conversion and no big deal if you keep your units
straight.
Section 6-3 Chemistry in a fishtank
do I want this? I skipped in 2006
An interesting series of titrations that take you through all the basics, as well as
throwing in a back titration. Basic problem Fish make NH3, which will kill the fish in a
closed tank. Luckily bacteria in the take convert NH3 to NO2- . Unfortunately NO2 is
also toxic, so yet a different bacteria converts to NO3- which is not toxic.
To monitor this system to need to be able to determine NO2- levels. There is an easy
spectrophotometric way (section 17-4) but since there are no NO2 salts that are primary
standards, how do we calibrate the system?
Solution is the two example calculations shown here:
Make a KMnO4 secondary standard by titrating against H2C2O4 primary standard
Make a NaNO2 standard by titrating against the standardized KMnO4
I won’t go over the numbers here, I’ll just give you an overview so you can work through
the problem
4
Because you are using a known amount of analyte to determine the molarity of
the KMnO4 the overall proess is revered from the way you have done it so far, so
lets go in reverse
3. Final units of analyte to moles of analyte
g Na2C2O4 6 moles Na2C2O4
2. mole of analyte to moles of titrant
(Remember to throw in a conversion for the reaction:
5 C2O42- + 2 MnO4- + 16H+ 6 10CO2 + 2 Mn2+ + 8H2O)
moles Na2C2O4 * (2/5)6 moles KMnO4
1. Moles of titrant to volume of titrant
moles KMnO4 6 M KMnO4 * V:
M = moles KMnO4/V
Now let’s throw in the numbers
Calibrating the KMnO4 secondary standard against the Na2C2O4 primary
standard:
Make standard solution of H2C2O4
Actually weigh out 3.299 g Na2C2O4
Adding excess H2SO4 to convert C2O42- to H2C2O4)
Solution is 3.299/134.0 moles in 1 l = .02462M
Remove a 25 ml aliquot so have
.02462 x .025 = 6.155x10-4 moles H2C2O4
From equation
6.155x10-4 moles H2C2O4×(2 mole MnO4-/5 mole H2C2O4)
= 2.462x10-4 mole KMnO4
Molarity = Moles / liters
= 2.462x10-4 mole KMnO4 / .02839 l KMnO4 (28.39 ml)
= 8.681x10-3 M
Now for the second titration using the equation
5NO2- + 2 MnO4- + 6H+ 6 5NO3- + 2Mn2+ + 3H2O
5
Pipet 25.00 mls of the KMnO4 solution and titrate with 34.76 ml of NaNO2
.025 L x 8.681x10-3 M KMnO4
= 2.170 x10-4 moles
From equation
2.170 x10-4 moles KMnO4 x (5 NO2-/2MnO4-)
= 5.425x10-4 mole NO25.425x10-4 mole NO2-/.03476 L
= .01561M NaNO2
6-4 Solubility Product
Now that you understand some of the basic calculations involved in a titration
calculation, lets change gears, and start to look at equilibrium expressions
Perhaps the simplest equilibrium system is the solubility product, Ksp
Calculating the Solubility of an Ionic Compound
The Ksp is used to describe the system where you have an ionic solid sitting on
the bottom of a beaker, and it is in equilibrium with the dissolved constituent ions in
solution ( a fancy way of saying a saturated solution).
Say we have solid PbI2 and it is in equilibrium with its ions Pb2+ and I-. The reaction is:
PbI2(s)W Pb2+ (aq) + 2I-(aq)
Ksp = [Pb2+][I-]2 = 7.9x10-9 (page 538 in back of book)
We typically use the Ksp to calculate the concentration of the ions in the solution. For
instance, If I add solid PbI2 to water, what is the maximum concentration of Pb2+ and Ithat I will find in the dissolved solution?
Well the solubility product tells us that:
[Pb2+][I-]2 = 7.9x10-9
So how do you come up with [Pb2+] and [I-]
Let’s start with the reaction tables that we used a little bit in Gen Chem
Initial
Final
PbI2(s)
solid
solid
W
Pb2+ (aq)
0
?
+
2I-(aq)
0
?
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What is stochiometric relationship between Pb2+ and I-?
[I-] = 2 [Pb2+]
so if for simplicity I make [Pb2+] = x
Final
solid
x
2x
and then plugging these values into the equilibrium expression
x (2x)2 = 7.9x10-9
4x3 = 7.9x10-9
x = cube root of (7.9x10-9/4)
=.0013
[Pb2+] = .0013, [I-] = .0026
If I want solubility in g/l
.0013 M PbI2 = .0013 moles/liter = (.0013 x 461 g/mol)/l
= ~.6 g/liter
one final note. Often use the Ksp to calculate solubility. It is often the major story, but
as I have shown you Analytical Chemists tend to be picky and so this is sometime not
entirely correct. So far we have assumed that the only way that the PbI2 can exist in
solution is in the form of its component ions. This is not always true some solids like
CaSO4 and exist in solution in an undissociated form hence a more proper equation for
this material is:
CaSO4(s) W CaSO4(aq) + Ca2+(aq) + SO42- (aq)
And the solubility equilibrium is a bit more complicated
even our simple PbI2 is more complicated, there can be other species called
‘complex ions’ that can also form
PbI2(s)W Pb2+ (aq) + 2I-(aq) + PbI+ + PbI3So things can get a bit tricky if you do them exactly, but for right now I won’t worry
about this level of complication. Maybe later
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The Common Ion Effect
We had the common ion effect in Gen Chem. What was it?
(Have two or more chemicals in solution, and the chemicals share a same
common ion)
For instance our PbI2 and add a second solution of NaI
Let’s look at solving this system
Looking up PbI2 and NaI in the back of the book. Have a Ksp for PbI2 but nothing
for NaI
Why?
NaI sodium salt should be completely soluble. If complete soluble then no Ksp
So if we have .030F NaI, what we realy have is .03M Na+ and .03M ISo let’s look at our reaction table again
PbI2
Initial
From PbI2(s) solid
From NaI
0
Initial total solid
W
Pb2+
0
0
0
+
2I0
.030
.030
But, as usual a little of the PbI2 will ionize, and as before I will call [Pb2+] X and by the
stoichiometery the I- that comes from PbI2 must be 2x
So our reaction table is:
RXN
solid
x
.030 + 2x
so our Ksp = 7.9x10-9 = x (.030+2x)2
= going to get messy
Can we make any assumptiuon to make this less messy? Think back to our
previous answer what was [I-] based on PbI2 alone? (.0026M)
With the common ion effect will the reaction shift to right or left
(Added a product, so will shift to left, back to reactant)
If it shifts that way then the concentration will be even lower!
How big is <.0026 compared to .03? Less than a tenth maybe much less
so I will assume .030+2x..03, solve the equation because it is simple this way,
then see if my assumption was OK
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so our Ksp = 7.9x10-9 = x (.030+2x)2 . x(.03)2
X=8.8x10-6
[Pb+] =8.8x10-6
[I-] = 1.76x10-5
This answer was easy to obtain, is the assumption that we used OK?
Is .03 + .0000176 . .03? Yes, so assumption OK
Either ion can be the common ion. What if we had our PbI2 in a solution of 1F
Pb(NO3)2?
(Nitrate salts are soluble so [Pb2+] = 1M
Now our net reaction table looks like
PbI2
W
Pb2+
Inital
PbI2
solid
0
Pb(NO3)2
0
1M
rxn
Solid
1+x
+
2I0
0
2x
Ksp = 7.9x10-9 = (1+x)(2x)2
Again this could get messy unless we assume that x is small so 1+x .1
7.9x10-9 = (2x)2
7.9x10-9/4 = x2
x = sqrt(7.9x10-9/4)
=.000044 = Pb+2
And 1+.00004 .1 was also a good assumption
6-5 Titration of a mixture & 6-6 Titration involving silver ion
Now let’s turn the Ksp reaction into a titration reaction. For this we will determine
the amount of Cl- in a sample by a precipitation reaction with Ag+ using the reaction:
Ag+(aq) + Cl- (ag)W AgCl(s) (the reverse of the Ksp reaction)
(K = 1/Ksp = 1/1.8x10-10 = 5.86x10+9)
The idea here is that since the Ksp of AgCl is low (1.8x10-10) as you form the salt
it will precipitate out of solution, removing Cl-. At the equivalence point you will have
added Ag+ exactly equal to Cl- and it will all have ppt’s out as a solid.
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How do you tell you are at the endpoint?
Well, each time you add some titrant you form some ppt, so the point where you
add titrant and don’t see a ppt would be one way of seeing the endpoint. But in
practice this is a lousy procedure. By the time you et to the E.P. you will have a
solution full of precipitated AgCl, so how can you tell that you aren’t making more ppt?
It will be virtually impossible.
For silver there are a couple of tricks (6-6 out of order and not in as much depth)
The Volhard method The Volhard method is essentially a backtitration method.
You add an amount of Ag+ that you know is an excess, and then you add some SCN- to
the solution to titrate the Ag+ . The indicator for this endpoint is Fe3+. When you add
one drop SCN- past the Ag+, you have excess SCN- and this reacts with the Fe3+ to form
FeSCN2+ which is red.
2006 did a demo spatula full of NaSCN, AgNO3 and Fe(NO3)3 in each of 3 test
tubes. Mixed to see endpoint color
The Fajans method In the Fajans method you add a dye that contains a charge so it likes to bind to + surfaces. The AgCl ppt that forms should be electrically
neutral, however when you have excess Cl- in the solution, these Cl- ions tend to
cluster on the ppt surface ready to bind, givng the surface of the crystal a negative
quality. Since our dye is also negative, nothing happens. After the E.P. we have
excess Ag+ in solution, and now these + ions tend to cluster by the surface, giving it a
+_ character, to our dye tries to bind, and we see a color change.
9-18-02 will try Demonstration 6-1 in class. In 2004 it worked well. Didn’t do in 2006
because they do it for themselves in lab!
Both of the above methods are specific for silver ions, although there may be similar
reactions for other species. Is there something that will work for all ppt reaction? Yes
using an electrode to follow the reaction. We will learn how electrodes work in Chapter
13. For now just think of the electrode as a way to measure concentration of an ion in
the solution using a special voltmeter. In fact you may have probably already done
this. When you measure the pH of a solution you are using an electrode to measure
the concentration of H+ With a little ingenuity you can make an electrode that will
measure the concentration of virtually any chemical you choose.
Figure 5-5 gives a set of 2 titration curves that are obtained using an electrode the first
(b) is the titration of a solution of I- with Ag+ (Ag+ is the titrant)
(a) is the titration curve of a mixture of I- and Cl- with Ag+
Notice in B it looks kind of like a normal titration curve the line is fairly flat for a while,
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and then it changes sharply at the equivalence point. Can yo expalin what is
happening in a?
2 equilvalence points, one for I- and one for ClWhy would you have 2 equivalence points?
Ksp AgCl
1.8x10-10
Ksp AgI
8.3x10-17
Which is first, which is second? AgI has smaller Ksp so it is less soluble so it will ppt
first so it is the 1st eq. Pt.
Bottom line - you can sometimes to titrations of mixtures and see different equivalence
points. You will need to think a bit about these calculations.