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Analytical Chemistry
Lecture 7
10-30-02
Today I am going to talk about four topics: the Exam, Titrations, Concentration Units and
the Homework assignment.
Reading for next week: Chapter 10 and the end of Chapter 6
Exam
Mean 75, Median 82, Mode 100,91,89,88,85,81
You found it pretty easy
22 = 2x2 = 4; watch sig figs 0.276  0.3  11.564%
what is the difference between 1.7 (±0.2) x 10-6 and 1.7 x 10-6 (±0.2)?
copying errors (check your work) 8 x 10-13  8 x 10-1
accuracy and precision
(draw) scale
Now an overview of titrations.
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Titrimetry includes a group of analytical methods that are based on determining the
quantity of a reagent of known concentration that is required to react completely with the
analyte.
Reaction can be of many types – solubility, acid/base, redox, complex formation.
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The alkalinity and Mohr titrations you are doing involve Volumetric tritimetry where
the volume of a solution of known concentration (the standard reagent) that is needed to
react completely with the analyte is measured.
Titrimetric methods are widely used for routine analyses because they are rapid,
convenient, accurate, and can even be automated.
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A standard solution is a reagent of exactly known concentration that is used to carry out
a titrimetric analysis.
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A titration is performed by slowly adding a standard solution from a buret or other
liquid dispensing device to a solution of the analyte until the reaction between the two is
judged complete.
The volume of reagent needed to complete the titration is determined from the difference
between the initial and final volume readings.
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The equivalence point in a titration is reached when the amount of added titrant is
exactly chemically equivalent to the amount of analyte in the sample.
The equivalence point of a titration is a theoretical point that cannot be determined
experimentally. Instead, we can only estimate its position by observing some physical
change associated with the condition of equivalence.
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This change is called the end point for the titration.
Every effort is made to ensure that any volume difference between the equivalence point
and the end point is small. Such differences do exist, however, as a result of
inadequacies in the physical changes and in our ability to observe them.
The difference in volume between the equivalence point and the end point is the titration
error. The titration error is a systematic error that we must be aware of and, if possible,
correct for.
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Indicators are often added to the analyte solution in order to give an observable physical
change (the end point) at or near the equivalence point. Typical indicator changes
include the appearance or disappearance of a color, a change in color, or the appearance
or disappearance of turbidity (usually solid formation).
We often use instruments to detect end points. These instruments respond to certain
properties of the solution that change in a characteristic way during the titration. Among
such instruments are voltmeters, ammeters, and ohmmeters; colorimeters; temperature
recorders; and refractometers.
A pH meter is a voltmeter monitoring a glass electrode that changes voltage with the
activity of H+ in solution.
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A primary standard is an ultrapure compound that serves as the reference material for a
titrimetric method of analysis.
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Important requirements for a primary standard are
1. High purity. Established methods for confirming purity should be available.
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2. Stable in air.
3. Absence of hydrate water so that the composition of the solid does not change with
variations in relative humidity.
4. Ready availability at modest cost.
5. Reasonable solubility in the titration medium.
6. Reasonably large molar mass so that the relative error associated with weighing the
standard is minimized.
Compounds that meet or even approach these criteria are very few, and only a limited
number of primary standard substances are available to the chemist.
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As a consequence, compounds that are less pure must sometimes be employed in lieu of a
primary standard. The purity of such a secondary standard must be established by
careful analysis.
In order to determine the amount of analyte by titrimetry, we must know the
concentration of the standard solution of titrant accurately.
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This is accomplished by the process of standardization in which the concentration of a
volumetric solution is determined by using it to titrate a known mass of a primary or
secondary standard (or an exactly known volume of another standard solution).
These are the key terms used in describing titrations. Now let's look at the calculations
involved.
There are two types of calculations:
1. Determine the Analyte Concentration
2. Follow concentrations during the titration process
When you undertake a titration or any analysis for that matter, whether in the lab or in the
homework, the first thing you should do is write down a balanced chemical equation for
the chemical reaction involved. The equation contains the key stoichiometry that you
will need to relate the analyte with the titrant.
Suppose we are to determine the amount of iron in iron ore by dissolving the sample in
acid, reducing the iron to Fe2+ and then titration with a standard solution of potassium
dichromate.
The reaction is an oxidation/reduction reaction and so I am going to give you the
balanced equation. We will review balancing redox reactions next quarter.
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Cr2O72- + 6 Fe2+ + 14 H+ == > 2 Cr3+ + 6 Fe3+ + 7 H2O
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Now suppose we started with a 0.4891-g sample of iron ore and it takes 36.92 mL of
0.02153 M K2Cr2O7 to get to the diphenylamine sulfonic acid end point. What is the
percent Fe2O3 (FW 159.69) (%w/w) in the ore?
What do we need to get an answer? % iron in sample is found from mass iron/mass
sample. So we need the mass of iron from the titration.
I usually write stoichiometry calculations in one equation so that I am sure that units all
cancel, and so I don’t get round-off errors. You may prefer to do the calculation one step
at a time. Remember however to be sure the units cancel and to carry extra significant
figures until the end.
You can even set up just the units without the numbers
mL dichromate x L/mL x mol dichromate/L x mol/Fe/mol dichromate x
mol Fe2O3/mol Fe x g Fe2O3/mol Fe2O3 = g Fe2O3
What other chemical information do we need?
Need formula mass of Fe2O3
Need ratio of moles Fe2+ to moles Fe2O3
So we want to get the grams of Fe2O3 in the sample and then divide by the mass of the
sample to get percent by weight.
What is the measured analytical quantity? – mL of dichromate
36.92 mL dichromate x 1L/1000mL x 0.02153 mol/L x 6 mol iron/1 mol dichromate x
1 mole iron oxide/2 mol iron x 159.69 g Fe2O3 /mol = 0.3808 g Fe2O3 sig figs?
0.3808 g / 0.4892 g sample x 100% = 77.86% w/w Fe2O3
Before we go further, let me a few words about solution concentrations.
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So far we have talked primarily about concentration in terms of Molarity, moles/liter.
Both moles and Liters are fundamental SI units. It is a good thing we use liters, because
a cubic meter would be very hard to carry around.
Another term is Formality, the number of moles of solute, regardless of chemical form,
per liter of solution (F).
Both molarity and formality express concentration as moles of solute per liter of solution.
There is, however, a subtle difference between molarity and formality. Molarity is the
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concentration of a particular chemical species in solution. Formality, on the other hand,
is a substance's total concentration in solution without regard to its specific chemical
form.
There is no difference between a substance's molarity and formality if it dissolves without
dissociating into ions. The molar concentration of a solution of glucose, for example, is
the same as its formality.
For substances that ionize in solution, such as NaCl, molarity and formality are different.
For example, dissolving 0.1 mol of NaCl in I L of water gives a solution containing 0.1
mol of Na+ and 0.1 mol of Cl-. The molarity of NaCl, therefore, is zero since there is
essentially no undissociated NaCl in solution.
The solution, instead, is 0.1 M in Na+ and 0.1 M in Cl-. The formality of NaCl, however,
is 0.1 F because it represents the total amount of NaCl in solution.
Unfortunately, the rigorous definition of molarity, for better or worse, is largely ignored
in the current literature, as it is by our textbook. When we state that a solution is 0.1 M
NaCl we understand it to consist of Na+ and Cl- ions.
Thus you must know something about the chemistry of the substances, i.e. salt is very
soluble and will be completely dissociated in an aqueous solution.
Molar concentrations are used so frequently that a symbolic notation is often used to
simplify its expression in equations and writing. The use of square brackets around a
species indicates that we are referring to that species' molar concentration. Thus, [Na+] is
read as the "molar concentration of sodium ions."
Normality is an older unit of concentration that, although once commonly used, is
frequently ignored in today's laboratories.
However, Normality is still used extensively in Standard Methods for the Examination
of Water and Wastewater, a major source of analytical methods for environmental
laboratories.
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Normality makes use of the chemical equivalent, which is the amount of one chemical
species reacting stoichiometrically with another chemical species.
You have encountered this in geochemistry already.
Note that this definition makes an equivalent, and thus normality, a function of the
chemical reaction in which the species participates. Although a solution of H3PO4 has a
fixed molarity, its normality depends on how it reacts.
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The number of equivalents, n, is based on a reaction unit, which is that part of a chemical
species involved in a reaction. In a precipitation reaction, for example, the reaction unit
is the charge of the cation or anion involved in the reaction; thus for the reaction
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Pb2+(aq) + 2 I-(aq) <= => PbI2 (s)
n = 2 for Pb2+ and n = 1 for I-. Thus each mole of lead ions has two equivalents and each
mole of iodide has one.
In an acid-base reaction, the reaction unit is the number of H+ ions donated by an acid or
accepted by a base. For the reaction between phosphoric acid and hydroxide
H3PO4(aq) + 3OH-(aq) <= => PO43- (aq) + 3 H2O (l)
n = 3 for phosphoric acid and n = 1 for hydroxide.
Thus a 1 M solution of phosphoric acid in the above reaction is 3 N.
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Normality is the number of equivalent weights (EW) per unit volume and,
like formality, is independent of speciation. An equivalent weight is defined as the
ratio of a chemical species' formula weight (FW) to the number of its equivalents
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EW = FW / n
Consequently, the following simple relationship exists between normality and molarity.
N = nxM
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Lets calculate the equivalent weight and normality for a solution of 1.0 M H3PO4 in the
following reactions:
H3PO4(aq) + 2 NH3 (aq) <= => HPO42- (aq) + 2 NH4+ (aq)
For phosphoric acid, the number of equivalents is the number of H+ ions donated to the
base. For this reaction the number of equivalents is 2. Thus, the calculated equivalent
weights and normalities are
FW
97.994
EW = n = 2
= 48.997
N = n x M = 2 X 1.0 M = 2 N
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So if we have a 1M solution of phosphoric acid on the shelve is it 3N or 2N?
It depends on the reaction you use it in.
So most modern chemists avoid using normality. As long as the equilivents are related
to the charges on redox reactions that normality works.
Finally we get to Molality. What is this? Mols/kg of solvent. Molality is important
when solutions are very concentrated such as the lake brines you have seen in Eastern
Washington.
Molarity, formality and normality are based on the volume of solution in which the solute
is dissolved.
Since density is a temperature dependent property of a solution’s volume, and thus its
molar, formal and normal concentrations, will change as a function of temperature. By
using the solvent’s mass in place of its volume, the resulting concentration becomes
independent of temperature.
Now let’s look at the homework assignment.
Problem Assignment due next Wednesday:
Problems 6-26; 9-17,23,29; 7-7,8,11
Let’s start with problem 9-17.
9-17 a) Why do many rivers in Box 9-1. lie on the line [HCO3-] = 2[Ca2+]?
According to Box 9-1, the source of calcium in the rivers is the mineral calcite, which
dissolves by reacting with carbon dioxide in the river waver according to the equation:
CaCO3(s) + CO2(aq) + H2O < == > Ca2+ + 2 HCO3If the predominate product is bicarbonate and not carbonate or carbonic acid,then the
mass balance of this reaction is [HCO3-] = 2[Ca2+]. Rivers on the line [HCO3-] = 2[Ca2+]
are saturated with calcite.
b) What is happening in rivers that lie above the line [HCO3-] = 2[Ca2+]?
Rivers above the line [HCO3-] = 2[Ca2+] contain more bicarbonate than Ca2+. These
rivers are not saturated with calcite.
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c) The Rio Grande lies below the line [HCO3-] = 2[Ca2+]. Propose a hypothesis for what
might be happening in this river.
The Rio Grande has more Ca2+ than expected from dissolved calcite. Therefore there is
probably another source of calcium ions than calcite. This source is likely to be more
soluble than calcite. For example, CaSO4 found in anhydrite and gypsum is 104 times
more soluble than calcite. Perhaps the Rio Grande flows over minerals like this that are
more soluble than calcite.