CHEM 201 – Quantitative Analysis – Lecture 1

Review of Basic Concepts/Solutions and their

Concentrations

The universe isn’t “too complex” for our understanding,

because it’s our understanding that determines our universe.

Principles/Topics Covered

 Steps in Quantitative Analysis

 Methods in Quantitative Analysis
 Chemical Concentrations
 Stoichiometry
 Chemical Equilibrium

Terms to Understand

Students should be familiar, but not memorize terms at end of

each chapter.

Homework/Problem Sheets

Students should work all homework problems and problem

sheets.
Analytical Chemistry – the separation, identification and
determination of relative amounts of components (analytes).

a) Qualitative Analysis – chemical identity of analyte.

b) Quantitative Analysis – amount of analyte.

Involves two measurements:

1) Measurement of some physical property
proportional to amount of analyte in sample;
2) Volume or mass of sample.

*Combined= concentration of analyte to be determined

-Analytical methods classified according to nature of physical

property measured.

Classical
1) Gravimetric – mass of analyte;
2) Volumetric (Titrimetric) – volume of solution
containing sufficient reagent to react completely with
analyte.

Instrumental
1) Spectroscopic – measures electromagnetic radiation
absorbed or emitted by analyte;
2) Electroanalytical – properties resulting from
oxidation/reduction behavior or analyte.
Questions/Discussions when choosing method

1) What level of accuracy and precision is required

2) Interferences
3) Number of samples

Sampling issues (environmental)

1) Must be representative of bulk sample, e.g. water body

2) May limit accuracy and precision
3) Contamination (need for in situ instrumentation)

Preparing a sample

1) Replicates – portions of sample treated identically

2) Homogeneity
3) Matrix considerations
4) Isolation – separation, interferences

Calibration & Measurement

-measurement of the physical property is directly proportional
to the concentration:

CA =kx

where, CA = concentration of analyte, k = proportionality

constant and x = physical property measured.
 Calibration Graph
7

4
Y

0
1 2 3 4 5 6
X

Solutions and their concentrations

Molar concentration (CA) - Molarity

CA = # mol solute
# L solution

Molality – concentration expressed as moles per substance per

kg of solvent (not total solution).

M = mol solute
kg solvent

Normality (N) = N=nM

Where n is the # of electrons donated or accepted by a species
in a chemical reaction.
Lecture 2- Experimental Errors & Statistical
Analysis

- no analysis is free of error or “uncertainty”

Goal: limit errors and define tolerable level or

limits

Analytical Procedure

2-3 replicates are performed and carried out

through the entire experiment

-results vary, must calculate “central” or best value

for data set.
Mean – “arithmetic mean”, average ( x )

Mean =

x

Median – arrange results in increasing or

decreasing order,

Rules:
For odd # values, median is middle value
For even # values, median is the average of the two
middle values
Ex: 19.4, 19.5, 19.6, 19.8, 20.1, 20.3

Mean =
= 19.4 + 19.5 + 19.6 + 19.8 + 20.1 + 20.3

x

n 6

= 19.78

19.6 + 19.8
Median = = 19.7
2

How do we determine error?

Accuracy – closeness of measurement to its true or

accepted value

Precision – agreement between 2 or more

measurements of the sample made in exactly the
same way
Absolute error (E) – difference between true and
measured value

E = xi – xt
where xi = experimental value
xt = true value

*Ex: See Fig. Iron data

x = 19.78 ppm Fe
xt = 20.00 ppm Fe

E = 19.78 – 20.00 ppm = -0.22 ppm Fe

(-) value too low, (+) value too high

Relative error – expressed as % or in ppt

x
xt
Er = i x 100 (as %)
t

x
xt
Er = i x 1000 (as ppt)
t
Several ways to describe precision

Standard Deviation (S)

n( x  x) 2
S=
i
i = 1 n 1

Variance (S2) = square of the standard deviation

n-1 = degrees of freedom

Another name for precision…..

Relative Standard Deviation (RSD)

RSD (ppt) = ( S ) x 1000

X

When express as a percent, RSD termed the coefficient of

variation (Cv).

Cv (%) = ( S ) x 100
X
Ex: Four measurements: 51.3, 55.6, 49.9 and 52.0 –
Calculate the %RSD of the experiment.

1.

2.

3.
Types of Errors in Chemical Analysis

1. Indeterminate (random) errors

- no identifiable cause
- can’t be eliminated
- random, yields both high and low values

2. Determinate (systematic) errors

- have a definite source, identifiable and corrected
a. Instrumental – imperfections in measuring device
b. Method – non ideal chemical or physical behavior
c. Personal – carelessness, personal limitations

Reporting Analytical Data

1. RSD or Cv, with # of data used, i.e. n=4

2. Significant figures

3. Report 90% or 95% confidence intervals

Significant figures – all the certain digits and the 1st

uncertain, min # of digits needed to write a value in
scientific notation.
Rules for determining sig. fig’s…

1. Disregard all initial zeros

0.03026
not significant – only mark with decimal place

2. Zeros are significant (1) in the middle of a # or (2) at the

end of a # on the right hand side of decimal.

104 0.0106 0.106 0.1060

Use of sig figs in computations

Answer can only have the same # of decimal places as that
carried by the # with the fewest.

3.4 + 0.020 + 7.310 = 10.730 = 10.7

*uncertainty lies in the tenth place of 3.4

Addition and subtraction – sci notation

1. Express all #’s with the same exponent
2. Align all #’s w/respect to the decimal point
3. Round off #’s according to the # of decimal places in the
# w/the fewest decimal places.
Ex:
1.62 x 105
+ 4.107 x 103
+ 0.984 x 106
----------------
Multiplication and division
* normally limited to # of digits in the # w/ fewest sig figs
* the power of 10 has no influence on the # of figs retained

3.26 x 10-5 4.3179 x 1012 34.60

x 1.78 x 3.6 x 10-19 ÷ 2.462 87
---------------- ------------------ -------------
5.80 x 10-5 1.6 x 10-6 14.05

Logarithms and antilogarithms

logarithm – composed of a characteristic and a mantissa.

Characteristic is the integer part, mantissa the decimal.

1. In a logarithm of a #, keep as many digits to the right of

the decimal (mantissa) as there are sig figs in original #

2. In an antilogarithm, keep as many digits as there are

digits to the right of the decimal in the original #.

log 339 = 2.530 How to write in sci. notation = 3.39 x 102

a. Rule 1
log 4.000 x 10-5 = -4.3979400 = -4.3979

b. Rule 2
antilog 12.5 = 3.16227 x 1012 = 3 x 1012
Rules for Rounding final data

1. If last digit of a # is 0-4, round down

2. If last digit of a # is 6-9, round up

3. If last digit of a # is 5, round to nearest even #

Don’t round any # until calculation is complete…round at

end.

Ex: mean of data set is 61.555 and std = 0.069

For the mean, do we round to 61.55 or 61.56?
Look at rule 3, we see round to nearest even # = 61.56 and
std = 0.07

Distribution of repeated measurements (large # of data)

Standard deviation gives the measure of the spread of a set

of results, but not shape of distribution.

X and S are reasonable estimates of mean and standard

deviation, for small # of samples.

population = large # of data when determining distribution

and comparing two results.

X = µ and S =

Where µ is the population mean,
= population std

*the exact value of µ for a population of data can never be
determined; it requires an infinite # of measurements to be
made.

However, we can set limits (confidence limits) = an

interval around x that probably contains µ

The magnitude of the confidence limits is termed the

confidence interval.
Comparison of Two Standard Deviations (precision) using
F-test

s12
Fcalculated =
s22

Rules: If Fcalculated > Ftable (95%) = significant

Are the two standard deviations sig. different?

Fcalculated = 0.001382/0.000142
= 93.1

Significant? Yes according to table

Treatment of Outliers

1. Reexamine for gross errors.

2. Estimate precision to be expected.

3. Repeat analysis if time and sufficient sample is

available.

4. If analysis can’t be repeated, perform Q-test.

5. If Q-test indicates retention of value, consider

reporting median.
Review of Last Lecture
-Types of Errors
-Significant Figures
-Mean, Standard Deviation, RSD, etc…
-Gaussian Curve
-Confidence Intervals
-Comparison of Standard Deviations with the F test
-Calibration Issues

Comparison of Means (Student’s t-test)

Rejection of Data (Q-test)
1
Calibration Cont…
Limit of Detection (LOD) - lowest amount of analyte in a sample which
can be detected but not necessarily quantitated as an exact value.

- mean of the blank sample plus 2 or 3 times the SD obtained on the blank
sample (i.e., LOD = meanblk + Zsblk)

LOD calculation - alternative

Data required:
(1) calibration sensitivity = slope of line through the signals of the
concentration standards including blank solution
(2) standard deviation for the analytical signal given by the blank solution

3x SD blank signals
LOD = slope of signals for std's
2
Calibration Cont…

Determining Concentration from Calibration Curve

Basic steps:
(1) Make a series of dilutions of known concentration for the
analyte.
(2) Analyze the known samples and record the results.
(3) Determine if the data is linear.
(4) Draw a line through the data and determine the line's slope
and intercept.
(5) Test the unknown sample in duplicate or triplicate. Use the
line equation to determine the concentration of the analyte: y =
mx + b

reading
intercept
Concanalyte = slope
3
 Chemometrics

Application of mathematical, statistical, graphical or symbolic methods

to maximize chemical information.

-However, this definition can be expanded to include:

biology (biometrics),
environmental science (environmetrics),
economics (econometrics)

-Two lines of development:

– experimental design: planning and performing experiments in a
way that the resulting data contains the maximum information
about stated questions.
– multivariate data analysis: utilizing all available data in the best
possible way. 4
Chemometrics

1. Basic Statistics
2. Pattern Recognition and Classification
3. Optimization and Experimental Design
4. Multivariate Calibration Techniques
5. Quality Assurance and Good Laboratory
Practice

5
Chemometrics
1. Basic Statistics Descriptive

Spectrophotometric measurement (Abs) of a sample

solution from 15 replicate measurements.

Measurement Value Measurement Value

1 0.3410 9 0.3430
2 0.3350 10 0.3420
3 0.3470 11 0.3560
4 0.3590 12 0.3500
5 0.3530 13 0.3630
6 0.3460 14 0.3530
7 0.3470 15 0.3480
8 0.3460
6
Chemometrics
Descriptive statistics for the spectrophotometric
measurements.
Parameter Value
Sample #, n 15
Mean 0.3486
Median 0.347
Std Dev 0.00731
RSD % 2.096
Std error 0.00189
Max value 0.363
Min value 0.335

Statistical Tests – Student’s t-test, F-test, tests for

outliers 7
8
 Chemometrics
Multivariate Dataset
2. Pattern Recognition and n x p (samples x variables)
Classification
Organization, Preparation &
Inspection

Preprocessing:
Centering, Scaling

HCA= HCA=
PCA
Samples Variables

Evaluations (selection of Relationship

Cluster Selection factors, scores and loadings) among variables

Reveal patterns within

dataset

Guidance to seek out

9
additional data (supporting)
Chemometrics
Hierarchical Cluster Analysis (HCA) – interrelationships
between samples presented in the form of a dendogram.

Each sample is initially considered as an individual cluster,

and the clusters are progressively combined using a measure
of similarity.

The use of HCA is to emphasize natural groupings. The

length of the branches connecting two clusters is inversely
related to their similarity: the longer the branch, the less the
similarity.

10
Chemometrics

At a chosen similarity value (0.46 here), we can define and color 4 clusters –
blue, red, green and orange - in the samples. These clusters are also clearly
visible in the projections of samples generated by Principal Component
Analysis. 11
 Chemometrics
Principle Component Analysis
(PCA) – a mathematical
manipulation of a data matrix
where the goal is to represent the
variation present in many samples
and/or variables using a small
number of “factors.”

The samples are plotted in a 3D

space, with the first three
components (or factors) defining
three axes, and sample points are
color-coded according to the
grouping in a cluster analysis.

12
Outliers

Treatment of Outliers

1. Re-examine for Gross Errors

2. Estimate Precision to be Expected
3. Repeat Analysis if Time and Sufficient Sample is Available
4. If Analysis can not be Repeated, Perform a Q-Test
5. If Q-Test Indicates Retention of Value, Consider Reporting
the Median

13
Chemometrics
3. Optimization and Experimental Design
Allow the experimenter to better understand and evaluate the
factors that influence a particular system by means of statistical
approaches.
The relationship between the various factors and response
within a given system can be shown mathematically as
follows:
y = f (x1, x2, x3…, xk)

where y is the response of interest in the system and x are the

factors that affect the response when their values change.

Applications:
Screening- the factors that influence the experiment are identified.
Optimization- the optimal settings or conditions for an experiment are found.
14
Chemometrics
Experimental Designs
1. Full Factorial Designs (two levels per factor)
2. Fractional Factorial Design
3. Latin Squares
4. Greco-Latin Squares
5. Response Surface Designs (more than two levels for one or more
factors)
6. Box-Behnken Designs
7. Mixture Designs
In general, the following types of factors can be distinguished: 1)
continuous, e.g. temperature; and, 2) discrete, e.g. experimenter.

Factors are considered to be independent if there is no relationship

between them and dependent if a relationship exists. 15
Chemometrics
Full Factorial Designs (two levels per factor)
Ex: The effects of reaction temperature and pH in determining the
spectrophotometric response (absorbance) of a standard analyte solution.

Figure shows a graphical definition of the experimental domain.

The best experimental points in the domain are located in the corners A,
B, C and D as follows:
A (40oC, pH 1); B (60 oC, pH 1); C (40oC, pH 3); D (60 oC, pH 3)

3 D
C
Factor 2 (pH)

1 A B

40 60 16
Factor 2 (Temperature oC)
Chemometrics
The four trials of experimental matrix are shown in the Table, with the
results of each experiment indicated in the response column and the
factor levels in the rows below the experimental matrix.

Expt. Temperatu pH Response

number re
A -1 -1 y1
B +1 -1 y2
C -1 +1 y3
D +1 +1 y4
Factor
levels
(-) 40oC pH 1
(+) 60oC pH 3

Note that -1 is used for the low level of each factor and +1 for the high level.
17
Chemometrics
If we introduce another variable (e.g. reagent concentration) in the
experiment, it is then possible to represent the factors as faces on
one or more cubes with the responses at the points.

G H

+1 C D

X3
E F
+1

X2
-1
A B -1

-1 X1 +1

18
Chemometrics
Mixture Designs
The independent factors are proportions of different
components of a blend and often measured by their portions,
which sum to 100% or normalized to 1, i.e.
N
 xi = 1 for xi
0
i=1

The design region for mixture proportions is termed a simplex. Simplex is

a simple optimization algorithm seeking the vector of parameters
corresponding to the global extreme (maximum or minimum) of any n-
dimensional function F(x1, x2,..,xn).

19
Chemometrics
Ex: If the optimization of two factors occurs, the simplex will be a triangle.
Points labeled 1, 2 and 3 define the first simplex with the worst response
found at point 3.
V
2

6 7
8
4 5

2 3
1

V
1

As the simplex continues along the path of the surface, points 1, 2 and 4
form a new simplex and the response is measured for the combination of
factor levels given by 4. 20
Chemometrics
Multivariate Calibration Techniques
Traditional univariate calibration techniques involve the use of a
single instrumental measurement to determine a single analyte.

In an ideal chemical measurement using high-precision

instrumentation, an experimenter may obtain measurements linearly
related to analyte concentration:
y

x 21
 Chemometrics
Limitations of Univariate Techniques
-Very sensitive to the presence of outlier points.

-Selectivity and interferences (chemical and physical) - causing

some degree of non-linearity.

-Univariate techniques are not well suited to the multitude of data

collected from instrumentation currently being used in the
analytical sciences.
y

x
22
Chemometrics
CLASSICAL LEAST SQUARES (CLS)

CLS approach is best applied to systems where the concentration of

every analyte in the sample is known and obeys a linear relationship
with measurement vectors.

For a single wavelength and a single analyte, this relationship (Beer’s

Law) can be explained mathematically by the following equation:

A
= 
bc

where A
= the absorbance at wavelength
; 
= molar absorption

coefficient at wavelength
in L mol-1 cm-1; b = cell pathlength (cm);
and c = concentration of the analyte (mol L-1).

23
Chemometrics
INVERSE LEAST SQUARES (ILS)

The dependent variable (concentration) is solved by calculating a

solution from multiple independent variables (responses at the selected
wavelengths).

In ILS, we can combine the absorptivity coefficient (
) and cell

pathlength (b) from Beer’s Law to form a single constant relationship
(matrix notation) with concentration:
c = P A

where P = the matrix of coefficients.

ILS used in complex mixtures where it is not necessary to obtain the

spectra of pure analytes present.
24
Chemometrics

Simple Complex
System under study

CLS ILS

DCLS ICLS MLR

PCR PLS

25
Chemometrics
Approach Advantages Disadvantages
CLS Used in estimating multivariate limits of Not useful for mixtures with
detection, often based directly on Beer’s components that interact
Law
Used for Moderately complex mixtures Requires knowledge of all
components in calibration
mixture
Wavelength selection is not necessarily Susceptible to baseline effects
required for calibration
Averaging effects make it less Interferences must be included
susceptible to noise in model

ILS Used in estimating multivariate limits of Wavelength selection can be

detection, often based directly on Beer’s difficult and time consuming
Law
Allows calibration of very complex Accurate calibration often
mixtures requires large numbers of
samples
Calculations are relatively fast Number of wavelengths used
in the model limited by the
26
number of calibration samples
 Laboratory Practice
Quality Assurance and Good Laboratory Practice
Validation – the assurance that an analytical procedure
provides reproducible and secure results.

Validation Criteria:
1. Precision
2. Dynamic Range
3. Trueness
4. Selectivity
5. Limit of Detection
6. Limit of Determination
7. Robustness
27
Laboratory Practice
Internal Quality Assurance

Control Samples:

1. Standard Solutions
2. Blank Samples
3. Real Samples
4. Synthetic Samples
5. Certified Standard Reference Materials

Control samples should be analyzed at least once or twice in each

series of analyses to monitor the accuracy of measurements.

28
Laboratory Practice

External Quality Assurance

Laboratory Intercomparison Studies:

1. Standardization of Analytical Procedures

2. Controlling the Analyses of a Laboratory
3. Preparation of Certified Reference Materials

29
 Lecture 4 - Gravimetric Analysis

1. Precipitation Methods – dissolved analyte converted to

sparingly soluble precipitate.

a. readily filtered
b. low solubility
c. converted to product of known composition (heat)

Ex. Excess of oxalic acid (H2C2O42-) added carefully to

measured volume of Ca2+.
(1) In basic sol’n:

Ca 2 +(aq) + C2O4 (aq)
CaC2O4 (s)

(2) CaC2O4(s) is collected in a filtering crucible then dried

(3) CaC2O4 ignited to produce calcium oxide:

CaC2O4 (s)  CaO(s) + CO(g) + CO2 ( g )

(4) CaO(s) cooled, weighed

(5) Original concentration of Ca2+ computed

2. Volatilization Methods

a. Analyte is volatilized at suitable temperature

b. Volatile product is collected and weighed

1
3. Precipitates – Particle Size & Filterability

a. Colloids – (d = 10-7 to 10-4 cm)

-invisible to naked eye
-not easily filtered, don’t settle out of solution

b. Particles – (0.10 mm or greater)

-spontaneously settle out of solution
-readily filtered and washed free of impurities
-more desirable

*Size of particles influenced by relative supersaturation

of the solutions in which is formed:

Q
S
Relative Supersaturation =
S
Where Q = concentration of solute, S = solute’s equilibrium
constant

-Precipitate solubility
-Temperature
-Reactant concentration
-Rate of reactant mixing

Q
S
If is large = small particles (colloids)
S

Q
S
If is small = crystalline solid likely
S

2
 c. Crystalline Formation

(1) Raising temperature (increases S)

(2) Using dilute solutions (minimizes Q)
(3) Slow addition of precipitating agent w/stirring
(minimizes Q)

4. Mechanism of Precipitate Formation

a. Nucleation – formation of a stable solid due to # of

atoms, ions or molecules join together, e.g.
formation on surface of suspended contaminants
(dust particles).
b. Particle Growth – growth on existing nuclei

Q
S
high – rate of nucleation increases
S

Q
S
low – particle growth dominates, excluding nucleation
S

*Nucleation dominates – results in a large # of very

fine particles

*Particle growth dominates – small # of larger particles

3
 5. Gravimetric Calculations

Wanalyte
% Analyte = x 100
Wsample

CaCl2 + 2AgNO3
2AgCl2 (s) + Ca(NO3)2

FW CaCl 1 mol CaCl

wt CaCl2 = wt AgCl x ( 2x 2)
FW AgCl 2 mol AgCl

Gravimetric Factor (F)

F – Relates mass of product to mass of analyte, stoichiometry

a(FW of substance A)
F=
b(FW of substance B)

where a and b are the coefficients of A and B, respectively

4
Ex. Calculate the % Phosphorus in a 0.3516 g detergent sample.
Final yield is 0.2161 g Mg2P2O4

Mass Analyte
%Analyte = x 100
Mass Sample

a. Mass P =
1 mol Mg P O 2 mol P 30.97g P
0.2161 g Mg 2P2O4 x 2 2 4 x x
222.57 g Mg P O 1 mol Mg P O 1 mol P
2 2 4 2 2 4
Mass product Gravimetric Factor

= 0.0614 g P

0.0614g P
%P= = 17.10 %
0.3516g sample
or…

a FW analyte
F= x
b FW sample

2 30.97g
F= x = 0.27833
1 222.57g

(0.2161g Mg P O )(0.27833)
%P= 2 2 7 x100 = 17.10%
0.3516g

5
Ex: A 10.00 mL solution containing Cl- was treated with excess
AgNO3 to precipitate 0.4368 g of AgCl. What was the molarity of
Cl- in the unknown?

Formula mass of AgCl = 143.321. A precipitate weighing 0.4368 g

contains:

0.4368g AgCl
= 3.048 x 10-3 mol AgCl
143.321g AgCl / mol AgCl

*Note: 1 mol of AgCl contains 1 mol of Cl-

3.048 x 10 - 3 mol AgCl

[Cl-] = = 0.3048 M
0.0100L

* IN CLASS PROBLEMS TO FOLLOW!

6
Ex1:Phosphate is precipitated from its solution with ammonium
molybdate, as (NH4)3[PMo12O40•xH20]. Since the precipitate does
not have a constant composition with regard to water content, it is
dissolved in ammonia and the molybdate is precipitated with
Pb(NO3)2, as PbMoO4.

a) What is the value of the gravimetric factor for the calculation of

%P?

b) If the final precipitate weighs 0.100 g, what is the weight of P in

the initial sample?

7
Ex2: A 0.2025 g sample consisting of only BaCl2 and KCl required
20.25 mL of 0.1200 M AgNO3 solution for the quantitative
precipitation of chloride. Calculate the %Ba and %K in the sample.

Ex3: A 0.4994 g sample of a hydrate of CuSO4 • xH2O, is heated

to a constant weight of 0.3184 g (total loss of water). Calculate the
value of x.

8
Ex4: In the gravimetric determination of sulfate in a 0.2841 g
sample of pure Na2SO4, a BaSO4 precipitate weighing 0.4604 g
was obtained. The weight of the precipitate was smaller than the
theoretical one, since some BaSO4 was converted to BaS during
the heating process.

a) Calculate the per cent of BaS in the precipitate (Hint: best to

solve algebraically).

b) The per cent error of the analysis (Hint: compare calculated

with weight stated in problem).

9
 Lecture 5 – Volumetric Analysis/Titrations

Methods for Establishing Concentration of

Standard Solutions

1. Direct – carefully weighed out sample into a

volumetric flask
2. Standardization - solution is standardized by:
a. titration with a weighed quantity of a
primary std
b. weighed quantity of secondary std
c. or measured volume of another std
solution

Stoichiometric Calculations in Volumetric

Analysis
1. amount A = # mol A = wtA( g )
fwA( g / mol )

2. amount A = # mol A = V (L) x CA (mol A /L)

* Solution preparation – lecture 1

* When primary std’s not available, use

standardization method
Volumetric titration – addition of a standardized
solution (titrant) from a buret (or other volumetric
device) to a solution of analyte until reaction
completed (equivalence point).

Key: Relate # moles of titrant to moles of analyte

Titration Curves in Titrimetric Methods
(precipitation)

End point = some physical change that occurs near

the equivalence point of a titration

Equivalence point - reaction b/t titrant and analyte

is complete.
Color change – most recognizable indicator
(p.367)

A + T
AT

In + T
InT

For change - [InT ] change by 1-2 orders of

[In]
magnitude

Titration curves – allow understanding of the basis

of end points and titration errors.

1. Sigmoidal – small region (0.1-0.5 mL)

surrounding the equivalence point
2. Linear – measurements made on both sides of
the equivalence point.
Using data near the equivalence point

Concentration changes during titration

Vol 0.1000 T (mL) Conc. A (mol/L) pA
0 1.00 x 10-1 1.00
5.00 6.67 x 10-2 1.18
10.00 4.29 x 10-2 1.37
24.00 2.04 x 10-3 2.69
24.90 2.00 x 10-4 3.70
24.99 2.00 x 10-5 4.70
25.00 2.24 x 10-6 5.65
25.01 2.50 x 10-7 6.60
25.10 2.50 x 10-8 7.60
26.00 2.50 x 10-10 9.60
40.00 1.67 x 10-10 9.78
45.00 1.25 x 10-10 9.90
50.00 1.00 x 10-10 10.00

0.12

0.1
Concentration A (M)

0.08

0.06

0.04

0.02

0
0 5 10 15 20 25 30
Volume T (mL)
Graphs of titration curves
10
9
8
7
6
5
4
3
2
1
0
0 10 20 30 40 50

10
9
8
7
6
5
4
3
2
1
0
0 10 20 30 40 50
Factors Influencing Endpoints

1. Concentrations- increases in analyte and

reagent concentrations enhance the change in
pA (equivalence region). Rule: change of 2 in
p-function.

2. Reaction completeness – end points improve

as reaction becomes more complete.
-product solubility large factor

Titration curves of mixtures

Direct Titration
-titrant reacts directly with the analyte:

amount analyte = amount titrant (moles or mmoles) x

reacting ratio

H3PO4 + 2 NaOH
Na2 HPO4 + 2H 2O

analyte titrant

Stoichiometric ratio: 2 moles NaOH needed to

react with 1 mole H3PO4

1H PO
amt of H3PO4 (mmol) = amt NaOH (mmol) x 3 4
2 NaOH

wt H3PO4 = amt H3PO4 x fw H3PO4

wtH PO
% H3PO4 = 3 4 x100
wt sample

****IN Class Calculations to Follow!

 Spectrophotometry – Fundamentals

1. Electromagnetic radiation – properties

Electromagnetic radiation
E

Photons

M
Propagation direction

Relationships

a. Frequency/Wavelength - νλ = c
where c = the speed of light in vacuum (2.998 x 108 m/s)

b. Energy/Frequency – E = hν
where h = Planck’s constant (6.626 x 10-34 J s-1), ν (Hz)

E = hc = hcν
λ

1
where ν = and called the wavenumber
ν
2. Electromagnetic Spectrum

Why is a red solution red?

e.g. FeSCN2+

Wavelength (nm) Color Complementary color

400-435 Violet Yellow-green

435-480 Blue Yellow
480-490 Green-blue Orange
490-500 Blue-green Red
500-560 Green Purple
 3. Absorption of Light

Term & Symbol* Definition Alt. Name & Symbol

Radiant power Energy of radiation Radiation intensity
P, P0 I, I0

Absorbance A Log P0/P Optical density D

Transmittance T P/ P0 Transmission T

Path length of ----- l,d

radiation b

Absorptivity a A/bc Extinction

coefficient k

Molar absorptivity ε A/bc Molar extinction

coefficient

* Recommended by American Chemical Society

 b
%T A
100 0.00
P0 P 90 0.05
80 0.10
Absorbing solution
(concentration c) 70 0.15
0.20
60
0.25
Transmittance T = P/ P0 50 0.30
0.35
40 0.40
Absorbance A = log P0/P = -logT = abc 0.45
0.50
30
0.60
20 0.70
0.80
Beer’s Law 10 1.00
1.50

a. Beer’s Law
A = abc(g L-1) or A = εbc(mol L-1)

Ex: A 7.50 x 10-5 M solution of KMnO4 has T of 36.4%

when measured in a 1.05-cm cell at 525nm.

(1) Calculate the absorbance of this solution:

(2) Calculate the molar absorptivity of KMnO4:

b. Limitations to Beer’s Law

(1). Real Limitations – high concentrated solutions,

concentrated electrolyte solutions (proximity alters molecular
absorption).

1.0 1 - 10 mg l
-1
NO 3 -N

0.8 r = 0.8791

0.6

0.4

0.2

0.0
0 2 4 6 8 10
-1
[NO 3-N] (mg l )

(2) Chemical Limitations – absorbing species participate in

association or dissociation reactions, e.g. weak acids in
concentrated solutions, complexation.

HNO + H O ←→ H O + NO −
2 2 3 2

(3) Instrumental Deviations – polychromatic radiation used

to measure absorbance, stray light.

- Use of filters, diffraction gratings

- Molar absorptivities must be equal
Stray radiation

P P +P
A = log o A = log o s
P P + Ps
where Ps = incident power of stray radiation

4. Absorption Spectra
2.6
2.4 523 nm
0.5 A.U.
2.2 1.0 A.U.
2.0 1.5 A.U.
1.8 2.0 A.U.
1.6
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0.0
400 450 500 550 600 650 700

Wavelength (nm)

- Plotting of spectral data

- λ maximum, nm

- Solutions of different concentrations

5. Theory of Molecular Absorption

-every molecular species has a unique set of energy states;

the lowest termed the ground state.

Excited State – energy of photon transferred to molecule:

Energy transition
Absorption
Matter
M + hν → M * (excited state)

M * → M + heat (relaxation)

a. Types of Molecular Transitions

(1) Electronic transition – electron between two orbitals

(UV-vis).
hν = Energy Difference of Orbitals

(2) Vibrational transitions – vibrational states

associated with bonds holding molecule. Involve
simultaneous:

(3) Rotational transitions – changes in rotational states,

about its center of gravity.
 Types of electronic transitions

σ to σ* Alkanes

σ to π* Carbonyl compounds

π to π* Alkenes, carbonyl compounds, alkynes, azo

compounds
η to σ * Oxygen, nitrogen, sulfur and halogen compounds

η to π* Carbonyl compounds

σ*
Empty levels
π*
η to σ* η to π*
η
π to π*
π Occupied levels
σ to σ*
σ

Highest occupied molecular orbital (HOMO)

Lowest unoccupied molecular orbital (LUMO)

E = h ν = h v/ λ = E (LUMO) – E (HOMO)
Therefore - Overall change in Energy with an orbital of a
molecule is:
E = Eelectronic + Evibrational + Erotational

E electronic ≈ 10 Evibrational
≈ 100 Erotational
 Emission Processes

b. Luminescence – emission of light from any excited state

of a molecule.
- more sensitive than absorption measurements

(1) Fluorescence – emission of photon during transition

between states with same spin quantum #’s (e.g.
S S )
1→ 0

(2) Phosphorescence - emission of photon during

transition between states with different spin
quantum #’s (e.g. T S )
1→ 0
6. Instrumentation

(a) Basic components

(1) Radiation source – Visible = tungsten filament

lamp; UV = deuterium and hydrogen lamps
(2) Monochromators/filters – Restricts wavelength to
offer narrow band of radiation
(3) Sample containers (cuvettes or cells) – glass, plastic
or quartz (1 cm)
(4) Phototubes/photomultiplier tubes – emit
photoelectrons after being irradiated, current is then
amplified and measured

Monochromator Detector

Radiation source Sample cell

Signal processor

- Single Beam Instruments

- Double Beam Instruments

- Array Detector Instruments

 Photodiode array
Source

Dispersed spectral
Focusing lens radiation

Sample cuvette

Grating
7. Scope - Application to absorbing species

Chromophore Example Solvent λmax

Alkenes C6H13CH=CH2 n-Heptane 177 nm
Alkynes CH2=CHCH=CH2 n-Heptane 196 nm
Nitro CH3NO2 Isooctane 280 nm
Nitrate C2H5ONO2 Dioxane 270 nm

Absorption by inorganic ions (η to σ *)

Carbonate, 217 nm; nitrite, 313 nm; azido, 230 nm

Absorption due to σ to σ*
Methane, 125 nm; ethane, 135 nm

Absorption of aromatic compounds

Compound Formula λmax

Benzene C6H6 256
Toluene C6H5CH3 261
Phenol C6H5OH 270
Styrene C6H5CH=CH2 282
8. Optimizing Experimental Conditions
a. Wavelength Selection
λmax.= 620
0.5 550 nm
A 0.4
bs
0.3
or 510 nm
b 0.2
a
nc 0.1
e
0.0

-0.1

-0.2
350 400 450 500 550 600 650 700

Wavelength (nm)

b. pH of Solution

0.12
0.10
0.08
0.06
0.04
0.02
0
0 2 4 6 8
PO4-P / mg l-1
c. Reagent Concentration

0.06
30 g l-1
0.05

0.04

0.03

0.02 15 g l-1

0.01

0
0 50 100 150
Time / s
d. Temperature
e. Interfering Substances

9. Applications
a. Health Sciences – 95% of all analyses are performed by
spectrophotometry.

b. Biological Sciences

c. Chemical & Environmental Sciences

Organic, inorganic systems and biochemical systems

10. Analysis of Mixtures

MixtureAbs = ε xb[ x] + ε b[Y ] + ......

Y

where ε = molar absorptivity of each species at wavelength

specified
b = cell pathlength

-measure A at more wavelengths then there are components in

mixture

Ex. Two-component mixture, at least three wavelengths. More is

better – increase in accuracy.
ABSORBANCE
0.5

0.4
IRON

0.3 CHROMIUM

0.2 COPPER

NICKEL
0.1
COBALT

0
300 350 400 450 500 550 600 650 700 750 800
WAVELENGTH / nm

11. Flow Injection Analysis

 0.020
0.018
0.016
0.014
Absorbance
0.012
0.010
0.008
0.006
0.004
0.002
0.000
-0.002

0 50 10 150 200 25 300

0 0
Time (sec)
Monoprotic Acid-Base Equilibria (CH
10)

{Chapter 10 – monoprotic acids

z A monoprotic acid can donate one proton.
z This chapter includes buffers; a way to ‘fix’ the pH.

{Chapter 11 – polyprotic acids

z A polyprotic acid can donate multiple protons.
z This chapter is just an extension of chapter 10.

{Chapter 12 – acid base titrations.

Strong acids and bases
Accounting for activity, calculate the pH of 0.10 M HCl.
z HCl is a strong acid, so it totally dissociates.
{ Table 6-2
{ The concentration of H+ will be 0.10 M.
z Solution:
{ The ionic strength of 0.10 M HCl is 0.10 M, at which the
activity coefficient of H+ is 0.83 (Table 8-1).
{ The pH is –log AH+
{ pH = -log[H+]γH+ = -log(0.10)(0.83)=1.08

z If you know [H+], you can always find [OH-] {or vice versa}.
{ Because [H+][OH-] = Kw =1.0 x 10-14
{ AND pH + pOH = 14.00
 Strong acids and bases
z We can neglect the concentration of H+ and OH- due to the auto-protolysis of
water only if the ‘extra’ H+ or OH- is much greater than 10-7.
What is the pH of 1.0 x 10-8 M M HCl?
z HCl is a strong acid, so it totally dissociates.
{ You memorized table 6-2 didn’t you?
{ The concentration of H+ will be 10-8 PLUS the H+ from water autoprotolysis.
{ An activity correction can be neglected here because the ionic strength is
very small.
z Solution:
[H+][OH-] = Kw
− 10 − 8 ± (10 − 8 ) 2 − 4 (1)( − 1 . 0 × 10 − 14 )
Let x be our unknown OH- x =
2 (1)
concentration. = 9 . 6 × 10 − 8 M or − 1 . 1 × 10 − 7 M
(10-8 + x)(x) = 1.0x10-14
Rearrange
Reject the negative solution.
x + (10 )x – (1.0 x 10 ) = 0
2 -8 -14
pH = -log[H+] = -log{10-8 + 9.6 x 10-8} = 6.97.
Use quadratic formula to solve for x:
 Weak acids and bases {review}
z Weak acid dissociation:
HA←
Ka
→H+ + A−
[H+ ][A− ]
Ka =
[HA]
z Weak base dissociation:
+ −
B + H 2O ←
Kb
→ BH + OH
[ BH + ][ OH − ]
Kb =
[B]
{ Remember, a base is a proton acceptor. Just because you see
OH- doesn’t imply base.
z Always!
Ka ⋅K b = K w
{ The conjugate base of a weak acid is a weak base.
{ The conjugate acid of a weak base is a weak acid.
Implies
Weak-acid equilibria
z Consider a weak acid HA that has a given Ka. Find the pH of the solution:
z What are the pertinent reactions?

HA←
Ka
→H+ + A−
z What is the charge balance? H O← Kw
→H+
+ OH−
2
{ [H+]=[A-] + [OH-]
z What is the mass balance?
{ Let’s call the formal concentration F.

Formal concentration is the total number of moles of a compound dissolved in a liter.

The formal concentration of a weak acid is the total amount of HA placed in the solution.

{ F = [A-] + [HA]
z Equilibria:

[H+ ][A− ]
Ka = Kw =[H+ ][OH− ]
[HA]
Weak-acid equilibria
z Even though called ‘weak’, any respectable acid will give an
[H+] concentration much greater than the [H+] concentration
due to water autoprotolysis.
{ In other words, if the [H+] from the acid dissociation is much greater than
the [H+] from the water dissociation then [A-] will be much greater than
[OH-].
z Because the ‘extra’ [H+] came from the HA dissociation.
{ The charge balance equation reduces to [H+] ¡ [A-].
{ This reduces a cubic equation to a quadratic equation.
z “I have trouble solving cubic equations.”
z Let [H+] = x, then:
{ Charge balance says that [H+] ¡ [A-] = x.
{ AND mass balance says that [HA] = F – [A-] = F – x.
z Plugging these results into the acid dissociation equilibria
gives:
[H+ ][A− ] (x)(x) x2
Ka = = =
[HA] F − x F − x
Weak-acid equilibria
z When dealing with a weak acid, you should immediately
realize that [H+] ¡ [A-] ¡ x
{ Unless the acid is very dilute or too weak.

+ − 2
[H ][A ] (x)(x) x
Ka = = =
[HA] F − x F − x
{ This results in using the quadratic formula!
{ In a solution of a weak acid, [H+] is derived almost entirely from
the weak acid, not from the H20 dissociation.

z Unless the acid is very dilute or too weak.

Weak-acid equilibria: A possible
approximation.
z The quadratic formula can always be used to solve weak acid
problems.
{ Unless the acid is very dilute or too weak.
z However, the problem is even easier if you can neglect x from
the denominator. 2 2
x x
Ka = ≈
F −x F
z This can ONLY be done if x is MUCH smaller than [HA].
{ How do I know if x is much smaller than [HA]?
z Make the approximation and solve the problem.
z If your answer supports your assumption then your answer is fine.
z Suppose you are given that [HA] is 0.1 M and you find [A-] to be
1x10-6, then you are safe to say that x = [A-] << [HA].
Fraction of dissociation

z The fraction of dissociation, α, is defined as

the fraction of an acid HA in the form of A-.
[A - ] x x
α = = =
[A - ] + [HA] x + (F − x) F
Weak-base equilibria
z The treatment of weak bases is almost the same as that
of weak acids.
B + H 2 O ←→
Kb
BH + + OH −
[BH + ][OH − ]
Kb =
[B]
z We suppose that nearly all of the OH- comes from the
reaction of B + H20 and little comes from the dissociation
of water.
z The formal concentration of base will be:
[B] = F - [BH+] = F – x

{ because F = [B] + [BH+]

Weak-base equilibria example
Find the pH of 0.10 M ammonia. {It’s not 13.}
z Pertinent reactions:
NH3 + H2O ) NH4+ + OH-
H2O ) H+ + OH-
z Woops, we have no Kb tables in our text.
{ But, Ka x Kb = Kw, and Ka for NH4+ is listed in our table at the back of
the book.
{ Kb=Kw/Ka.
z To find the pH of 0.10 M NH3, we set up and solve the equation

+
[NH 4 ][OH − ] Kw 10 −14.00 −5
= Kb = = = 1.75 × 10
[NH 3 ] K a 5.70 ×10 −10

{ If we let x = [NH4 ], the x also = [OH ] through stoichiometry.

+ -

{ Also, [NH3] = F – x where F = 0.10 M.

( x)( x) x2
= = 1.75 ×10 −5
F- x 0.1 − x
 Weak-base equilibria example
continued
Find the pH of 0.10 M ammonia. {It’s not 13.}
z Let’s assume x << 0.1 to avoid a quadratic equation.
x2
= 1.75 ×10 −5
0.1
x 2 = 1.75 ×10 −6
[OH − ] = x = 1.75 ×10 −6 = 1.32 ×10 −3 M
{ 1.32 x 10-3 is not << 1 x 10-1, so we should probably not make that
assumption.
z Using the quadratic formula.

x2 − 1.75 ×10 −5 ± (1.75 ×10 −5 ) 2 − 4(1)(−1.75 ×106 )

= 1.75 ×10 −5 x=
2(1)
0.1 − x
= 1.31×10 −3 M or − 1.32 (throw out negative solution)
x 2 = (1.75 ×10 −5 )(0.1 − x)
We get a slightly different answer, but the
x 2 = (1.75 ×10 −5 )(0.1) − ( x)(1.75 ×10 −5 ) difference is in our first uncertain digit.
x 2 + ( x)(1.75 ×10 −5 ) − (1.75 × 10 −6 ) = 0 We would have been fine making the
assumption after all!
Weak-base equilibria example
continued
Find the pH of 0.10 M ammonia. {It’s not 13.}
z Find the pH now that we know [OH-] = x = 1.31 x10-3 M.
pOH = -log(1.31 x 10-3) = 2.88
pH = 14.00 – pOH = 14.00 – 2.88 = 11.12

z The solution is less basic than if the ammonia was totally dissociated.
 Lecture 8 – Acid/Base Equilibria and Titrations

Monoprotic Systems

Ka + ][A−]
HA ← → H + + A - Ka = [ H
[HA]
Mass Balance = F = [HA] + [A-]

[HA] [H + ]
Fraction in form HA: αHA = =
F [H + ] + K a

[A-] [K a ]
-
Fraction in the form A : αA = =
F [H + ] + K
a
Diprotic Systems

K
1 → H + + HA -
H A ←
2
K
2 → H + + A 2-
HA ← 


Mass Balance:

F = [H2A] + [HA-] + [A2-]

K K K
F = [H2A] + 1 H A + 1 2 [H A]
+
[H ] 2 [H + ]2 2

K K K
F = [H2A] + (1 + 1 + 1 2)
[H + ] [H + ]2
 H A
2 [H + ]2
Fraction in form H2A: αH2A = =
F [H + ]2 + [H + ]K + K K
1 1 2

[HA-] K [H + ]
Fraction in form HA-: αHA- = = 1
F [H + ]2 + [H + ]K + K K
1 1 2

[A 2 - ] K K
2-
Fraction in form A : αA2- = = 1 2
F [H ] + [H + ]K + K K
+ 2
1 1 2

See Figure 11-3 in book, pg. 217

Common Titration curves

Strong acid, strong base Strong base, strong acid

Weak acid, strong base Strong acid, weak base Weak acid, weak base
Strong acid with strong base

1. Before the equivalence point: excess OH-

2. At the equivalence point: pH = 7.00

3. After the equivalence point: excess H

Titration of 50.0 mL of 0.0200M KOH with 0.100M HBr.

1. Before EP (e.g. v=3.00 mL)

2. At EP (i.e. v=10.00mL)

3. After EP (i.e. v=10.50 mL)

Titration of Weak Acid with Strong Base

1. Before any base is added: weak acid

2. Before the equivalence point: buffer

3. At the equivalence point: All HA converted to HA-

4. After the equivalence point: excess of OH-

Titration of 50.00 mL MES with 0.1000 M NaOH. MES is 2-(N-
morpholino) ethanesulfonic acid, weak acid with pKa = 6.15

1. Before base is added (i.e. v=0) Weak acid calculation

2. Before EP (e.g. v=3.00 mL) Buffer

3. At EP (i.e. v=10.00mL) Solution of weak base A

4. After EP (e.g. v+10.10 mL) Excess of OH-

Titration of weak base with strong acid (B + H+ → BH+)

Reverse of the titration of a weak acid with a strong base:

1. Before acid is added (i.e. v= 0)- Calculation of weak base

B + H 2 O ←→

Kb
BH + + OH

*when Va (vol of added acid) = 0

2. Before EP – Buffer containing B & BH+

 [B] 
pH = pKa (for BH+) + log  + 

 [BH ] 

3. At EP – Solution of weak acid BH+ (when Va=Ve)

BH+ ↔ B + + H +

4. After EP – excess of H, neglect contribution of weak acid

Note that the pH at EP is below 7.00 because the solution contains

the weak acid BH+
*Titration curve depends on the acid dissociation constant of HA and
on the concentrations of reactants.

-HA becomes a weaker acid or the concentration of analyte and

titrant decrease…..inflection deceases.
Titrations in diprotic systems

Titration of 10.0 mL of 0.100 M base (pKb1 = 4.00, pKb2 = 9.00) with

0.100 M HCl.

6 different regions, with two equivalence points (C&E)

Points B and D are the half-neutralization points, whose pH values

equal pKa2 and pKa1
Point A – (i.e. v = 0) Solution of weak base B

K
→ BH + + OH -
b1
B + H O ← 
2

Point B – before 1st EP (e.g. v=1.5 mL) Buffer sol’n with B and BH+
Point C – at 1st EP (e.g. Ve1 = 10.0 mL) Sol’n with all B converted to
the intermediate form BH+

Point D – between EP1 and EP2 (e.g. v=17.2 mL) Buffer containing
BH+ and BH2+
Point E – at EP2 (i.e. Ve2 = 20.0 mL) Sol’n with all BH+ converted to
the weak acid BH2+
 Summary: Acid-base Titrations

To determine the points on a curve for:

a. strong acid/strong base (problem 12-2)

b. weak acid/strong base (problem 12-6)

c. weak base/strong acid (problem 12-12)

d. diprotic systems (problem 12-25)

Read sections 12-5 to 12-8