CHEMISTRY 59-320
ANALYTICAL CHEMISTRY
Fall - 2010
Lecture 7
The flow chart for Lab 1
• Step 1: Check your inventory list
(< 10 minutes).
• Step 2: Student A weights sodium thiosulfate and sodium carbonate
(using regular weighting method),
while your partner student B checks and prepares (cleans)
glasswares
(<15 minutes).
• Step 3: Student A moves on to prepare sodium thiosulfate solution,
while student B uses weight-by-difference to weight 3
potassium iodate samples
(<25 minutes).
• Step 4: Student B moves on to prepare potassium solution, sets up
the titration buret and titrates the potassium solution.
At the same time, student A uses weight-by-difference to
weight Vitamin C, dissolves it, filters the solution, and sets
up titration buret (< 70 minutes).
• Step 5: Student A continues his/her titration of vitamin C solution,
while student B starts calculating standardization etc (30
minutes).
• Ste 6: clean your glassware (15 minutes)
• Most importantly, everyone practices all the
skills!!
4-7: The method of least squares
(Regression Analysis)
• For most chemical analysis, the establishment of
a calibration curve is desired so that unknown
quantity can be explained.
• Most often we wish to work in a region where the
calibration curve is a straight line.
• Using the method of least squares to draw the
best line through experimental data points (of
the standard samples).
• The best scenario is some points lie above and
some lie below the straight line.
Find the equation for the line
The straight line model
y  mx  b
in which m is the slope and b is the y-intercept
For Lines through the origin
y  mx
• This is a general assumation that the experimental
response is less certain than the quantity of analytes (x
values).
•Because we minimize the squares of the deviations (di), it is
called the method of least squares.
The method of least squares takes
the best fitting model by minimizing the quantity,
n
S  S (  )   ( yobs   xi )2
i 1
A plot of S as a function of Beta produces a minimum with
a constant least square estimate for beta “m”.
After “m” is known, you have all the calculated values
yi  mxi
The difference between these two values is the residual,
and the sum of the squares of the residuals is also a minimum value.
n
S R   ( yobs  yi )
i 1
2
Finding the slope and intercept with
a spreadsheet
Calculate the standard deviation of
the slope (sm) and the intercept (sb)
Where D is calculated through equation 4-18.
One can also use the confidence instead of standard deviation to present the
Slope and intercept.
After class exercise: Using spreadsheet to calculate Sy, Sm and Sb (see page 68)
4-8 calibration curves
• Standard solutions:
• Blank solutions:
•It is not reliable to extrapolate any calibration
curve, linear or nonlinear.
Measure standards in the entire concentration
range of interests.
Chapter 5 Quality assurance
and Calibration method
The need for quality assurance
•
•
Quality assurance is what we do to get the right answer for our purpose .
There is no point to waste more resources to obtain more precise
answer if it is not necessary.
5-1 Basics of Quality assurance
• Use objectives: The purpose for which results will be used. This is a
critical step in quality assurance.
• Specifications: It includes sampling requirements, accuracy and
precision, rate of false results, selectivity, sensitivity, acceptable
blank values, recovery of fortification, calibration checks, quality
control samples.
• Quality assurance begins with sampling: collecting a representative
sample.
• A false positive says that the concentration exceeds the legal limit
when, in fact, the concentration is below the limit.
• A false negative says that the concentration is below the limit when
it is actually above the limit.
•
•
•
•
Specificity
Linearity
Range
Sensitivity
• method blank: A
sample without
deliberately added
analyte.
5.2 Method validation
• To prove that an analytical method is acceptable
for its intended purposes.
• It includes
1. specificity
2. linearity
3. accuracy
4. precision
5. range
6. limits of detection and quantitation
7. robustness.
Detection limit
• detection limit: The smallest
quantity of analyte that is
“significantly different” from a
blank.
• The detection limit is often
taken as the mean signal for
blanks plus 3 times the
standard deviation of a lowconcentration sample.
• If you have a recorder trace
with a signal plus adjacent
baseline noise, the detection
limit is sometimes taken as
twice the peak-to-peak noise
level or 10 times the rootmean-square noise level
(which is 1/5 of the peak-topeak noise level). Also called
lower limit of detection.
• where ysample is the signal
observed for the sample
and m is the slope of the
linear calibration curve.
• A signal that is 10 times
as great as the noise is
defined as the lower limit
of quantitation, or the
smallest amount that can
be measured with
reasonable accuracy.
5.3 Standard Addition
• In standard addition,
known quantities of
analyte are added to
the unknown. From
the increase in signal,
we deduce how much
analyte was in the
original unknown.
• This method requires
a linear response to
analyte.
Standard-Addition
kVs cs kVx cx
S

Vt
Vt
x – unknown
s – standard solution
k – proportionality constant
• Serum containing Na+ gave a signal of 4.27 mV
in an atomic emission analysis. Then 5.00 mL of
2.08 M NaCl were added to 95.0 mL of serum.
This spiked serum gave a signal of 7.98 mV.
Find the original concentration of Na+ in the
serum.
• Solution the final concentration of Na+ after
dilution with the standard is [X]f = [X]i(V0/V) =
[X]i(95.0 mL/100.0 mL). The final concentration
of added standard is [S]f = [S]i(Vs/V) = (2.08
M)(5.00 mL/100.0 mL) = 0.104 M. Standard
addition equation becomes
•
The total volume is V = V0 + Vs and the
5.4 Internal standard
• An internal standard is a
known amount of a
compound, different from
analyte, that is added to
the unknown.
• Signal from analyte is
compared with signal
from the internal standard
to find out how much
analyte is present.
• S5-5. Internal standard. A mixture containing 80.0 nM iodoacetone
(designated A) and 64.0 nM p-dichlorobenzene (designated B) gave
the relative detector response (peak area of A)/(peak area of B) =
0.71 in a chromatography experiment. A solution containing an
unknown quantity of A plus 930 nM B gave a relative detector
response (peak area of A)/(peak area of B) = 1.21. Find the
concentration of A in the unknown.
• Solution: