Data Analysis and Presentation

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Chapter 5- Calibration Methods and Quality Assurance EXCEL – How To Do 1- least squares and linear calibration curve/function (chapter 4) 2- standard addition (chapter 5) 3-internal standard (addition) (chapter 5)

External standard

1- least squares and linear calibration curve/function (chapter 4)

CHAPTER 05: Opener B

CHAPTER 05: Table 5.1

Calibration function

Q: How the selected instrument responds to the change in quantity (concentration, M ppm ) of the measured analyte?

A: Measure the instrument responses (peak hights, peak areas etc.) from known but different concentrations of the analyte (say 0, 5,10, 15 20 ppm standards). Then construct the response diagram figure (response vs. concentration) with the response or calibration function through the points.

GO TO EXCELL or your calculator

Calibration Data: Answer 2

MFYI: Method of Least Squares/Linear regression

The method of least squares assumes that the errors in the y values are substantially greater than the errors in the x values. A second assumption is that the uncertainties in all of the y values are similar.

Derivation of the Least Squares Method (Equation for a straight line) y = mx + b Vvertical deviation = di i = yi–y = yi–(mxi+b) SSome of the deviations are positive and some are negative. To minimize the magnitude of the deviations irrespective of their signs, we square the deviations to create positive numbers.

OLD SLIDE: Caffeine again: first step calibration

If points are out of line: error

If one assumes that the calibration curve is

linear

– there are physical reasons for that. There is a linear function that can best represent the response of the instrument.

There are methods based on the propagation of error that can help us calculate the best fit: linear regression of the straight-line calibration curves

CALIBRATION: Instrument used to measure signals must be calibrated, must have the calibration relation . Example (linear): y = m [x ]+ b (general algebraic form for linear equation) Often given in other similar forms: S measured (total

signal

) =k n A (concentration) + S reag (

signal

from reagents, from blank )

Calibration

is the determination of that relation: it is used for the determination of analyte by using standards and blanks to determine a relationship (function) between concentrations and assay responses.

Validation

implies that other labs have approved the analytical method of analysis and produced similar results.

Standards

are materials containing accurately known concentrations of the desired analyte.

Standardization is

the process of determining relationship between the measured signal and the amount of analyte (fittting for k in linear relations).

Blanks

are solutions containing all added reagents except the sample analyte.(method, reagent and field blanks)

Controls

can be an alternate sample, in which the contents are well known.

Primary and Secondary reagents

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Calibration Data: Answer 2

CHAPTER 04: Unnumbered Table 4.3

CHAPTER 04: Unnumbered Table 4.4

CHAPTER 04: Figure 4.12

CHAPTER 04: Figure 4.14

IMPORTANT : BLANK !!

Before you start measuring standards, you need to see the signal from a sample without analyze.

BLANK: everything but analyte

CHAPTER 04: Table 4.7

EXAMPLE: CORRECTION FOR nonzero BLANK Your blank is 0.001 (for zero concentration of analyte) Your signals are as follows Signal Corrected signal for blank (!) concentration 0.001

0.003

0.006

0.009

0.001-0.001

0.003-0.001

0.006-0.001

0.009—0.001

0

0.0287

0.0601

0.0889

2- standard addition (chapter 5)

Is such external standardization the only way to calibrate and acquire the accurate and precise values?

• No, there are also other methods that include addition of analyte or some other material into the sample (aliquot)

Possible problems with external standardization

Spike

• Sometimes (often) the response to analyte is affected by something else in the sample, which we call MATRIX.

• SPIKE is a known quantity of analyte added directly to the sample to verify if the response to analyte is the same as that expected from pure sample observed in the calibration curve.

2. Standard addition

SStandard addition

: Known quantities of analyte are added to the unknown, and the increased signal lets us deduce how much analyte was in the original unknown. Typically we use the method of standard addition

when unknown sample matrix

is sufficiently complex.

The blank and standards are not representative of the unknown sample and lead to analysis error. This method requires a linear response to analyte.

For a single spike , one trial [X] i [X] i / [X] f + [S] f = I x / I s+x = unknown initial concentration of analyte I x = signal from first solution [S] f = concentration of standard in second solution [X] f I s+x = diluted concentration of analyte = signal from second solution

CHAPTER 05: Equation 5.7

EExample 2 A blood serum sample containing sodium ion gives a signal of 4.27 mV on a light intensity meter in an atomic emission experiment. The Na + concentration in the serum is then increased by 0.104 M by a standard addition, without significantly diluting the sample. This "spiked" serum sample gives a signal of 7.98 mV in atomic emission. Find the original concentration of Na + in the serum.

[X] i I x = unknown initial concentration of Na = 4.27 mV + [S] f = 0.104 M [X] f = diluted concentration of analyte = [X] i I s+x = 7.98 mV [X] i = 0.120 M

FYI: (2) Sample  direct addition S sample = S spike a b a b

FYI: Accurate description (1)Sample  separate aliquots (portions) S sample = S spike a b S sample = signal sample C S = concentration spiked C A = concentration analyte a b

Example 3. WWe use standard addition methods when our sample has matrix effects which are hard to duplicate in a blank. Let's again find the amount of Pb +2 in a blood sample. A 5.00 mL blood sample containing lead yielded a signal of 0.712 units. Afterwards, the sample was spiked with 5.00  L of 1560 ppb Pb +2 standard. (note: a small addition) This spiked solution gave a reading of 1.546 units. Find the concentration of Pb +2 .

or [C] i / [C] f + [Cs] f = S x / S s+x [X] i / [X] f + [S] f = I x / I s+x Be careful with volume!

C A =1.33ppb

3. Internal Standard

• In addition to the analyte you measure add another similar compound (not analyte ) that has similar response ( sensitivity ) as the analyte.

Internal Standard is sometimes added to an unknown sample.

The reason may be to verify the signal response in situations where instrument response varies slightly from run to run.

For example, an analysis is preformed on different days or different instruments or under different operating conditions. Typically, the internal standard resembles the analyte. Let's say we are separating isomers of octane and determining their concentration on a gas chromatograph (GC). To verify how much is present we might add a known amount of cyclohexane as an internal standard

Internal Standard

• • An

internal standard

is a known amount of a compound, different from the analyte, that is added to the unknown sample. A Standard mixture of analyte and standard is prepared before hand. Internal standards are desirable whenever losses of sample are likely to occur during handling or analysis. They are also used to calibrate the instrument when same analysis is done on different days. Use the following relation:

Internal Standard

• SA(signal due to analyte) SIS(signal due to int. stand) =k A =k IS CA CIS

CHAPTER 05: Equation 5.11

Example Example on internal standard: Note we are not diluting the sample significantly! A lab uses Cu +2 Pb +2 as a internal standard (1.75 ppm) and Cu +2 for a Pb +2 analysis in blood. Known solution standards for (2.25 ppm) give readings such that the signal ratio of Pb to Cu is 2.37. A blood sample was spiked with the Cu standard (2.25 ppm). We assume that the blood sample volume does not change significantly. The ratio of the readings for Pb to Cu in the unknown spiked solution is 1.80. Based on the response for the known amount of Cu +2 added to a blood sample determine, the amount of Pb +2 present.

First measurement S A (signal due to analyte) = k A C A C A S IS (signal due to int. stand) = k IS C IS = K C IS Second measurement C IS S A C A = K S IS K=3.05

C A =1.33ppb Pb 2+