Art PowerPoints Harris: Quantitative Chemical Analysis, Eight Edition CHAPTER 05: QUALITY ASSURANCE AND CALIBRATION METHODS

CHAPTER 05: Opener A Measurement of Lead in River Need for a blind test! The Institute for Reference Materials and Measurements – in Belgium

CHAPTER 05: Opener B Results from National Measurement Institutes

CHAPTER 05: Table 5.1

CHAPTER 05: Equation 5.1a Terms 1.False positive vs. false negative 2.Selectivity (or specificity) 3.Sensitivity = slope of calibration curve = change in signal / change in analyte concentration 4. Blanks 1) Method blank 2) Reagent blank 3) Field blank 5. Matrix 6. Spike (or fortification)

CHAPTER 05: Unnumbered Figure 5.1 Assessment-Control Chart Warns us when a monitored property strays dangerously far from an intended target value. The process should be shut down if the following things occur. 1 observation outside the action line. 2 out of 3 consecutive measurements bet. the warning and action lines 7 consecutive meas. all above or below the center line. 6 consecutive meas. all increasing or all decreasing 14 consecutive points alternating up and down An obvious nonrandom pattern

CHAPTER 05: Figure 5.1 Method Validation 1.Specificity: the ability of an analytical method to distinguish analyte from everything else that might be in the sample. 2.Linearity

CHAPTER 05: Figure 5.1 Is electrophoresis specific to cefotaxime?

CHAPTER 05: Unnumbered Figure 5.2 Accuracy Precision Range Limits of Detection and Quantitation Signal detection limit: y dl = y blank + 3s Calibration line: y sample – y blank = m x sample conc. Detection limit: min. detectable conc. = 3s/m Lower limit of quantitation = 10s/m Robustness

CHAPTER 05: Unnumbered Figure 5.2 Example. Detection Limit Signals from seven replicate samples: 5.0, 5.0, 5.2, 4.2, 4.6, 6.0, 4.9 nA Reagent blanks: 1.4, 2.2, 1.7, 0.9, 0.4, 1.5, 0.7 nA Slope m = nA/ μ M (a)Find the signal detection limit and the minimum detectable conc. (b)What is the conc. of analyte in a sample that gave a signal of 7.0 nA? Solution Blank: Average = y blank = nA Sample: Standard deviation = s = nA (a)Signal detection limit, y dl = y blank + 3s = (3)(0.5 6 ) = nA Minimum detectable concentration = 3s/m = (3)(0.5 6 nA)/(0.229 nA/ μ M) = 7. 3 μ M (b) y sample – y blank = m x concentration  Concentration = (7.0 nA – nA) / nA/ μ M = μ M

CHAPTER 05: Figure 5.4 Standard Addition Matrix effect Different groundwaters have different matrix effect. An ordinary calibration curve is not working.

CHAPTER 05: Equation 5.7 Standard Addition Method Equation

CHAPTER 05: Equation 5.7 Example. Standard Addition Serum containing Na + gave a signal of 4.27 mV in an AES mL of 2.08 M NaCl were added to 95.0 mL of serum. This spike sample gave a signal of 7.98 mV. Find [Na + ] in the serum. Solution [X] f = [X] i (V 0 /V) = [X] i (95.0 mL/100.0 mL) [S] f = [S] i (V S /V) = (2.08 M)(5.00 mL/100.0 mL) = M

CHAPTER 05: Equation 5.7 Graphic Procedure-successive standard addition b/m = -[X] i Plot y vs x gives the slope, m and x- intercept, -b/m.

CHAPTER 05: Figure 5.5 [X] I = 2.89 mM

CHAPTER 05: Equation 5.11 Internal Standard F: response factor

CHAPTER 05: Equation 5.11 Example. A solution containing M X and M S gave peak areas of A X = 423 and A S = 347. To analyze the unknown, 10.0 mL of M S were added to 10.0 mL of unknown, and the mixture was diluted to 25.0 mL in a vol. flask. This mixture gave the chromatogram below. A X = 553 and A S = 582. Find the [X] in the unknown. Solution: 423/ = F(347/0.0666)  F = [S] = (0.146 M)(10.0/25.0) = M A X /[X] = F(A S /[S]) 553/[X] = (582/0.0584)  [X] = M Considering dilution, [X] in the unknown is M x (25/10) = M

CHAPTER 05: Figure 5.10 Efficiency in Experimental Design -maximum information in the fewest number of trials Example: Find unknown acid concentrations, [A], [B], and [C] with uncertainties. Solution: If we perform three experiments for each solution, we need nine experiments. More efficient way is to determine acid concentration of mixtures of A, B, and C.

CHAPTER 05: Figure 5.10

CHAPTER 05: Equation equations are needed to solve 3 unknowns. Five equations with three unknowns give us uncertainty in the unknowns. 5-3 = 2 is degrees of freedom.

Homework problems of chapter 5-15, 16, 24, 25, 30, 31