04. Standardization Methods

Question 1

Differentiate single-point and multiple-point calibration

Answer 1

1. Single-point
   → calibration based on one known standard
   → quick, lower accuracy, assumes linearity
2. Multiple-point
   → calibration using multiple standards to generate a calibration curve
   → more accurate, detects non-linearity

Question 2

Standardization method equation

Answer 2

Sa = KaCa

Sa → signal due to analyte a
Ka → sensitivity constant
Ca → concentration of analyte in the sample

Question 3

______ are solutions with known concentrations of the analyte, prepared separately from the sample being analyzed.

Answer 3

External standards

Question 4

A spectrophotometric method is used to determine the concentration of Pb²⁺ in blood. A standard solution with Pb²⁺ concentration of 1.75 ppb gives an absorbance of 0.474. What is the concentration of Pb²⁺ in a blood sample that gives an absorbance of 0.361 under the same conditions?

Answer 4

Sa = KaCa

Ka = S_std/C_std
Ka = 0.474/1.75 ppb⁻¹
Ka = 0.271 ppb⁻¹

Ca = S_samp/Ka
Ca = 0.361/0.271 ppb⁻¹
Ca = 1.33 ppb

Question 5

A second spectrophotometric method for the quantitative analysis of Pb²⁺ in blood yields a calibration curve defined by the equation:

S_std = 0.296 C_std + 0.003

What is the concentration of Pb²⁺ in a blood sample if the measured signal is 0.397?

Answer 5

S = 0.296C + 0.003
0.397 = 0.296C + 0.003
C = (0.397 - 0.003)/0.296
C = 1.33

Question 6

Single-point standardization leads to a determinate (systematic) error in an analyte’s reported concentration if _____

Answer 6

it is incorrectly assumed that Ka is constant.

Question 7

A multiple-point standardization should include at least ____ standards, although more are preferable.

Answer 7

three standards or more

Question 8

An _____ is a reference compound that is chemically and physically similar to the analyte but distinct from it.

It is added in a constant amount to all samples and standards containing the analyte.

Answer 8

internal standard

Question 9

Derive the K and Ca equation for internal standard method

Answer 9

Sa = Ka Ca → analyte
S_is = K_is C_is →internal standard

get the ratio:
Sa/S_is = Ka Ca / K_is C_is
Sa/S_is = K (Ca/C_is)

equate to K:
K = (Sa/S_is)_std (C_is/Ca)

equate to Ca:
Ca = (Sa/S_is)_samp (C_is/K)

Question 10

A spectrophotometric method for the quantitative analysis of Pb²⁺ in blood uses Cu²⁺ as an internal standard. A standard solution containing 1.75 ppb Pb²⁺ and 2.25 ppb Cu²⁺ produces a signal ratio (Sa/S_is)_std = 2.37.

A blood sample, spiked with the same concentration of Cu²⁺, gives a signal ratio (Sa/S_is)_samp = 1.80. What is the concentration of Pb²⁺ in the blood sample?

Answer 10

a → Pb²⁺
istd → Cu²⁺

(Sa/S_is)_std = Ka Ca/K_is C_is
2.37 = K 1.75/2.25
K = 3.047

(Sa/S_is)_samp = K Ca/C_is
1.80 = 3.047 (Ca/2.25)
Ca = 1.33 ppb

Question 11

When matrix effects are significant or the sample matrix is unknown, external standardization may not yield accurate results unless matrix matching is performed.

What standardization method is applied?

Answer 11

standard addition method

Question 12

Standard addition equations for (a) with dilution (b) without dilution?

Answer 12

a. with dilution
Sa/S_spike = [Ca(Vi/Vf)] / [Ca(Vi/Vf)] + [Cs(Vs/Vf)

b. without dilution
Sa/S_spike = Ca / [Ca(Vi/Vi + Vs)] + [Cs(Vs/Vi+Vs)]

Question 13

A 1.00 mL blood sample is diluted to 5.00 mL and yields an absorbance of 0.193.

A second 1.00 mL blood sample is spiked with 1.00 μL of a 1560- ppb Pb²⁺ standard solution, diluted to 5.00 mL, and gives an absorbance of 0.419. What is the concentration of Pb²⁺ in the original blood sample?

Answer 13

Vi = 1.00 mL
Vf = 5.00 mL
Sa = 0.193

Vi = 1.00 mL
Vf = 5.00 mL
Vs = 1.00 μL (10⁻⁶L/μL) (mL/10⁻³L) = 10⁻³ mL
Cs = 1560 ppb
S_spike = 0.419

Sa/S_spike = [Ca(Vi/Vf)] / [Ca(Vi/Vf)] + [Cs(Vs/Vf)
0.193/0.419 = [Ca(1/5)] / [Ca(1/5)] + [1560(10⁻³/5)
Ca = 1.33 pbb

Question 14

A 5.00 mL blood sample yields an absorbance of 0.712.

After spiking the sample with 5.00 μL of a 1560-ppb Pb²⁺ standard, the absorbance increases to 1.546. What is the concentration of Pb²⁺ in the original blood sample?

Answer 14

Vi = 5.00 mL
Sa = 0.712

Vs = 5.00 μL (10⁻⁶L/μL) (mL/10⁻³L) = 5x10⁻³ mL
Cs = 1560 ppb
S_spike = 1.546

Sa/S_spike = Ca/[Ca(Vi/Vi+Vs)]+[Cs(Vs/Vi+Vs)]
0.712/1.546 = Ca/[Ca(5/5+5x10⁻³)]+[1560(5x10⁻³/5+5x10⁻³)]
Ca = 1.33 ppb

Question 15

A chemist is analyzing caffeine in energy drinks using HPLC. A calibration curve is prepared by using known concentrations of pure caffeine dissolved in water. The curve is then used to determine caffeine concentration in the drink sample. What type of quantification method is being used?

Answer 15

External standard method

Question 16

A chemist is analyzing a complex wastewater sample with many unknown interfering substances. The sample matrix cannot be easily replicated. To improve accuracy, which standardization method should the chemist use?

Answer 16

Standard addition method

Question 17

During an experiment, the instrument response fluctuates slightly due to unstable source intensity. The analyst wants to compensate for these fluctuations by adding a compound similar to the analyte to both samples and standards. Which method is being applied?

Answer 17

Internal standard method

Question 18

When it is best to use external standards for calibration of instrument response?

I. When the analysis is significantly affected by matrix effect.
II. When there are no interference effects from matrix components in the analyte solution.
III. When analyzing a large number of samples with uniform composition.

A. I only
B. II only
C. I and III
D. II and III

Answer 18

D. II and III

Question 19

When performing calculations using the standard additions method, which of the following is true?

A. The volume of the standard added can be ignored because it is usually very small.
B. The volume of the standard added must be included in the volume of the sample solution.
C. The volume of the standard added can be ignored because it is accounted for in the blank correction.
D. The volume of the standard added can be ignored because volumes are not used in calculations.

Answer 19

B. The volume of the standard added must be included in the volume of the sample solution.

Question 20

An unknown sample produces a signal of 10.0 mV. After adding 1.00 mL of a 0.0500 M standard solution to 100.0 mL of the unknown, the signal increases to 14.0 mV. Calculate the concentration of the analyte in the original unknown.

A. 1.21 mM
B. 5.26 mM
C. 9.18x10⁻⁴ M
D. 7.14x10⁻⁵ M

Answer 20

A. 1.21 mM

Standard Addition:
Sa/S_spike = Ca / [Ca(Vi/Vi + Vs)] + [Cs(Vs/Vi+Vs)]
Ca = 1.21 mM

Question 21

The table below presents chromatographic data for the analysis of benzene using ethylbenzene as an internal standard. Based on this data, what is the concentration of benzene in the sample?

concentration of benzene μg/mL:
std: 50 | sample: -
concentration of ethylbenzene μg/mL:
std: 10 | sample: 10
peak area of benzene, mV-s:
std: 2,500 | sample: 2,500
peak area of ethylbenzene, mV-s:
std: 1,000 | sample: 500

A. 10 μg/mL
B. 25 μg/mL
C. 100 μg/mL
D. 500 μg/mL

Answer 21

C. 100 μg/mL

Internal Standard:
Sa/S_is = K (Ca/C_is)
(Sa/S_is)_std = K (Ca/C_is)
2500/1000 = K (50/10)
K = 0.5

(Sa/S_is)_sample = K (Ca/C_is)
2500/500 = 0.5 (Ca/10)
Ca = 100 μg/mL