91Ó°ÊÓ

Atomic emission measurements were made to determine sodium in a blood serum sample. The following emission intensities were obtained for standards of \(5.0\) and \(10.0 \mathrm{ng} / \mathrm{mL}\) and for the serum sample. All emission intensities were corrected for any blank emission. The mean value for the blank intensity \(\left(c_{\mathrm{Na}_{2}}=0.0\right)\) was \(0.000\) with a standard deviation of \(0.0071\) (arbitrary units). $$ \begin{array}{cc} c_{\mathrm{N} a}, \mathrm{ng} / \mathrm{mL} & \text { Emission Intensity } \\ \hline 5.0 & 0.51 \\ 5.0 & 0.49 \\ 5.0 & 0.48 \\ 10.0 & 1.02 \\ 10.0 & 1.00 \\ 10.0 & 0.99 \\ \text { Serum } & 0.71 \\ \text { Serum } & 0.77 \\ \text { Serum } & 0.78 \\ \hline \end{array} $$ (a) Find the mean emission intensity values for the \(5.0\) - and \(10.0-\mathrm{ng} / \mathrm{mL}\) standards and for the serum sample. Find the standard deviations of these values. (b) Find the least squares best line through the points at \(c_{\mathrm{Na}}=0.0,5.0\), and \(10.0 \mathrm{ng} / \mathrm{mL}\). Find the calibration sensitivity and the analytical sensitivity. *(c) Find the detection limit for \(k\) values of 2 and 3 . To what level of confidence do these correspond? (d) Find the concentration of \(\mathrm{Na}\) in the serum sample and the standard deviation of the concentration.

Step by step solution

01

Calculate Mean and Standard Deviation of Emission Intensities

For each concentration (5.0 ng/mL, 10.0 ng/mL, and Serum), calculate the mean emission intensity by averaging the given values. Also, calculate the standard deviation for each set of values using the formula \( \sigma = \sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i - \bar{x})^2} \), where \( N \) is the number of observations, \( x_i \) are the individual measurements and \( \bar{x} \) is the mean value. For 5.0 ng/mL: - Mean = \( \frac{0.51 + 0.49 + 0.48}{3} = 0.493 \)- Standard Deviation = \( \sqrt{\frac{(0.51 - 0.493)^2 + (0.49 - 0.493)^2 + (0.48 - 0.493)^2}{2}} = 0.015 \)For 10.0 ng/mL:- Mean = \( \frac{1.02 + 1.00 + 0.99}{3} = 1.003 \)- Standard Deviation = \( \sqrt{\frac{(1.02 - 1.003)^2 + (1.00 - 1.003)^2 + (0.99 - 1.003)^2}{2}} = 0.015 \)For Serum:- Mean = \( \frac{0.71 + 0.77 + 0.78}{3} = 0.753 \)- Standard Deviation = \( \sqrt{\frac{(0.71 - 0.753)^2 + (0.77 - 0.753)^2 + (0.78 - 0.753)^2}{2}} = 0.037 \)

02

Determine the Least Squares Calibration Line

To find the calibration line in the form \( y = mx + b \), where \( y \) is the emission intensity and \( x \) is the concentration, use the known points for 0.0, 5.0, and 10.0 ng/mL:- Points: (0, 0), (5.0, 0.493), (10.0, 1.003)- Calculate the slope \( m \) using \( m = \frac{(X - \bar{X})(Y - \bar{Y})}{(X - \bar{X})^2} \)- Calculate the intercept \( b \) from \( b = \bar{Y} - m\bar{X} \)Substituting and calculating gives us:- Slope \( m = \frac{(5-5)(0.493-0.5) + (10-5)(1.003-0.5)}{(0-5)^2 + (10-5)^2} = 0.1 \)- Intercept \( b = 0.5 - 0.1 \times 5 = -0.5 \). So, the line is \( y = 0.1x - 0.5 \)

03

Evaluate Calibration Sensitivity and Analytical Sensitivity

- **Calibration Sensitivity**: This is given by the slope of the calibration line, \( m = 0.1 \).- **Analytical Sensitivity**: This is calculated using \( \frac{\text{sensitivity}}{\text{standard deviation of blank}} = \frac{0.1}{0.0071} \approx 14.08 \)Thus, Calibration Sensitivity is 0.1 and Analytical Sensitivity is approximately 14.08.

04

Calculate Detection Limit for k Values of 2 and 3

The detection limit can be calculated using the formula \( DL = k \times \text{standard deviation of blank} / m \), where \( k \) is the confidence factor and \( m \) is the calibration sensitivity.For \( k = 2 \):- \( DL = 2 \times 0.0071 / 0.1 = 0.142 \)- This corresponds to about 95% confidence.For \( k = 3 \):- \( DL = 3 \times 0.0071 / 0.1 = 0.213 \)- This corresponds to about 99% confidence.

05

Determine Concentration of Na in Serum Sample

Using the calibration line equation \( y = 0.1x - 0.5 \) and the mean emission intensity for the serum (0.753), solve for \( x \):- \( 0.753 = 0.1x - 0.5 \)- \( x = \frac{0.753 + 0.5}{0.1} = 12.53 \text{ ng/mL} \)To find the standard deviation of the concentration, use the formula\[ \sigma_x = \frac{\sigma_y}{m} \]- \( \sigma_x = \frac{0.037}{0.1} = 0.37 \text{ ng/mL} \)Therefore, the concentration of Na in the serum sample is 12.53 ng/mL with a standard deviation of 0.37 ng/mL.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Detection Limit

The detection limit in analytical chemistry is pivotal, serving as the minimum concentration of an element that can be reliably distinguished from the background noise. To determine it, analysts often evaluate the standard deviation of the blank sample, which contains no analyte, to measure the inherent noise of the system.

The detection limit (\( DL \)) is calculated by multiplying the confidence factor \( k \), typically values of 2 or 3, by the standard deviation of the blank, and dividing by the calibration sensitivity. This formula is represented as:
\[ DL = \frac{k \times \text{standard deviation of blank}}{\text{calibration sensitivity}} \]Key Points:

• Using \( k = 2 \) gives a 95% confidence level, indicating a 95% probability that the signal is distinct from the noise.
• Using a \( k = 3 \) factor ensures a higher confidence of 99%, meaning a stronger assurance of detectability.

This method effectively helps ensure that even minimal concentrations won't be missed during analysis.

Calibration Curve

A calibration curve acts like a roadmap in atomic emission spectroscopy. It relates the emission intensity directly to the concentration of an element for accurate sample analysis. Constructing a calibration curve involves plotting known concentrations versus their corresponding emission intensities.

The underlying equation to describe this relationship is a linear equation in the form \( y = mx + b \), where \( y \) is the emission intensity, \( x \) is the concentration, \( m \) is the slope (calibration sensitivity), and \( b \) is the intercept. To find the least squares best line:

• Calculate the slope \( m \) by measuring how changes in concentration affect intensity.
• Determine the intercept \( b \), which fits the line to the data set more precisely.

This straightforward model allows unseen sample concentrations to be predicted from their emitted light intensity.

Analytical Sensitivity

Analytical sensitivity is distinct from calibration sensitivity, as it's a measure of how well an instrument can detect small changes in concentration. It is essential in determining the lowest concentration that can be measured with reliability.

This sensitivity is calculated by dividing the calibration sensitivity by the standard deviation of the blank:
\[ \text{Analytical Sensitivity} = \frac{\text{Calibration Sensitivity}}{\text{Standard Deviation of Blank}} \]Its importance:

• Allows us to understand and compare the performance of different instruments or methods.
• Facilitates improvement of analytical methods to increase sensitivity and, thus, reliability.

Higher analytical sensitivity signifies a method's ability to detect subtle concentration changes, enhancing the quality of analytical results.

Standard Deviation

In the realm of analytical measurements, standard deviation provides insight into the precision of the obtained emission intensities. It signifies how measurements for a set of observations, such as emission intensities at various concentrations, deviate from their mean.

The standard deviation (\( \sigma \)) is calculated as follows:
\[ \sigma = \sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i - \bar{x})^2} \]
where \( N \) is the number of measurements, \( x_i \) are individual measurements, and \( \bar{x} \) is their average.

• A lower standard deviation indicates measurements are close to the mean, which implies high precision.
• A larger standard deviation signals greater spread in the measurements, suggesting potential issues with consistency.

Standard deviation is a core tool for evaluating experimental data quality and ensuring reliable conclusions.