91Ó°ÊÓ

The following atomic absorption results were obtained for determinations of \(\mathrm{Zn}\) in multivitamin tablets. All absorbance values are corrected for the appropriate reagent blank \(\left(c_{\mathrm{Zn}}=0.0 \mathrm{ng} / \mathrm{mL}\right)\). The mean value for the blank was \(0.0000\) with a standard deviation of \(0.0047\) absorbance units. $$ \begin{array}{cc} c_{\text {Zn. }}, \mathrm{ng} / \mathrm{mL} & A \\ \hline 5.0 & 0.0519 \\ 5.0 & 0.0463 \\ 5.0 & 0.0485 \\ 10.0 & 0.0980 \\ 10.0 & 0.1033 \\ 10.0 & 0.0925 \\ \text { Tablet sample } & 0.0672 \\ \text { Tablet sample } & 0.0614 \\ \text { Tablet sample } & 0.0661 \\ \hline \end{array} $$ (a) Find the mean absorbance values for the \(5.0\) - and \(10.0-\mathrm{ng} / \mathrm{mL}\) standards and for the tablet sample. Find the standard deviations of these values. (b) Find the least-squares best line through the points at \(c_{\mathrm{Zn}}=0.0,5.0\), and \(10.0 \mathrm{ng} / \mathrm{mL}\). Find the calibration sensitivity and the analytical sensitivity. (c) Find the detection limit for a \(k\) value of 3 . To what level of confidence does this correspond? (d) Find the concentration of \(\mathrm{Zn}\) in the tablet sample and the standard deviation in the concentration.

Step by step solution

01

Calculate Mean and Standard Deviation for Standards and Tablet Samples

For the concentration of 5.0 ng/mL, calculate the mean absorbance by averaging the three measurements: \[ \text{Mean}_{5.0} = \frac{0.0519 + 0.0463 + 0.0485}{3} = 0.0489 \]Calculate the standard deviation using:\[ \sigma_{5.0} = \sqrt{\frac{(0.0519-0.0489)^2 + (0.0463-0.0489)^2 + (0.0485-0.0489)^2}{3-1}} = 0.0028 \]Repeat this for 10.0 ng/mL:\[ \text{Mean}_{10.0} = \frac{0.0980 + 0.1033 + 0.0925}{3} = 0.0979 \]\[ \sigma_{10.0} = \sqrt{\frac{(0.0980-0.0979)^2 + (0.1033-0.0979)^2 + (0.0925-0.0979)^2}{3-1}} = 0.0054 \]For tablet samples:\[ \text{Mean}_{\text{Tablet}} = \frac{0.0672 + 0.0614 + 0.0661}{3} = 0.0649 \]\[ \sigma_{\text{Tablet}} = \sqrt{\frac{(0.0672-0.0649)^2 + (0.0614-0.0649)^2 + (0.0661-0.0649)^2}{3-1}} = 0.0029 \]

02

Determine the Least-Squares Regression Line

Using the mean absorbance values, compute the slope \( m \) and intercept \( b \) of the least-squares line \( A = mc_{\mathrm{Zn}} + b \). This can be done using the formulae:Slope \( m \) = \( \frac{n\sum x_i y_i - \sum x_i \sum y_i}{n\sum x_i^2 - (\sum x_i)^2} \) Intercept \( b \) = \( \frac{\sum y_i \sum x_i^2 - \sum x_i \sum x_i y_i}{n\sum x_i^2 - (\sum x_i)^2} \) Where each \( x_i \) is the concentration (0.0, 5.0, 10.0) and \( y_i \) is the mean absorbance (0.0000, 0.0489, 0.0979). Calculations yield:\( m \approx 0.00979 \), \( b \approx 0.00005 \) (rounded to sig. figures).

03

Find Calibration and Analytical Sensitivity

Calibration sensitivity is equivalent to the slope of the regression line: \( S_c = m = 0.00979 \) absorbance/ng·mL.Analytical sensitivity \( S_a \) is defined as \( S_a = \frac{m}{\sigma_y} \), where \( \sigma_y \) is the standard deviation of the response (y-intercept of the regression line).Assuming the standard deviation of the blank from Step 1: \( \sigma_y = 0.0047 \)\( S_a = \frac{0.00979}{0.0047} \approx 2.08 \) ng/mL.

04

Determine Detection Limit

The detection limit (DL) is calculated using: DL = \( 3 \times \sigma_{\text{blank}} / S_c \). With \( \sigma_{\text{blank}} = 0.0047 \) and \( S_c = 0.00979 \),DL = \( \frac{3 \times 0.0047}{0.00979} \approx 1.44 \) ng/mL. This corresponds to a confidence level of approximately 99%.

05

Calculate Concentration of Zn in Tablet Samples

Use the regression equation \( A = mc_{\mathrm{Zn}} + b \) to find the concentration. Substitute each tablet's absorbance into the equation.For mean tablet absorbance \( A = 0.0649 \), solve for \( c_{\mathrm{Zn}} \):\[ 0.0649 = 0.00979 \times c_{\mathrm{Zn}} + 0.00005 \]\( c_{\mathrm{Zn}} \approx \frac{0.0649 - 0.00005}{0.00979} \approx 6.63 \) ng/mL.To find standard deviation in concentration, use propagation of error based on standard deviation of absorbance.

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Key Concepts

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

Analytical Sensitivity

Analytical sensitivity is a crucial measure in atomic absorption spectroscopy that describes how well a method can distinguish between small differences in analyte concentrations. It is especially important in methods where precise quantification of low concentration analytes is required. This parameter is often used to compare the responsiveness of different analytical methods for trace analysis.

In simple terms, analytical sensitivity is calculated by taking the calibration sensitivity and dividing it by the standard deviation of the signal's noise. The calibration sensitivity, represented by the slope (\( S_c = m \)) of the calibration curve, indicates how much the signal changes per unit change in concentration.

• An increase in slope signifies higher analytical sensitivity.
• The standard deviation of the noise represents the variability in the measurement.

Thus, analytical sensitivity is given by\( S_a = \frac{m}{\sigma_y} \), where \( \sigma_y \) is the standard deviation of the response, usually influenced by the blank sample's noise.

In the provided solution, analytical sensitivity was calculated as \( S_a \approx 2.08 \) ng/mL, which means that the method can differentiate between zinc concentrations effectively due to its ability to provide distinct signal changes even for small concentration differences.

Detection Limit

The detection limit in the context of atomic absorption spectroscopy defines the smallest concentration of an analyte that can be reliably distinguished from zero. It is critical for ensuring that your method can detect trace amounts of a substance in a sample.

To calculate the detection limit, you use the formula:\( DL = \frac{k \times \sigma_{\text{blank}}}{S_c} \), where \( k \) is a multiplier that determines the level of confidence, typically set to 3 for a standard confidence level of 99%. \( \sigma_{\text{blank}} \) is the standard deviation of the blank, and \( S_c \) is the slope of the calibration line. This calculation takes into account both the signal variability and the sensitivity of the analytical method.

In this specific scenario, the detection limit was calculated as 1.44 ng/mL, indicating that concentrations below this value might not be accurately measured with confidence using this method. The value corresponds to a high confidence level, ensuring that when zinc is detected above this threshold, one can be 99% certain that it is present in the sample.

Least-Squares Regression

Least-squares regression is a statistical method used in constructing a calibration curve in atomic absorption spectroscopy. The goal is to find the best-fit line through a set of data points, minimizing the sum of the squares of the vertical deviations from each data point to the line.

For atomic absorption, this method helps in establishing a relationship between the concentration of the analyte (in this case, Zn) and the absorbance measured. The mathematical equation for the line is generally of the form \( A = mc_{\mathrm{Zn}} + b \), where:

• \( m \) represents the slope of the line, indicating calibration sensitivity.
• \( b \) is the y-intercept, showing background absorbance when the concentration is zero.

The calculations for slope \( m \) and intercept \( b \) involve using the summed concentrations and absorbance values to minimize error. The solution provided a least-squares line with \( m \approx 0.00979 \) and \( b \approx 0.00005 \), demonstrating a minimal background effect and reliable calibration curve.

This method is foundational in analytical chemistry, allowing analysts to predict concentrations from an absorbance measure accurately.