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Chapter 11: Problem 28

Sittampalam and Wilson described the preparation and use of an amperometric sensor for glucose. \(^{27}\) The sensor is calibrated by measuring the steady- state current when it is immersed in standard solutions of glucose. A typical set of calibration data is shown here. \begin{tabular}{cc} [glucose] \((\mathrm{mg} / 100 \mathrm{~mL})\) & current (arb. units) \\ \hline 2.0 & 17.2 \\ 4.0 & 32.9 \\ 6.0 & 52.1 \\ 8.0 & 68.0 \\ 10.0 & 85.8 \end{tabular} A \(2.00-\mathrm{mL}\) sample is diluted to \(10 \mathrm{~mL}\) in a volumetric flask and a steady-state current of 23.6 (arbitrary units) is measured. What is the concentration of glucose in the sample in \(\mathrm{mg} / 100 \mathrm{~mL}\) ?

Short Answer

Expert verified
The glucose concentration in the sample is approximately 13.75 mg/100 mL.

Step by step solution

01

Observe Calibration Data

The first step is to examine the calibration data provided. The glucose concentrations and their corresponding current readings are given as pairs: (2.0 mg/100 mL, 17.2), (4.0 mg/100 mL, 32.9), (6.0 mg/100 mL, 52.1), (8.0 mg/100 mL, 68.0), and (10.0 mg/100 mL, 85.8). These pairs indicate that as the concentration increases, the current reading also increases.
02

Determine Calibration Curve Equation

Next, we must establish a relationship between the concentration of glucose and the sensor current. Assuming a linear relationship, the equation of a line is needed: \( y = mx + c \), where \( y \) is the current, \( m \) is the slope, \( x \) is the glucose concentration, and \( c \) is the y-intercept. We calculate the slope (\( m \)) using two data points, for example, (2.0, 17.2) and (10.0, 85.8): \( m = \frac{85.8 - 17.2}{10.0 - 2.0} \approx 8.55 \).
03

Calculate Y-Intercept

To find the y-intercept, substitute one of the data points into the linear equation. Using the point (2.0, 17.2), we substitute: \( 17.2 = 8.55 \cdot 2.0 + c \). Solving for \( c \), we get \( c = 17.2 - 17.1 = 0.1 \). Thus, our linear equation becomes \( y = 8.55x + 0.1 \).
04

Determine Diluted Sample Concentration

Now, substitute the measured current (23.6 arb. units) into the calibration equation to find the concentration: \( 23.6 = 8.55x + 0.1 \). Solving for \( x \), we find \( x \approx \frac{23.5}{8.55} \approx 2.75 \). This value represents the concentration of glucose in the diluted solution in mg/100 mL.
05

Adjust for Initial Sample Volume

Finally, convert the concentration found to reflect the original sample's concentration. The original 2.00 mL sample was diluted to a 10.0 mL solution. Thus, the initial concentration \( C_i \) can be calculated: \( C_i = 2.75 \times \frac{10.0}{2.0} \). Therefore, \( C_i = 2.75 \times 5 \approx 13.75 \) mg/100 mL.

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Key Concepts

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

Glucose Concentration Measurement
Glucose concentration measurement involves determining the amount of glucose present in a solution, usually expressed in mg/100 mL. In this scenario, an amperometric sensor is used, which measures the steady-state current proportional to the glucose concentration. This sensor is first calibrated using standard glucose solutions, enabling it to accurately measure unknown samples. Calibration involves preparing solutions with known glucose concentrations and measuring the resultant current. Once the calibration data is obtained, it can be used to determine the glucose concentration in an unknown sample by comparison. The accuracy of the measurement relies heavily on the precision of the calibration process, highlighting the importance of meticulous preparation of standard solutions and careful current measurements.
Linear Calibration Curve
A linear calibration curve is a tool used to establish a quantitative relationship between two variables, in this case, glucose concentration and sensor current. By plotting the current against the concentration, we can generate a straight line, described by the equation:
  • \( y = mx + c \)
Here, \( y \) is the current, \( x \) is the glucose concentration, \( m \) is the slope, and \( c \) is the y-intercept. This linear relationship implies that as the glucose concentration increases, the measured current increases proportionally. Calculating the slope involves selecting two data points and applying the formula:
  • \( m = \frac{(current_2 - current_1)}{(concentration_2 - concentration_1)} \)
Once the slope is determined, we find the y-intercept \( c \) by substituting any data point's values into the line equation. This linear equation helps predict the glucose concentration when an unknown current measurement is provided. The closer the data adheres to the line, the more reliable the predictions will be.
Dilution Factor
The dilution factor accounts for the effect of diluting a sample on its concentration. It鈥檚 crucial when measuring concentrations for small sample volumes that need to be expanded for easier handling and measurement. In this scenario, a 2.00 mL glucose solution is diluted to a final volume of 10 mL. The dilution factor is calculated as the ratio between the final volume and the original volume:
  • \( Dilution\ Factor = \frac{V_{final}}{V_{original}} \)
Here, the dilution factor is \( \frac{10}{2} = 5 \). When calculating the original concentration of a sample from its diluted state, you multiply the found concentration by this factor. For example, if the diluted concentration measured is 2.75 mg/100mL, the original concentration is obtained by:
  • \( Original\ Concentration = Diluent\ Concentration \times Dilution\ Factor \)
This calculation corrects for the effect of dilution, ensuring the concentration value reflects the undiluted sample鈥檚 true concentration. Understanding and applying dilution factors accurately is critical to obtaining precise analytical results.

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Most popular questions from this chapter

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