/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 A solution with a "true" absorba... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 24: Problem 25

A solution with a "true" absorbance \(\left[A=-\log \left(P_{0} / P\right)\right]\) of \(2.10\) was placed in a spectrophotometer with a stray light percentage \(\left(P_{J} / P_{0}\right)\) of \(0.75\). What absorbance \(A^{\prime}\) would be measured? What percentage error would result?

Short Answer

Expert verified
Apparent absorbance A' is lower than 2.10 due to stray light, resulting in a positive percentage error.

Step by step solution

01

Understanding Absorbance and Stray Light

Absorbance (A) is calculated using the formula \( A = -\log\left(\frac{P_0}{P}\right) \). This formula assumes there is no stray light affecting the measurements. However, if there is stray light, the intensity of light measured becomes \( P + P_J \) due to the additional light component. The stray light percentage given is \( \frac{P_J}{P_0} = 0.75 \).
02

Calculate Corrected Intensity

To find the corrected intensity (P_J), use the stray light percentage: \( P_J = 0.75 \times P_0 \). Then, calculate the effective transmitted intensity as \( P + P_J \). Given the true absorbance of 2.10, calculate P using \( 10^{-2.10} = \frac{P}{P_0} \). Therefore, \( P = P_0 \times 10^{-2.10} \).
03

Calculate Apparent Absorbance A'

Substituting P from the previous step, calculate A' (the apparent absorbance) considering the stray light effect:\[ A' = -\log\left(\frac{P_0}{P + P_J}\right). \] Substituting the values \( P_J \) and \( P \), calculate \( A' \).
04

Calculate Percentage Error

Percentage error is calculated by the formula: \( \text{Percentage Error} = \left| \frac{A - A'}{A} \right| \times 100\%. \) Substitute A = 2.10 and the calculated A' to find the error.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Absorbance
Absorbance is a crucial concept when working with spectrophotometry. It helps us understand how much light a substance absorbs at a particular wavelength. The formula for absorbance is given by \( A = -\log\left(\frac{P_0}{P}\right) \). Here, \( P_0 \) is the intensity of light before it passes through the sample, and \( P \) is the intensity after. A higher absorbance value indicates that more light is being absorbed by the sample. This is important because it relates directly to the concentration of the substance in the solution under study.

When performing experiments, it's often assumed there's no external interference, like stray light, affecting the results. But, as we will see, stray light can impact exact measurements and should not be overlooked. Understanding absorbance allows us to quantify and compare how different samples interact with light.
Stray Light
Stray light refers to any unintended or 'extra' light that enters the spectrophotometer detector alongside the light that passes through the sample. In the context of spectrophotometry, this can affect the accuracy of measurement. Stray light, usually coming from reflections or internal scattering, impacts the intensity reading that reaches the detector.

For example, if a system set up for high accuracy has stray light, it means that instead of measuring only the light passing directly through the sample, we are also measuring the light coming from other sources. In our exercise, this stray light was quantified by the percentage \( \left( \frac{P_J}{P_0} = 0.75 \right) \). To mitigate its effect, the correct intensity (P_J) is used. The adjusted readings would factor that in by adding it to \( P \), refining our absorbance calculations. Hence managing and correcting for stray light is essential for an accurate spectrophotometric analysis.
Percentage Error
When conducting experiments, it's important to evaluate the accuracy of your results. Percentage error offers a way to do this by comparing your measured value to a known or theoretical one. The formula for percentage error is \( \text{Percentage Error} = \left| \frac{A - A'}{A} \right| \times 100\% \).

This formula helps you understand the degree of deviation between the true value (\( A \)) and the apparent value (\( A' \)) measured, especially when factors like stray light are in play.
By calculating percentage error, one can evaluate the reliability of an experiment's outcomes and figure out whether adjustments or calibrations might be necessary to improve accuracy in future measurements. This tool is invaluable for maintaining rigorous standards in laboratory work.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The complex formed between \(\mathrm{Cu}(\mathrm{I})\) and 1,10phenanthroline has a molar absorptivity of \(7000 \mathrm{~L}\). \(\mathrm{cm}^{-1} \mathrm{~mol}^{-1}\) at \(435 \mathrm{~nm}\), the wavelength of maximum absorption. Calculate (a) the absorbance of a \(6.17 \times 10^{-5} \mathrm{M}\) solution of the complex when measured in a \(1.00-\mathrm{cm}\) cell at \(435 \mathrm{~nm}\). (b) the percent transmittance of the solution in (a). (c) the concentration of a solution that in a \(5.00-\mathrm{cm}\) cell has the same absorbance as the solution in (a). (d) the path length through a \(3.13 \times 10^{-5} \mathrm{M}\) solution of the complex that is needed for an absorbance that is the same as the solution in (a).

A solution containing the complex formed between \(\mathrm{Bi}(\mathrm{III})\) and thiourea has a molar absorptivity of \(9.32 \times 10^{3} \mathrm{~L} \mathrm{~cm}^{-1} \mathrm{~mol}^{-1}\) at \(470 \mathrm{~nm}\). (a) What is the absorbance of a \(5.67 \times 10^{-1} \mathrm{M}\) solution of the complex at \(470 \mathrm{~nm}\) in a \(1.00-\mathrm{cm}\) cell? (b) What is the percent transmittance of the solution described in (a)? (c) What is the molar concentration of the complex in a solution that has the absorbance described in (a) when measured at \(470 \mathrm{~nm}\) in a \(2.50-\mathrm{cm}\) cell?

Nitrite is commonly determined by a colorimetric procedure using a reaction called the Griess reaction. In this reaction, the sample containing nitrite is reacted with sulfanilimide and \(\mathrm{N}\)-(1-Napthyl) ethylenediamine to form a colored species that absorbs at \(550 \mathrm{~nm}\). Using an automated flow analysis instrument, the following results were obtained for standard solutions of nitrite and for a sample containing an unknown amount: \begin{tabular}{rc} \multicolumn{1}{c}{ Solution } & Absorbance at \(550 \mathrm{~nm}\) \\ \hline \(2.00 \mu \mathrm{M}\) & \(0.065\) \\ \(6.00 \mu \mathrm{M}\) & \(0.205\) \\ \(10.00 \mu \mathrm{M}\) & \(0.338\) \\ \(14.00 \mu \mathrm{M}\) & \(0.474\) \\ \(18.00 \mu \mathrm{M}\) & \(0.598\) \\ Unknown & \(0.402\) \\ \hline \end{tabular} (a) Find the slope, intercept, and standard deviation of the calibration curve. (b) Construct the calibration curve. (c) Determine the concentration of nitrite in the sample and its standard deviation.

What are the units for absorptivity when the path length is given in centimeters and the concentration is expressed in *(a) parts per million? (b) micrograms per liter? *(c) mass-volume percent? (d) grams per liter?

. One common way to determine phosphorus in urine is to treat the sample after removing the protein with molybdenum (VI) and then reducing the resulting 12-molybdophosphate complex with ascorbic acid to give an intense blue-colored species called molybdenum blue. The absorbance of molybdenum blue can be measured at \(650 \mathrm{~nm}\). A 24-hour urine sample was collected, and the patient produced \(1122 \mathrm{~mL}\). in 24 hours. A \(1.00 \mathrm{~mL}\) aliquot of the sample was treated with Mo(VI) and ascorbic acid and diluted to a volume of \(50.00 \mathrm{~mL}\). A calibration curve was prepared by treating \(1.00 \mathrm{~mL}\) aliquots of phosphate standard solutions in the same manner as the urine sample. The absorbances of the standards and the urine sample were obtained at \(650 \mathrm{~nm}\) and the following results obtained: \begin{tabular}{lc} Solution & Abserbance at \(650 \mathrm{~nm}\) \\ \hline \(1.00 \mathrm{ppm} \mathrm{P}\) & \(0.230\) \\ \(2.00 \mathrm{ppm} \mathrm{P}\) & \(0.436\) \\ \(3.00 \mathrm{ppm} \mathrm{P}\) & \(0.638\) \\ \(4.00 \mathrm{ppm} \mathrm{P}\) & \(0.848\) \\ Urine sample & \(0.518\) \\ \hline \end{tabular} (a) Find the slope, intercept, and standard error in \(y\) of the calibration curve. Construct a calibration curve. Determine the concentration number of phosphorus in ppm in the urine sample and its standard deviation from the least-squares equation of the line. Compare the unknown concentration to that obtained manually from a calibration curve. (b) What mass in grams of phosphorus was eliminated per day by the patient? (c) What is the phosphate concentration in urine in \(\mathrm{mM}\) ?
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.