/*! 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 15 A photometer with a linear respo... [FREE SOLUTION] | 91Ó°ÊÓ

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A photometer with a linear response to radiation gave a reading of \(498 \mathrm{mV}\) with the solvent in the light path and \(256 \mathrm{mV}\) when the solvent was replaced by an absorbing solution. The photometer was set to zero with no light striking the detector. Calculate (a) the percent transmittance and absorbance of the absorbing solution. (b) the expected transmittance if the concentration of absorber is one half that of the original solution. (c) the transmittance to be expected if the light path through the original solution is doubled.

Short Answer

Expert verified
(a) T% ≈ 51.41%, A ≈ 0.289; (b) T% ≈ 71.58%; (c) T% ≈ 26.92%

Step by step solution

01

Calculate Transmittance Percentage (T%)

The percent transmittance is calculated using the formula:\[ T\% = \left( \frac{I}{I_0} \right) \times 100\% \]where \(I\) is the intensity with the solution (256 mV) and \(I_0\) is the intensity with the solvent (498 mV). Thus,\[ T\% = \left( \frac{256}{498} \right) \times 100\% \approx 51.41\% \].
02

Calculate Absorbance (A)

Absorbance is calculated using the formula:\[ A = -\log_{10}\left( \frac{I}{I_0} \right) \]Using \(I = 256\) mV and \(I_0 = 498\) mV, we have:\[ A = -\log_{10}\left( \frac{256}{498} \right) \approx 0.289 \].
03

Calculate Transmittance at Half Concentration

According to the Beer-Lambert Law, absorbance is directly proportional to concentration. If the concentration is halved, absorbance is halved, thus:\[ A_{\text{half}} = \frac{0.289}{2} \approx 0.1445 \]Transmittance can be recovered from absorbance using:\[ T\%_{\text{half}} = 10^{-0.1445} \times 100\% \approx 71.58\% \].
04

Calculate Transmittance for Doubled Light Path

Doubling the path length can be represented as doubling the absorbance (as per Beer-Lambert Law, path length and concentration are directly proportional):\[ A_{\text{double}} = 0.289 \times 2 = 0.578 \]Thus, the corresponding percent transmittance is:\[ T\%_{\text{double}} = 10^{-0.578} \times 100\% \approx 26.92\% \].

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Key Concepts

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

Transmittance
Transmittance is a measure of how much light passes through a substance. It's expressed as a percentage of the original light that makes it through. Imagine you have a flashlight and a piece of tinted glass. How much light from the flashlight can pass through the glass? That's transmittance in action!

Mathematically, transmittance is given by the formula:
  • \[ T\% = \left( \frac{I}{I_0} \right) \times 100\% \]
Where:
  • \( I \) is the intensity of light through the solution.
  • \( I_0 \) is the initial light intensity with just the solvent (no absorbing material).
Using the numbers from our exercise, \( I = 256 \text{ mV} \) and \( I_0 = 498 \text{ mV} \), we calculate:
  • \[ T\% = \left( \frac{256}{498} \right) \times 100\% \approx 51.41\% \]
This means a bit over half of the initial light is making it through the absorbing solution. Lower transmittance means less light is getting through, suggesting the solution is quite good at absorbing the incoming light.
Absorbance
Absorbance tells us how much light gets absorbed by a sample rather than passing through it. This is a crucial concept in photometry as it helps determine the concentration of a substance in a solution.

Absorbance is calculated using the logarithmic formula:
  • \[ A = -\log_{10}\left( \frac{I}{I_0} \right) \]
Where:
  • \( I \) is the light intensity with the solution.
  • \( I_0 \) is the original light intensity with just the solvent.
In the given problem, with \( I = 256 \text{ mV} \) and \( I_0 = 498 \text{ mV} \), the absorbance is:
  • \[ A = -\log_{10}\left( \frac{256}{498} \right) \approx 0.289 \]
A higher absorbance value implies a stronger capacity of the solution to capture and hold light energy. It's essential for quantifying concentrations, as absorbance is directly proportional to how much of the absorbing substance is present.
Beer-Lambert Law
The Beer-Lambert Law provides a relationship between the absorbance of light by a solution and its concentration. Imagine it like a bridge connecting these two properties.

The law suggests that absorbance (\( A \)) is proportional to both:
  • Concentration (\( c \)) of the absorbing species.
  • Path length (\( l \)), which is the distance the light travels through the solution.
Using the formula:
  • \[ A = \varepsilon c l \]
Where:
  • \( \varepsilon \) is the molar absorptivity.
This relationship is highly useful for calculating unknown concentrations in solutions. In our exercise, if concentration is halved, absorbance is also halved, leading to a recalculated transmittance:
  • \[ T\%_{\text{half}} = 10^{-0.1445} \times 100\% \approx 71.58\% \]
If the light path is doubled, another common scenario:
  • \[ A_{\text{double}} = 0.289 \times 2 = 0.578 \]
  • \[ T\%_{\text{double}} = 10^{-0.578} \times 100\% \approx 26.92\% \]
This law helps in understanding how changes in concentration or path length affect light absorption, crucial for practical applications in labs.

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Most popular questions from this chapter

Describe the origin of shot noise in a spectrophotometer. How does the relative uncertainty vary with concentration if shot noise is the major noise source?

Describe the differences between the following and list any particular advantages possessed by onc over the other. (a) hydrogen and deuterium discharge lamps as sources for ultraviolet radiation. (b) filters and monochromators as wavelength selectors. (c) photovoltaic cells and phototubes as detectors for electromagnetic radiation. (d) photodiodes and photomultiplier tubes. (e) double-beam-in-space and double-beam-in-time spectrophotometers. (f) spectrophotometers and photometers. (g) single-beam and double-beam instruments for absorbance measurements. (h) conventional and multichannel spectrophotometers.

A compound \(X\) is to be determined by UV-visible spectrophotometry. A calibration curve is constructed from standard solutions of \(X\) with the following results: \(0.50 \mathrm{ppm}, A=0.24 ; 1.5 \mathrm{ppm}, A=0.36 ; 2.5 \mathrm{ppm}, A=0.44 ; 3.5 \mathrm{ppm}, A=0.59\); \(4.5 \mathrm{ppm}, A=0.70\). A solution of unknown \(\mathrm{X}\) concentration had an absorbance of \(A=0.50\). Find the slope and intercept of the calibration curve, the standard error in Y, the concentration of the solution of unknown X concentration, and the standard deviation in the concentration of \(X\). Construct a plot of the calibration curve and determine the unknown concentration by hand from the plot. Compare it to that obtained from the regression line.

Why does a deuterium lamp produce a continuum rather than a line spectrum in the ultraviolet?

The following data were taken from a diode-array spectrophotometer in an experiment to measure the spectrum of the Co(II)-EDTA complex. The column labeled \(P_{\text {solution }}\) is the relative signal obtained with sample solution in the cell after subtraction of the dark signal. The column labeled \(P_{\text {sulicu1 }}\) is the reference signal obtained with only solvent in the cell after subtraction of the dark signal. Find the transmittance at each wavelength. and the absorbance at cach wavelength. Plot the spectrun of the compound. \begin{tabular}{ccc} Wavelength, nm & \(\boldsymbol{P}_{\text {solvent }}\) & \(\boldsymbol{P}_{\text {solution }}\) \\ \hline 350 & \(0 .(12689\) & \(0.002560\) \\ 375 & \(0.006326\) & \(0 .(0) 5995\) \\ 400 & \(0.016975\) & \(0.015143\) \\ 425 & \(0.035517\) & \(0.031648\) \\ \(4.50\) & \(0.062425\) & \(0.024978\) \\ 475 & \(0.095374\) & \(0.019073\) \\ 500 & \(0.140567\) & \(0.023275\) \\ 525 & \(0.188984\) & \(0.0137448\) \\ \(5.0\) & \(0.26 .310 .3\) & \(0.088537\) \\ 575 & \(0.318361\) & \(0.200872\) \\ 600 & \(0.394600\) & \(0.278072\) \\ 625 & \(0.477018\) & \(0.363525\) \\ 650 & \(0.564295\) & \(0.468281\) \\ 675 & \(0.655066\) & \(0.611062\) \\ 700 & \(0.739180\) & \(0.704126\) \\ 725 & \(0.813694\) & \(0.777466\) \\ 750 & \(0.885979\) & \(0.863224\) \\ 775 & \(0.945083\) & \(0.921416\) \\ 800 & \(1.00(k) 00\) & \(0.977237\) \\ \hline \end{tabular}

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