/*! 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 19 The concentration of the barbitu... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 19

The concentration of the barbiturate barbital in a blood sample is determined by extracting \(3.00 \mathrm{~mL}\) of blood with \(15 \mathrm{~mL}\) of \(\mathrm{CHCl}_{3}\). The chloroform, which now contains the barbital, is extracted with \(10.0 \mathrm{~mL}\) of \(0.45 \mathrm{M} \mathrm{NaOH}(\mathrm{pH} \approx 13)\). A 3.00-mL sample of the aqueous extract is placed in a 1.00 -cm cell and an absorbance of 0.115 is measured. The \(\mathrm{pH}\) of the sample in the absorption cell is then adjusted to approximately 10 by adding \(0.50 \mathrm{~mL}\) of \(16 \% \mathrm{w} / \mathrm{v} \mathrm{NH}_{4} \mathrm{Cl}\), giving an absorbance of 0.023 . When \(3.00 \mathrm{~mL}\) of a standard barbital solution with a concentration of \(3 \mathrm{mg} / 100 \mathrm{~mL}\) is taken through the same procedure, the absorbance at \(\mathrm{pH} 13\) is 0.295 and the absorbance at a \(\mathrm{pH}\) of 10 is 0.002. Report the mg barbital/100 mL in the sample.

Short Answer

Expert verified
The sample concentration is approximately 88.23 mg/100 mL.

Step by step solution

01

Determine Absorbance Changes

Identify how the absorbance of the sample changes with pH adjustment. The sample absorbance decreases from 0.115 at pH 13 to 0.023 at pH 10. Similarly, the standard's absorbance decreases from 0.295 to 0.002.
02

Calculate Absorbance Ratio for Sample

The absorbance ratio due to pH change for the sample is calculated as follows:\[\text{Ratio (Sample)} = \frac{0.023}{0.115} = 0.2\]
03

Calculate Absorbance Ratio for Standard

Similarly, calculate the absorbance ratio for the standard:\[\text{Ratio (Standard)} = \frac{0.002}{0.295} \approx 0.0068\]
04

Determining Barbital Concentration

Calculate the concentration of barbital in the sample by comparing the absorbance ratios:\[\text{Concentration Ratio} = \frac{0.2}{0.0068} \approx 29.41\]This ratio implies that the concentration of barbital in the sample is 29.41 times higher than that in the standard.
05

Adjust for Standard Concentration

Convert the concentration ratio to mg barbital per 100 mL by multiplying with the standard concentration. Since the standard concentration is 3 mg/100 mL, the sample concentration is:\[\text{Sample Concentration} = 29.41 \times 3 \approx 88.23 \text{ mg/100 mL}\]

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 Measurement
Absorbance measurement is a crucial technique in analytical chemistry for determining the concentration of compounds like barbital. This process involves using a spectrophotometer, which measures how much light is absorbed by a solution at a specific wavelength. The higher the absorbance, the more concentrated the solution.

The exercise involves measuring the absorbance of barbital in blood at two different pH levels. Initially, at a pH of 13, an absorbance of 0.115 is recorded. When the pH is adjusted to 10, the absorbance drops to 0.023. This change indicates how barbital's absorption characteristics vary with pH, which can be used to deduce the concentration of barbital by comparing these values to a standard solution.
pH Adjustment
pH adjustment is an essential step in many chemical analyses because the solubility and absorbance of substances often depend on the pH of their environment. In this exercise, the pH of the barbital solution was modified twice: first, the solution was made highly basic (pH 13) using sodium hydroxide, and later, the pH was lowered to 10 using ammonium chloride.

Each pH level affects the ionization state of barbital, influencing its absorbance. By adjusting the pH, we can optimize the conditions for accurate measurement and ensure that the absorbance data reflects only the changes in barbital concentration.
Barbital Concentration Determination
Determining the concentration of barbital in a sample involves comparing its absorbance with that of a standard solution. In this problem, absorbance measurements were taken before and after pH adjustment for both the sample and a standard solution.

The changes in absorbance give insight into the concentration differences. By calculating the absorbance ratios for both the sample and the standard, the concentration ratio is determined. This ratio is then multiplied by the known concentration of the standard to find the concentration of barbital in the sample, yielding a result of approximately 88.23 mg per 100 mL.
Extraction Techniques
Extraction techniques are fundamental in isolating specific analytes from complex matrices, such as blood. In the scenario presented, chloroform was used to extract barbital from the blood sample.

These techniques rely on the differing solubility of substances in various solvents. Barbital, being more soluble in chloroform than in blood, is efficiently separated into the organic layer during extraction. This separation allows for subsequent analysis and quantification in a simpler mixture.
Chloroform Extraction
Chloroform extraction is a common method employed to separate compounds from aqueous solutions due to its effective solvating properties. In this exercise, it serves to isolate barbital from blood by mixing a specific volume of chloroform with the sample.

Following the extraction, barbital dissolves preferentially in chloroform, creating an organic layer that can be separated from the aqueous phase. This step is crucial for eliminating other components that may interfere with further analyses, like absorbance measurement.

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

EDTA forms colored complexes with a variety of metal ions that may serve as the basis for a quantitative spectrophotometric method of analysis. The molar absorptivities of the EDTA complexes of \(\mathrm{Cu}^{2+}, \mathrm{Co}^{2+}\), and \(\mathrm{Ni}^{2+}\) at three wavelengths are summarized in the following table (all values of \(\varepsilon\) are in \(\left.\mathrm{M}^{-1} \mathrm{~cm}^{-1}\right).\) $$ \begin{array}{cccc} \text { metal } & \varepsilon_{462.9} & \varepsilon_{732.0} & \varepsilon_{378.7} \\ \hline \mathrm{Co}^{2+} & 15.8 & 2.11 & 3.11 \\ \mathrm{Cu}^{2+} & 2.32 & 95.2 & 7.73 \\ \mathrm{Ni}^{2+} & 1.79 & 3.03 & 13.5 \end{array} $$ Using this information determine the following: (a) The concentration of \(\mathrm{Cu}^{2+}\) in a solution that has an absorbance of 0.338 at a wavelength of \(732.0 \mathrm{nm}\). (b) The concentrations of \(\mathrm{Cu}^{2+}\) and \(\mathrm{Co}^{2+}\) in a solution that has an absorbance of 0.453 at a wavelength of \(732.0 \mathrm{nm}\) and 0.107 at a wavelength of \(462.9 \mathrm{nm}\) (c) The concentrations of \(\mathrm{Cu}^{2+}, \mathrm{Co}^{2+},\) and \(\mathrm{Ni}^{2+}\) in a sample that has an absorbance of 0.423 at a wavelength of \(732.0 \mathrm{nm}, 0.184\) at a wavelength of \(462.9 \mathrm{nm}\), and 0.291 at a wavelength of \(378.7 \mathrm{nm}\). The pathlength, \(b\), is \(1.00 \mathrm{~cm}\) for all measurements.

The following data is recorded for the phosphorescent intensity of several standard solutions of benzo[a] pyrene. $$ \begin{array}{cc} \text { [benzo[a]pyrene] }(\mathrm{M}) & \text { emission intensity } \\ \hline 0 & 0.00 \\ 1.00 \times 10^{-5} & 0.98 \\ 3.00 \times 10^{-5} & 3.22 \\ 6.00 \times 10^{-5} & 6.25 \\ 1.00 \times 10^{-4} & 10.21 \end{array} $$ What is the concentration of benzo[a] pyrene in a sample that yields a phosphorescent emission intensity of \(4.97 ?\)

A chemical deviation to Beer's law may occur if the concentration of an absorbing species is affected by the position of an equilibrium reaction. Consider a weak acid, HA, for which \(K_{\mathrm{a}}\) is \(2 \times 10^{-5}\). Construct Beer's law calibration curves of absorbance versus the total concentration of weak acid \(\left(C_{\text {total }}=[\mathrm{HA}]+\left[\mathrm{A}^{-}\right]\right),\) using values for \(C_{\text {total }}\) of \(1.0 \times 10^{-5}, 3.0 \times 10^{-5}, 5.0 \times 10^{-5}, 7.0 \times 10^{-5}, 9.0 \times 10^{-5}, 11 \times 10^{-5}\), and \(13 \times 10^{-5} \mathrm{M}\) for the following sets of conditions and comment on your results: (a) \(\varepsilon_{\mathrm{HA}}=\varepsilon_{\mathrm{A}^{-}}=2000 \mathrm{M}^{-1} \mathrm{~cm}^{-1} ;\) unbuffered solution. (b) \(\varepsilon_{\mathrm{HA}}=2000 \mathrm{M}^{-1} \mathrm{~cm}^{-1} ; \varepsilon_{\mathrm{A}^{-}}=500 \mathrm{M}^{-1} \mathrm{~cm}^{-1} ;\) unbuffered solution. (c) \(\varepsilon_{\mathrm{HA}}=2000 \mathrm{M}^{-1} \mathrm{~cm}^{-1} ; \varepsilon_{\mathrm{A}^{-}}=500 \mathrm{M}^{-1} \mathrm{~cm}^{-1} ;\) solution buffered to a \(\mathrm{pH}\) of 4.5 Assume a constant pathlength of \(1.00 \mathrm{~cm}\) for all samples.

The stoichiometry of a metal-ligand complex, \(\mathrm{ML}_{n}\), is determined by the mole-ratio method. A series of solutions are prepared in which the metal's concentration is held constant at \(3.65 \times 10^{-4} \mathrm{M}\) and the ligand's concentration is varied from \(1 \times 10^{-4} \mathrm{M}\) to \(1 \times 10^{-3} \mathrm{M}\). Using the following data, determine the stoichiometry of the metal-ligand complex. $$ \begin{array}{cccc} \text { [ligand] (M) } & \text { absorbance } & \text { [ligand] (M) } & \text { absorbance } \\ \hline 1.0 \times 10^{-4} & 0.122 & 6.0 \times 10^{-4} & 0.752 \\ 2.0 \times 10^{-4} & 0.251 & 7.0 \times 10^{-4} & 0.873 \\ 3.0 \times 10^{-4} & 0.376 & 8.0 \times 10^{-4} & 0.937 \\ 4.0 \times 10^{-4} & 0.496 & 9.0 \times 10^{-4} & 0.962 \\ 5.0 \times 10^{-4} & 0.625 & 1.0 \times 10^{-3} & 1.002 \end{array} $$

Kawakami and Igarashi developed a spectrophotometric method for nitrite based on its reaction with 5,10,15,20 -tetrakis( 4 -aminophenyl) porphrine (TAPP). As part of their study they investigated the stoichiometry of the reaction between TAPP and \(\mathrm{NO}_{2}^{-}\). The following data are derived from a figure in their paper. \({ }^{29}\) $$ \begin{array}{ccc} {[\mathrm{TAPP}](\mathrm{M})} & {\left[\mathrm{NO}_{2}^{-}\right](\mathrm{M})} & \text { absorbance } \\ \hline 8.0 \times 10^{-7} & 0 & 0.227 \\ 8.0 \times 10^{-7} & 4.0 \times 10^{-8} & 0.223 \\ 8.0 \times 10^{-7} & 8.0 \times 10^{-8} & 0.211 \\ 8.0 \times 10^{-7} & 1.6 \times 10^{-7} & 0.191 \\ 8.0 \times 10^{-7} & 3.2 \times 10^{-7} & 0.152 \\ 8.0 \times 10^{-7} & 4.8 \times 10^{-7} & 0.127 \\ 8.0 \times 10^{-7} & 6.4 \times 10^{-7} & 0.107 \\ 8.0 \times 10^{-7} & 8.0 \times 10^{-7} & 0.092 \\ 8.0 \times 10^{-7} & 1.6 \times 10^{-6} & 0.058 \\ 8.0 \times 10^{-7} & 2.4 \times 10^{-6} & 0.045 \\ 8.0 \times 10^{-7} & 3.2 \times 10^{-6} & 0.037 \\ 8.0 \times 10^{-7} & 4.0 \times 10^{-6} & 0.034 \end{array} $$ What is the stoichiometry of the reaction?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

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.