/*! 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} Q5-22 P Why is it desirable in the metho... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Q5-22 P (page 116)

Why is it desirable in the method of standard addition to add a small volume of concentrated standard rather than a large volume of dilute standard?

Short Answer

Expert verified

  • The large volume of dilute standard will alter the concentration of the matrix and will possibly result to matrix effect.

  • The addition of diluted standards may also cause a difficulty in detection.

  • The large volume of dilute standard may also affect the sensitivity of the instruments being used.

  • The measurements made will be nearly similar to each other when only a small volume of concentrated standard is added.

Step by step solution

01

Step 1:Concentrated and dilute standard definition

  • A standard solution is a solution that has a precisely known concentration of an element or compound in analytical chemistry.
  • A dilution is a solution that decreases the concentration of a solute by adding additional solvent to a more concentrated solution (stock solution).
02

Key points

There are a couple of reasons why it is desirable to add only a small volume of concentrated standard rather than a large volume of dilute standard:

  • The large volume of dilute standard will alter the concentration of the matrix and will possibly result to matrix effect.

  • The addition of diluted standards may also cause a difficulty in detection.

  • The large volume of dilute standard may also affect the sensitivity of the instruments being used.

  • The measurements made will be nearly similar to each other when only a small volume of concentrated standard is added.

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!

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

Standard addition graph. Students performed an experiment like that in Figure 5-7 in which each flask contained 25.00mLof serum, varying additions of 2.640MNaCIstandard, and a total volume of 50.00mL.


(a) Prepare a standard addition graph and find [Na+]in the serum.

(b) Find the standard deviation and 95%confidence interval for [Na+].

Internal standard. A solution was prepared by mixing 5.00mLof unknown elementXwith2.00mLof solution containingrole="math" localid="1654777035083" 4.13渭gof standard elementSper millilitre, and diluting to10.0mL. The signal ratio in atomic absorption spectrometry was (signal fromX)/ (signal fromS)=0.808. In a separate experiment, with equal concentrations ofXandS, (signal fromX)/signal fromS)=1.31. Find the concentration ofXin the unknown.

Let C stand for concentration. One definition of spike recovery is%recovery=Cspikedsample-CunspikedsampleCaded脳100

An unknown was found to contain10.0渭驳of analyte per liter. A spike ofrole="math" localid="1654945506958" 5.0渭驳/Lwas added to a replicate portion of unknown. Analysis of the spiked sample gave a concentration of15.3渭驳/LFind the percent recovery of the spike.

What is the difference between a calibration check and a performance test sample?

Detection limit. In spectrophotometry, we measure the concentration of an analyte by its absorbance of light. A low-concentration sample was prepared and nine replicate measurements gave absorbances of 0.0047,0.0054,0.0062,0.0060,0.0046,0.0056,0.0052,0.0044, and 0.0058. Nine reagent blanks gave values of 0.0006,0.0012, 0.0022,0.0005,0.0016,0.0008,0.0017,0.0010, and 0.0011.

a) Find the absorbance detection limit with equation 5-3.

b) The calibration curve is a graph of absorbance versus concentration. Absorbance is a dimensionless quantity. The slope of the calibration curve is m=2.24x104M-1Find the concentration detection limit with Equation 5-5.

(c) Find the lower limit of quantitation with Equation 5-6.
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Physical 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.