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

An unknown was found to contain10.0μ²µof analyte per liter. A spike ofrole="math" localid="1654945506958" 5.0μ²µ/Lwas added to a replicate portion of unknown. Analysis of the spiked sample gave a concentration of15.3μ²µ/LFind the percent recovery of the spike.

Short Answer

Expert verified

The range from96%to115%then106% is unacceptable. Something in your method or techniques needs improvement

Step by step solution

01

define the spike recovery

A spike, also called a fortification, is a known quantity of analyte added to a sample to test whether the response to the spike is the same as that expected from a calibration curve.

Spiked samples are analysed in the same manner as unknowns.

For example, if drinking water is found to contain 10.0μ²µof nitrate, a spike of5.0μ²µ/Lcould be added. Ideally, the concentration in the spiked portion found by analysis will be15.0μ²µ/L.If a number other than 15.0μ²µ/Lis found, then the matrix could be interfering with the analysis.

%recovery=Cspikedsample-CunspikedsampleCaded×100

02

Find the percentage.

The percent of the spike found by analysis is

%recovery=15.3μg/L-10.0μg/L5.0μg/L×100=106%

If the acceptable recovery is specified to be in the range from96%to115%then106% is unacceptable. Something in your method or techniques needs improvement.

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Most popular questions from this chapter

In a murder trial in the 1990 s, the defendant's blood was found at the crime scene. The prosecutor argued that blood was left by the defendant during the crime. The defense argued that police "planted" the defendant's blood from a sample collected later. Blood is normally collected in a vial containing the metal-binding compound EDTA (as an anticoagulant) at a concentration of ~4.5mMafter the vial is filled with blood. At the time of the trial, procedures to measure EDTA in blood were not well established. Even though the amount of EDTA found in the crime-scene blood was orders of magnitude below ~4.5mM

, the jury acquitted the defendant. This trial motivated the development of a new method to measure EDTA in blood.

(a) Precision and accuracy. To measure accuracy and precision of the method, blood was fortified with EDTA to known levels.

Accuracy=100×meanvaluefound-knownvalueknownvaluePrecision=100×standarddeviationmean=coeffcientofvariation

For each of the three spike levels in the table, find the precision and accuracy of the quality control samples.

(b) Detection and quantitation limits. Low concentrations of EDTA near the detection limit gave the following dimensionless instrument readings: 175,104,164,193,131,189,155,133,151, and 176. Ten blanks had a mean reading of 45 - 1 . The slope of the calibration curve is1.75×199M-1. Estimate the signal and concentration detection limits and the lower limit of quantitation for EDTA.

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Europium is a lanthanide element found at parts per billion levels in natural waters. It can be measured from the intensity of orange light emitted when a solution is illuminated with ultraviolet radiation. Certain organic compounds that bindEu(III)are required to enhance the emission. The figure shows standard addition experiments in which10.00mLof sample and20.00mLcontaining a large excess of organic additive were placed in 50-mL volumetric flasks. Then Eu(III) standards (0,5.00,10.00,or15.00mL) were added and the flasks were diluted to50.0mLwithH2O. Standards added to tap water contained0.152ng/mL(ppb) of Eu(III), but those added to pond water were 100 times more concentrated (15.2 ng/mL).


(a) Calculate the concentration of Eu(III)(ng/mL) in pond water and tap water.

(b) For tap water, emission peak area increases by.4.61units when 10.00mL of 0.152ng/mL standard are added. This response is4.61 units/1.52ng = 3.03units per ng ofEu(III). For pond water, the response is12.5units when10.00mLof15.2ng/mLstandard are added, or0.0822units per ng. How would you explain these observations? Why was standard addition necessary for this analysis?

Standard addition. An unknown sample of Ni2+gave a current of 2.36μ´¡in an electrochemical analysis. When 0.500mLof solution containing role="math" localid="1654761474124" 0.0187MNi2+was added to 25.0mLof unknown, the current increased to 3.79μ´¡.

(a) Denoting the initial, unknown concentration as [Ni2+], write an expression for the final concentration, [Ni2+]f, after role="math" 25.0mLof unknown were mixed with 0.500mLof standard. Use the dilution factor for this calculation.

(b) In a similar manner, write the final concentration of added standard Ni2+, designated as [S]f.

(c) Find[Ni2+]in the unknown.

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