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Chapter 0: Q5-2 TY (page 1)

Find the minimum detectable concentration if the average of the blanks is 1.05nAand 0.63nA.

Short Answer

Expert verified

The minimum detectable concentration was calculated as 8.3μM.

Step by step solution

01

Concept used.

Detection limit: -

The concentration of an analyte that gives a signal with a standard deviation three times that of a blank signal.

=3sm

Minimum detectable concentration

Where,

S=Standard deviation.

m=Slope of linear calibration curve.

02

Find the angle θ.

Given,

Standard deviation=s=0.63nA.

Slope of the calibration curve for higher concentrationrole="math" localid="1663342361783" =m=0.229nA/μM

=3sm

Minimum detectable concentration

Minimum detectable concentration

=30.630.229

=8.3μM

Hence, the minimum detectable concentration was calculated as =8.3μM.

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