Question

Find the minimum detectable concentration if the average of the blanks is $\mathbf{1}\mathbf{.}{\mathbf{0}}_{\mathbf{5}}\mathit{n}\mathbf{A}$and $\mathbf{0}\mathbf{.}{\mathbf{6}}_{\mathbf{3}}\mathit{n}\mathbf{A}$.

Short answer

The minimum detectable concentration was calculated as $8.3\mu \mathrm{M}$.

Step-by-step solution

Concept used.

Detection limit: -

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

$=\frac{3_{\mathrm{s}}}{\mathrm{m}}$

Minimum detectable concentration

Where,

$\mathrm{S}=$Standard deviation.

$\mathrm{m}=$Slope of linear calibration curve.

Find the angle θ.

Given,

Standard deviation$=\mathrm{s}=0.63\mathrm{nA}$.

Slope of the calibration curve for higher concentrationrole="math" localid="1663342361783" $=\mathrm{m}=0.229\mathrm{nA}/\mathrm{\mu M}$

$=\frac{3_s}{\mathrm{m}}$

Minimum detectable concentration

Minimum detectable concentration

$=\frac{3\left(0.63\right)}{0.229}$

$=8.3\mu \mathrm{M}$

Hence, the minimum detectable concentration was calculated as $=8.3\mu \mathrm{M}$.