Calibration data for an electrochemical determination of glucose are in the foll

Question

Calibration data for an electrochemical determination of glucose are in the following table. Because these data span six orders of magnitude, if plotted on a linear graph most of the points will be near the origin. A logarithmic plot overcomes this difficulty, leading to a amore even distribution of the data. Plot log(nA) as a function of log(mu g/mL). Determine the equation of the line in the form log(nA) = m times log(mu g/mL) + b. m =____b =____What concentration of glucose corresponds to a signal of 72.3 nA?

Explanation / Answer

Graph will be

y = 0.9378x + 1.5783

So , m = 0.3378

intercept = 1.578

b) The value of nA = 72.3 nA

so log 72.3 = 1.8591

Putting this value in the equation y = 0.9378x + 1.5783

1.8591 = 0.9378 X log(conc) + 1.5783

0.2808 = 0.9378 log(conc)

log (conc) = 0.2994

Concentration = 1.9925 micrograms / mL

Glucose Current Log concentration Log nA 0.00813 0.416 -2.09 -0.38 0.0331 1.55 -1.48 0.19 0.109 4.74 -0.96 0.68 0.267 11 -0.57 1.04 0.992 37.2 0.00 1.57 3.01 107 0.48 2.03 9.82 324 0.99 2.51 33.9 1020 1.53 3.01 123 3470 2.09 3.54 363 9550 2.56 3.98

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