Logarithmic calibration curve. Calibration data spanning five orders of magnitude for an electrochemical determination of p-nitrophenol are given in the table. (The blank has already been subtracted from the measured current.) If you try to plot these dataon a linear graph extending from 0 to $\mathbf{310}\mathbf{\mu g}\mathbf{/}\mathbf{mL}$and from 0 to $\mathbf{5260}\mathbf{nA}$, most of the points will be bunched up near the origin. To handle data with such a large range, a logarithmic plot is helpful.

Overwhatrangeisthelog-logcalibrationlinear?

(a) Make a graph of log (current) versus log( concentration). Over what range is the log-log calibration linear?

(b)FindtheequationoftheLine

InTheform $\mathbf{log}\left(current\right)\mathbf{=}\mathit{m}\mathbf{\times }\mathbf{log}\left(concentration\right)\mathbf{+}\mathit{b}$

(c) Find the concentration of p-nitrophenol corresponding to a signal of 99.9nA.

(d) Propagation of uncertainty with logarithm. For a signal of 99.9nA, log (concentration) and its standard uncertainty turn out to be $\mathbf{0}\mathbf{.}\mathbf{68315}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{04522}$. With rules for propagation of uncertainty from Chapter 3, find the uncertainty in concentration.

Step by step solution

01

log-log calibration

(a) For this problem, we're going to use a worksheet. Enter the data, then form two more columns with the logarithmic values, and insert a scatter graph.

Notice that the log-log calibration is linear all over.

02

Equation of the line

(b) To find the equation of the line in the form $\log \left(\mathrm{current}\right)=\mathrm{m}·\log \left(\mathrm{concentration}\right)+\mathrm{b}$, we must edit the obtained graph and put a Linear Trendline. The equation of the line is$\log \left(\mathrm{current}\right)=0.9692·\log \left(\mathrm{concentration}\right)+1.3389.$

03

Value of concentration

(c)

Now that we know the equation of the line, we can calculate the value of the concentration by knowing the value of the current:

$$\begin{gathered} \log \left(\mathrm{current}\right)=0.9692·\log \left(\mathrm{concentration}\right)+1.3389 \\ =\frac{\log \left(\mathrm{current}\right)-1.3389}{0.9692} \\ =0.6817. \\ \mathrm{Concentration}={10}^{0.6817} \\ =4.808\mathrm{\mu g}/\mathrm{mL}. \end{gathered}$$

04

Propagation of uncertainty

(d)

The propagation of uncertainty with the logarithm is:

$\begin{array}{r}y=\log x\to e_y=\frac{1}{\ln 10}\frac{e_x}{x}\approx 0.43429\frac{e_x}{x}\\ ={10}^x\to \frac{e_y}{y}=(\ln 10)e_x\approx 2.3026e_x\end{array}$

In this case, we use the second equation as we are calculating the concentration from the logarithm.

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