Q37P Logarithmic calibration curve. C... [FREE SOLUTION]

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Chapter 4: Q37P (page 93)

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 $310 渭驳 / mL$ and from 0 to $5260 nA$ , 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 $\log (c u r r e n t) = m 脳 \log (c o n c e n t r a t i o n) + b$
(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 $0.68315 卤 0.04522$ . With rules for propagation of uncertainty from Chapter 3, find the uncertainty in concentration.

Short Answer

Expert verified

(a)The log-log calibration is linear all over.

(b)The equation of the line is:

$\log (current) = 0.9692 路 \log (concentration) + 1.3389 .$

(d) Propagation of uncertainty with logarithm is:

localid="1667561817319" $\begin{aligned} y = \log x 鈫� e_{y} = \frac{1}{\ln 10} \frac{e_{x}}{x} 鈮� 0.43429 \frac{e_{x}}{x} \\ = 10^{x} 鈫� \frac{e_{y}}{y} = (\ln 10) e_{x} 鈮� 2.3026 e_{x} \end{aligned}$

Step by step solution

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.

Equation of the line

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

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:

$\log (current) = 0.9692 路 \log (concentration) + 1.3389 = \frac{\log (current) - 1.3389}{0.9692} = 0.6817 . Concentration = 10^{0.6817} = 4.808 渭驳 / mL .$

Propagation of uncertainty

(d)

The propagation of uncertainty with the logarithm is:

$\begin{aligned} y = \log x 鈫� e_{y} = \frac{1}{\ln 10} \frac{e_{x}}{x} 鈮� 0.43429 \frac{e_{x}}{x} \\ = 10^{x} 鈫� \frac{e_{y}}{y} = (\ln 10) e_{x} 鈮� 2.3026 e_{x} \end{aligned}$