Chapter 4, Problem 4.37P

(a)

Interpretation Introduction

Interpretation:

A graph of $\text{log(current)}$ versus $\text{log(concentration)}$ has to be drawn and also the range at the log-log calibration gets linear has to be given.

Concept Introduction:

Calibration curves:

The response of an analytical methods to the known quantities of an analyte is shown by a calibration curve.

Standard solution are solutions that contains a known concentrations of analyte.

Blank solutions are solutions that has all the reagents and solvents used in analysis, but will not contain added analyte.

(b)

Interpretation Introduction

Interpretation:

A equation of line in the form of $\log \text{(current)}=m\times \log \text{(concentration)}+b$ has to be found out.

Concept Introduction:

Calibration curves:

The response of an analytical methods to the known quantities of an analyte is shown by a calibration curve.

Standard solution are solutions that contains a known concentrations of analyte.

Blank solutions are solutions that has all the reagents and solvents used in analysis, but will not contain added analyte.

Linear Calibration Curve:

The equation of linear calibration line is:

$y(\pm s_y)=[m(\pm u_m)]x+[b(\pm u_b)]$

(c)

Interpretation Introduction

Interpretation:

The concentration of p-nitrophenol corresponding to a signal of $99.9\text{ nA}$ has to be found out.

Concept Introduction:

Linear Calibration Curve:

The equation of linear calibration line is:

$y(\pm s_y)=[m(\pm u_m)]x+[b(\pm u_b)]$

Where,

$m$ is the slope

$y$ is the intercept

(d)

Interpretation Introduction

Interpretation:

Using propagation of uncertainty from chapter 3, the uncertainty in concentration has to be found out.

Concept Introduction:

Exponents and logarithm:

Consider, $y=\log x$ .

Here, the absolute uncertainty in $y(e_y)$ is proportional to the relative uncertainty in x, which is $e_x/x$

$\begin{array}{l}\text{Uncertainty for logarithm:}\\ y=1\text{og}x\\ \Rightarrow e_y=\frac{1}{\text{ln 10 }x}\frac{e_x}{x}\\ \approx 0.43429\frac{e_x}{x}\end{array}$