Question

Assume that the absorption signal shown in Figure 9-7 were obtained for $2-\mathrm{\mu L}$aliquots of standards and sample. Calculate the concentration in parts per million of lead in the sample of canned orange juice.

Short answer

The concentration of lead in the juice is$0.046\pm 0.094\mathrm{ppm}$.

Step-by-step solution

Step-1 Given Information

The lead concentration in parts per million (ppm) in a canned orange juice sample needs to be determined.

Step-2: Explanation

As per the given figure, for the standards $0.2,0.1,0.05,$the absorbance values are data-custom-editor="chemistry" $0.35,0.19,0.09$respectively.

Around 0.1 is the absorbance of the sample.

Calculating the sample concentration using excel:

The standard deviation can be calculated as follows:

$\text{SD=}\frac{\mathrm{Standard}\mathrm{error}\mathrm{in}y}{\mathrm{slope}}\times \sqrt{\frac{1}{\mathrm{replicates}}+\frac{1}{\mathrm{standards}}+\frac{{(\mathrm{sample}\mathrm{absorbance}-\mathrm{average}\mathrm{ratio})}^2}{({\mathrm{slope}}^2\times S_{\mathit{X}\mathit{X}})}}$

Putting the values,

$$\begin{gathered} \text{SD=}\frac{0.01069045}{1.51428571}\times \sqrt{\frac{1}{1}+\frac{1}{3}+\frac{{(0.1-0.20666667)}^2}{(1.{51428571}^2\times 0.01166667)}} \\ =7.06\times {10}^{-3}\sqrt{1+0.33+\frac{0.011378}{0.0267524}} \\ =9.354\times {10}^{-3}\approx 0.094 \end{gathered}$$

The concentration of lead in the juice isdata-custom-editor="chemistry" $0.046\pm 0.094\mathrm{ppm}$