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Erioglucine is a blue-colored dye that absorbs its complementary color, red, in aqueous solution around 645 nm. Unfortunately, the local distilled water supply used to make solutions is consistently contaminated with a trace amount of a metal cation that also absorbs 645 nm light. Suppose then, a control sample of 0.0552 M erioglucine (aq) has an absorbance of 0.331 and that a distilled water sample in a similar cuvette has an absorbance of 0.019. Determine the concentration of an erioglucine (aq) sample that has an absorbance of 0.217. 10 pts.

The answer is .0350 but I wanted to see how to get that answer. You have to subtract .019 from the absorbance of the control and the sample

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danialamin

Answer:

For the control experiment: [tex]0.0552 M[/tex] of aqueous solution of erioglaucine has absorbance of [tex]0.035 M\\[/tex]

From Lambert-Beer's law we know:

[tex]Absorbance = e\cdot c \cdot l\\[/tex]

Here; e is the molar absorptivity coefficient of erioglaucine

l = length of cuvette in which the solution is taken = [tex]1 cm[/tex]

A sorbance by the erioglaucine = total absorbance - absorbance by distilled [tex]water = 0.331-0.019 = 0.312[/tex]

So; by putting the values in the above equation; we get:

[tex]0.312 = 0.0552 M \cdot e\cdot~1cm\\[/tex]

So; [tex]e = \frac{0.331}{0.0552} M^{-1}cm^{-1}= 5.65 M^{-1}cm^{-1}[/tex]

The molar absorptivity coefficient of erioglaucine is [tex]5.65 M^{-1}cm^{-1}\\[/tex]

The absorbance of erioglaucine in distilled water (contaminated with metal ions) is: [tex]0 .217[/tex]

The absorbance of distilled water is [tex]0.019[/tex]

So; absorbance of erioglaucine itself is : [tex]0.217-0.019 = 0.198[/tex]

Again using Lambert Beer law; we get:

[tex]A= e\cdot c\cdot l[/tex]

[tex]0.198 = 5.65 M^{-1}cm^{-1} \cdot c \cdot 1 cm[/tex] [tex]( c = concentrartion)[/tex]

c = 0.198/5.65 M = 0.035 M

The concentration of the erioglaucine is [tex]0.035 M\\[/tex]

Explanation:

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