Question

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

Answer 1

Answer:

For the control experiment: $0.0552 M$ of aqueous solution of erioglaucine has absorbance of $0.035 M$

From Lambert-Beer's law we know:

$Absorbance = e\cdot c \cdot l$

Here; e is the molar absorptivity coefficient of erioglaucine

l = length of cuvette in which the solution is taken = $1 cm$

A sorbance by the erioglaucine = total absorbance - absorbance by distilled $water = 0.331-0.019 = 0.312$

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

$0.312 = 0.0552 M \cdot e\cdot~1cm$

So; $e = \frac{0.331}{0.0552} M^{-1}cm^{-1}= 5.65 M^{-1}cm^{-1}$

The molar absorptivity coefficient of erioglaucine is $5.65 M^{-1}cm^{-1}$

The absorbance of erioglaucine in distilled water (contaminated with metal ions) is: $0 .217$

The absorbance of distilled water is $0.019$

So; absorbance of erioglaucine itself is : $0.217-0.019 = 0.198$

Again using Lambert Beer law; we get:

$A= e\cdot c\cdot l$

$0.198 = 5.65 M^{-1}cm^{-1} \cdot c \cdot 1 cm$ $( c = concentrartion)$

c = 0.198/5.65 M = 0.035 M

The concentration of the erioglaucine is $0.035 M$

Explanation: