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

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 = (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\\`

Step-by-step explanation: