You want to determine the concentration of a dye in an unknown solution using vi

Question

You want to determine the concentration of a dye in an unknown solution using visible spectroscopy. The dye is known to exhibit maximum absorbance at 573 nm, with a molar absorptivity (elementof) of 84, 300 L/(mol middot cm). The original solution is too concentrated to give an absorbance reading below 1.00 at lambda_max so it is diluted by transferring 10.00 mL to a 25.00 mL volumetric flask and diluting to the mart. The dilute solution then exhibits an absorbance of 0.748 at lambda_max. What is the concentration of the dye in the original solution?

Explanation / Answer

Assume the concentration of stock solution is x mol/ml, so 10 ml contains 10xmol. When made up to 25 ml becomes 10x mol/25 ml

According to Beer -Lambert Law

A= e* C* l, where

A=Absorbance C= Concentration l=Length of cuvette e= Molar extinction coefficient

Given for dilute solution A=0.748 C= ? l= 1cm (say) e= 84300 L/(mol*cm)=84300 *10^3 ml/(mol*cm)

C=A/el

=0.748/{84300 *10^3 ml/(mol*cm) * 1cm}

= 8.87* 10^-9 mol/ml

which is equal to 10x mol/25 ml

ie, 10x mol/25 ml=8.87* 10^-9 mol/ml

solving for x, x=2.21*10^-8

So, concentration of stock solution=2.21*10^-8 mol/ml= 2.21*10^-5 mol/L

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