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Absorbance proportional to concentration of i2

How to calculate the concentration of I2 if proportional to the absorbance

Question 3b unified chemistry set 2
Absorbance=Concentration*Path Length*Extinction coefficient
Original post by c_rebeca
How to calculate the concentration of I2 if proportional to the absorbance

Question 3b unified chemistry set 2


In that question the 1cm3 of 1 mol dm-3 iodine is diluted with the other two reactants so the initial concentration of iodine is actually 0.01 mol dm-3. This corresponds to the absorbance of 0.80. As you are told that the relationship between iodine concentration is proportional, you just have to use ratios to find the iodine concentration that has an absorbance of 0.18
1.00cm3 of 1.00moldm–3 I2(aq) 49.5cm3 of 1.00moldm–3 CH3COCH3(aq) 49.5cm3 of 1.00moldm–3 HCl(aq)c1v1=c2v2 (I2 before dilution = I2 after dilution with other reactants)c2=c1v1/v2c2=(1.00x1.00)/(49.5 49.5 1.00) = 0.01 moldm^-3 (at 0.80 absorbance)So at 0.18 absorbance, as absorbance is proportional to concentration of I2, use ratios:0.80/0.18=4.440.01/4.44=2.25x10^-3 moldm^-3 (at 0.18 absorbance).
The question gives us:
1.00cm3 of 1.00moldm–3 I2(aq)
49.5cm3 of 1.00moldm–3 CH3COCH3(aq)
49.5cm3 of 1.00moldm–3 HCl(aq)

So use the equation:
c1v1=c2v2 (I2 before dilution = I2 after dilution with other reactants)

c2=c1v1/v2
c2=(1.00x1.00)/(49.5 49.5 1.00) = 0.01 moldm^-3 (at 0.80 absorbance)

So at 0.18 absorbance, as absorbance is proportional to concentration of I2, use ratios:

0.80/0.18=4.44

0.01/4.44=2.25x10^-3 moldm^-3 (at 0.18 absorbance)

(Sorry I don't know what happened to the formatting).
(edited 1 year ago)

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