Question

When measured with a 1.00-cm cell, a 0.00000433M solution of species A exhibited absorbances of 0.133 and 0.815 at 450 and 730nm, respectively. A 0.00000286M solution of species B gave absorbances of 0.735 and 0.102 under the same conditions. When solutions of A and B of different concentrations were measured using a 2.50-cm cell, the following absorbance data were obtained: 0.410 at 450nm and 0.965 at 730nm. The calculated molar concentration values for these conditions are?

Answer 1

Ans:

( In the question data of solution B must be reverse then problem solved)

According to Beer-Lambert law,. A=€.l.c

Where, A= absorbance, l= path length in cm, c= concentration in molarity and €= molar absorptivity constant.

For solution A at 450nm,

A=€.l.c

0.133=€ X 1 X 0.00000433

€=3.0715 X 10⁴

For solution A at 730nm,

0.815=€ X 1 X 0.00000433

€=1.882 X 10⁵

For solution B at 450nm

0.102=€ X 1 X 0.00000286

€= 3.566 X 10⁴

For solution B at 730nm

0.735=€ X 1 X 0.00000286

€= 2.569 X 10⁵

So according to above data

Average value of € at 450nm = 3.31875 X 10⁴

Average value of € at 730nm = 2.2255 X 10⁵

Molar concentration of solution A at 450nm using 2.5 cm cell,

C= 0.410/(3.31875 X 10⁴ X 2.5)

= 0.00000494M

Molar concentration of solution B at 730nm using 2.5 cm cell,

C=0.965/(2.2255 X 10⁵ X 2.5)

= 0.000001734M