Question

When measured in a 1.00 cm cuvet, a 8.50 x 10-5 M solution of species A exhibited absorbances of 0.129 and 0.764 at 475 nm and 700 nm, respectively. A 4.65 x 10-5 M solution of species B gave absorbances of 0.567 and 0.083 at 475 nm and 700 nm, respectively. Both species were dissolved in the same solvent, and the solvent's absorbance was 0.005 and 0.000 at 475 nm and 700 nm, respectively, in either a 1 or 1.25 cm cuvet. Calculate the concentrations of A and B in a solution that yielded the following absorbance data in a 1.25 cm cuvet: 0.502 at 475 nm and 0.912 at 700 nm.

(a) cB = 7.65 x 10-5 Molar and cA = 2.37 x 10-5 molar

(b) cB = 2.37 x 10-5 molar and cA = 7.65 x 10-5 molar

(c) cB = 4.65 x 10-5 molar and cA = 8.50 x 10-5 molar

Answer 1

Use Beer’s law.

A = ε*C*l

where A is the absorbance of a solution having concentration C and travelling through a path length of l.

For species A, we have

475 nm:

0.129 = ε_1,A*(8.50*10^-5 M)*(1.00 cm)

====> ε_1,A = (0.129)/(8.50*10^-5 M)(1 cm) = 1517.65 M^-1.cm^-1

700 nm:

0.764 = ε_2,A*(8.50*10^-5 M)*(1.00 cm)

====> ε_2,A = (0.764)/(8.50*10^-5 M)(1 cm) = 8988.23 M^-1.cm^-1

For species B, we have

475 nm:

0.567 = ε_1,B*(4.65*10^-5 M)*(1.00 cm)

====> ε_1,B = (0.764)/(4.65*10^-5 M)(1 cm) = 16430.11 M^-1.cm^-1

700 nm:

0.083 = ε_2,B*(4.65*10^-5 M)*(1.00 cm)

====> ε_2,B = (0.083)/(4.65*10^-5 M)(1 cm) = 1784.95 M^-1.cm^-1

The absorbance of the solvent at 475 nm and 700 nm are 0.005 and 0.000 respectively. The absorbance of the mixture of A and B are 0.502 and 0.912 at 475 nm and 700 nm respectively. Therefore, the corrected absorbances are

475 nm: 0.502 – 0.005 = 0.497

700 nm: 0.912 – 0.000 = 0.912

Let C_A and C_B be the concentrations of the species A and B in the mixture. Therefore,

0.497 = (1517.65 M^-1.cm^-1)*C_A*(1.25 cm) + (16430.11 M^-1.cm^-1)*C_B*(1.25 cm)

= (1897.0625 M^-1)*C_A + (20537.6755 M^-1)*C_B ……(1)

0.912 = (8988.23 M^-1.cm^-1)*C_A*(1.25 cm) + (1784.95 M^-1.cm^-1)*C_B*(1.25 cm)

= (11235.2875 M^-1)*C_A + (2231.1875 M^-1)*C_B ……..(2)

(1)*11235.2875 – (2)*1897.0625 gives

0.497*11235.2875 – 0.912*1897.0625 = (20537.6755 M^-1)*C_B*11235.2875 – (2231.1875 M^-1)*C_B*1897.0625

=====> 3853.8169 = 226513986.7*C_B

=====> C_B = 3853.8169/226513986.7

=====> C_B =1.70*10^-5

The concentration of species B in the mixture is 1.70*10^-5 M.

Put C_B = 1.70*10^-5 M in (1) and get

0.497 = (1897.0625 M^-1)*C_A + (20537.6755 M^-1)*(1.70*10^-5 M)

=====> 0.497 = (1897.0625 M^-1)*C_A + 0.349

=====> (1897.0625 M^-1)*C_A = 0.148

=====> C_A = 7.80*10^-5 M.

The concentration of species A in the mixture is 7.80*10^-5 M.

Option (b) is closest to the values obtained and is therefore, the correct answer.