WIth steps and explanation written on paper please! Thank you! Map Sapling Learn

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WIth steps and explanation written on paper please! Thank you!

Map Sapling Learning macmillan learning EC M-1cm-1 EB Spectroscopic data for three organic dyes, A, B, and C, are shown in the table. A solution containing a mixture of these three dves in a 1.000-cm cuvet had absorbances of 0.5343 at 450 nm, 0.4084 at 480 nm and 0.6816 at 520 nm. Find the concentrations of the three dyes in the solutions. For each solution, the zero is set with a blank EA M-1 m-1 3 200 5 900 27 500 | (nm) 450 480 520 | M-1cm-1 | 10 900 19 300 9 40010 100 12 200 3 500 Number Number Number MI [B]

Explanation / Answer

Ans. Let [A] = A M    ; [B] = B M      , and [C] = C M

# Beer-Lambert’s Law, A = e C L             - equation 1,              

where,

                                    A = Absorbance

                                    e = molar extinction coefficient (M-1cm-1)

                                    p = path length (in cm)

                                    C = concentration

# At 450 nM

Total Abs = Abs of A + Abs of B + Abs of C

Or, 0.5343 = (3200 M-1 cm-1 x A M x 1.0 cm) + (10900 M-1 cm-1 x B M x 1.0 cm) + (19300 M-1 cm-1 x C M x 1.0 cm)

Or, 0.5343 = 3200A + 10900B + 19300C            

Hence, 3200A + 10900B + 19300C = 0.5343                - equation 1

# At 480 nM (calculation similar to 450 nm)

            5900A + 9400B + 10100C = 0.4084                   - equation 2

# At 520 nM

            27500A + 12200B + 3500C = 0.6816                 - equation 3

#1. Comparing (equation 1 x 5900) – (equation 2 x 3200)

            18880000A + 64310000B + 113870000C = 3152.37

        (-) 18880000A + 30080000B + 32320000C = 1306.88

                        34230000B + 81550000C = 1845.49                  - equation 4

#2. Comparing (equation 1 x 27500) – (equation 3 x 3200)

            88000000A + 299750000B + 530750000C = 14693.25

     (-) 88000000A + 39040000B + 11200000C = 2181.12

                        260710000B + 519550000C = 12512.13                       - equation 5

#3. Comparing (equation 4 x 260710000) – (equation 5 x 34230000)

            8.9241 x 1015B + 2.1261 x 1016C = 4.8114 x 1011

      (-) 8.9241 x 1015B + 1.7784 x 1016C = 4.2829 x 1011

                                      3.4767 x 1015C = 5.2847 x 1010

                Or, C = (5.2847 x 1010) / (3.4767 x 1016C)

            Hence, C = 1.5200 x 10-5 = 0.0000152

Hence, [C] = C M = 1.5200 x 10-5 M

#4. Putting the value of C in equation 4-

            34230000B + 81550000 x (1.52 x 10-5) = 1845.49

            Or, 34230000B + 1239.5972 = 1845.49

            Or, 34230000B = 1845.49 - 1239.5972 = 605.8928

            Or, B = 605.8928 / 34230000

            Hence, B = 1.7701 x 10-5

Hence, [B] = B M = 1.7701 x 10-5 M

# 5. Putting the values of B and C in equation 1-

3200A + 10900B + 19300C = 0.5343                   

            Or, 3200A + 10900 (1.52 x 10-5)+ 19300 x (1.7701 x 10-5) = 0.5343

            Or, 3200A + 0.16568 + 0.6416293 = 0.5343

            Or, 3200A = 0.5343 – 0.5073 = 0.0270

            Or, A = 0.0270 / 3200

            Or, A = 8.4375 x 10-6

Hence, [A] = A M = 8.4375 x 10-6 M

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