An unknown solution containing a mixture of two compounds is being analyzed by U

Question

An unknown solution containing a mixture of two compounds is being analyzed by UV/visible spectrometry in order to determine the amounts of compounds A and B in the sample. Both compounds absorb to some extent at 475 and 630 nm. The calibration curve data obtained during the experiment were analyzed using linear regression analysis. The linear equations for each compound at the various wavelengths are provided in the table below. Results of Linear Regression Analysis Pathlength (cm) 75 nm 630 nm Compound A (y -1.534 x y-6.136 x 1.00 O Compound By = 1.855 y 879 x 1.00 The absorbance of the unknown solution was 0.632 at 475 nm and 0.575 at 630 nm. Calculate the concentration of compound A in the unknown sample. Report your answer to 3 sig figs using decimal notation. Use leading zeros. Do not include units. (HINT - in order to do this calculation, you will need to set up and solve simultaneous equations. You will then have the concentrations for both A and B. Report only the concentration requested. There is a video posted to Blackboard which covers this activity.)

Explanation / Answer

Ans. In the graph, Y-axis indicates absorbance and X-axis depicts concentration. The linear regression equation in in form of “ y = mx + x”                    , where-

            y = Y-axis value, depicts absorbance

            x = X-axis value, depicts concertation with respect to its absorbance

            m = slope ; (y/x) ration.

# Let, in the sample-

[A] = A            ;

[B] = B

# At 475 nm:

Absorbance due to A, y1 = 1.534x

                                    Or, y1 = 1.534A                                           

;[note that x represents concertation of the specified absorbance]

Similarly, absorbance due to B, y2 = 1.855B

Now,

            Total absorbance at 475 nm = y1 + y2

            Or, 0.632 = 1.534 A + 1.855 B

            Or, 1.534 A + 1.855 B = 0.632      - equation 1

# At 630 nm:

Absorbance due to A, y3 = 6.136 A

Absorbance due to B, y4 = 0.873 B

Total absorbance at 630 nm = y3 + y4 = 6.136 A + 0.873 B

            Or, 6.136 A + 0.873 B = 0.575      - equation 2

# Comparing (equation 1 x 6.136) minus (equation 2 x 1.534)

            9.41262 A + 11.3823 B = 3.87795

      (-) 9.41262 A + 1.33918 B = 0.88205

                                    Or, 10.0431 B = 2.9959

                                    Or, B = 2.9959 / 10.0431 = 0.2983

Hence, [B] = B = 0.2983

# Putting the value of B in equation 1-

            1.534 A + 1.855 x 0.2983 = 0.632

            Or, 1.534 A = 0.632 – 0.55335

            Or, A = 0.07865 / 1.534 = 0.05127

Hence, [A] = A = 0.05127 = 0.0513

# Result: [A] = A = 0.05127 = 0.0513

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