2. A student is interested in the concentration of blue dye in his favorite b kn

Question

2. A student is interested in the concentration of blue dye in his favorite b knows that spectroscopy can be used to determine the concentration of dye molecules in a solution. After making several solutions of known concentration, the student measured the absorbance of each of the solutions and prepared the following calibration curve. Blue Dye Calibration Curve : 0 60 R2 0.9998 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.00E+00 2.00E-06 4.00E-06 6.00E-06 8.00E-06 1.00E-05 1.20E-05 1.40E-O5 Dye Concentration (M) The student then finds that if 20.0 mL of his favorite beverage is diluted to a volume of 100.0 mL, the diluted solution has an absorbance of 0.730. Determine the concentration of blue dye in the undiluted beverage.

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.

That is, according to the trendline (linear regression) equation y = 68663x + 0.0 obtained from the graph, 1 absorbance unit (1 Y = Y) is equal to 068663 units on X-axis (concentration).

#Step 1: Given, abs of unknown aliquot = 0.730

Putting the value y = 0.730 in trendline equation-

            0.730 = 68663x

            Or, x = 0.730 / 68663

            Or, x = 1.0632 x 10-5

Hence, [Dye] in unknown aliquot = 1.0632 x 10-5 M

# Step 2: Preparation of unknown aliquot:

Let [Dye] in undiluted beverage = X M

20.0 mL of undiluted beverage is diluted to a final volume of 100.0 mL to prepare the unknown aliquot.

Now,

            Using C1V1 (undiluted beverage) = C2V2 (unknown aliquot)

            Or, X M x 20.0 mL = C2 x 100.0 mL

            Or, C2 = (X M x 20.0 mL) / 100.0 mL = 0.2X M

So, [Dye] in unknown aliquot = 0.2X M

Step 3: Comparing [Dye] in #step 1 and #step 2-

            0.2X M = 1.0632 x 10-5 M

            Or, X = 1.0632 x 10-5 M / 0.2M = 5.3158 x 10-5

Therefore, [Dye] in undiluted beverage = X M = 5.3158 x 10-5 M

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