Question

In: Chemistry

10 mL sample solution that contains arsenic each is filled into 5 100 mL-volumetric flasks. Then...

10 mL sample solution that contains arsenic each is filled into 5 100 mL-volumetric flasks. Then into the second volumetric flask until the fifth volumetric flask each added 5, 10, 15, and 20 mL standard solution 22.1 ppm of arsenic. Then water is added to all volumetric flasks until the volume reached the 100 mL mark. Then the absorbance of each solution is measured, with the data below. Calculate the content of As in sample.

Standard (mL) Absorbance
0.00 0.156
5.00 0.195
10.00 0.239
15.00 0.276
20.00 0.320

Solutions

Expert Solution

The total volume of solution in each test tube is 100 mL. 5 mL, 10 mL, 15 mL and 20 mL of 22.1 ppm As standard is added to each of the test tubes labelled 2-5. Since the 1st test tube does not contain any added standard, we can assume that 0 mL of 22.1 ppm As was added.

Find out the concentration of the added As in each of test tubes 2-5. Use the dilution equation

M1*V1 = M2*V2

where M1 = original concentration of As standard added = 22.1 ppm; V1 = volume of As standard added and V2 = final volume of the solution.

For the solution in test tube #2, we have, M2 = (5 mL)*(22.1 ppm)/(100 mL) = 1.105 ppm.

Prepare the following table as below.

Standard (mL)

Concentration of added As (ppm)

Absorbance

0.00

0.000

0.156

5.00

1.105

0.195

10.00

2.210

0.239

15.00

3.315

0.276

20.00

4.420

0.320

Plot the absorbance vs concentration of added standard as below.

Use the regression equation to find out the concentration of As in the original sample. Assume the absorbance is 0.00 when no standard is added. Put y = 0 and obtain

0 = 0.037x + 0.1554

====> -0.037x = 0.1554

====> x = 0.1554/(-0.037) = -4.200

Since concentration cannot be negative, hence, the concentration of As in the diluted sample is 4.20 ppm (ans).

The sample solutions were prepared by taking 10.00 mL of the original sample and diluting to 100.0 mL; therefore, the dilution factor is (100.0 mL)/(10.00 mL) = 10.00.

Concentration of As in the original sample = (concentration in dilute solution)*(dilution factor) = (4.20 ppm)*(10.0) = 42.0 ppm (ans).


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