Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Atomic emission spectroscopy and the method of standard addition are used to det

ID: 530110 • Letter: A

Question

Atomic emission spectroscopy and the method of standard addition are used to determine the Na^+ concentration and its uncertainty in an unknown sample. Increasing amounts of a 1.75 mu g/mL standard Na^+ solution are added to a series of volumetric flasks containing 10.00 mL of the unknown solution. The volumetric flasks are then diluted to a final volume of 200.0 mL. Atomic emission spectroscopy is then used to measure the emission intensity of each solution. Based on the data in the table above, determine the original concentration of Na^+ as well as its uncertainty in the unknown sample. Report the uncertainty with three significant figures.

Explanation / Answer

I can only tell you the concentration of the unknown. You haven’t mentioned the glassware; so I cannot talk about the uncertainty.

Use the dilution law:

M1*V1 = M2*V2

M1 = concentration of added standard; V1 = volume of added standard; M2 = concentration of the standard in the final solution; V2 = final volume of the solution.

Use the second entry,

(5 mL)*(1.75 µg/mL) = (200 mL)*M2

===> M2 = (5*1.75 µg/mL)/(200) = 0.04375 µg/mL.

Fill in the following table.

Standard Volume (mL)

Final volume (mL)

Concentration of standard Na+ (µg/mL)

0.00

200.00

0.00000

5.00

200.00

0.04375

10.00

200.00

0.08750

15.00

200.00

0.13125

20.00

200.00

0.17500

Plot a graph as below:

Use the regression equation to find out the concentration of the unknown in the final volume of the sample.

Put absorbance y = 0 in the regression equation.

0 = 3444.6x + 344

===> 3444.6x = -344

===> x = (-344)/(3444.6) = -0.0998 -0.1000

Ignore the negative sign and take the concentration of Na+ in the final solution = 0.1000 µg/mL.

The final solution was 200.00 mL and the volume of the original solution taken = 10.00 mL. The dilution factor = (200.00 mL)/(10.00 mL) = 200.00

The original concentration = (0.1000 µg/mL)*(200.00) = 20.00 µg/mL (ans).

Standard Volume (mL)

Final volume (mL)

Concentration of standard Na+ (µg/mL)

0.00

200.00

0.00000

5.00

200.00

0.04375

10.00

200.00

0.08750

15.00

200.00

0.13125

20.00

200.00

0.17500