1. We are not going to do Ch 29.2: Method Validation Seminar today CHE 315 – Lecture 9 9/19/05

2. Example: Exercise 5-B An unknown sample of Ni 2+ gave a current of 2.36 μA in an electrochemical analysis. When 0.500 mL of a solution containing 0.0287 M Ni 2+ was added to 25.0 mL of unknown, the current increased to 3.79 μA. Find [Ni] i in the unknown.

3. V i /V f = dilution factor I x = 2.36 μA I x+S = 3.79 μA [S] f = 0.0287 M Ni 2+ x 0.5 mL/25.5 mL [S] f = 0.00056 M Ni 2+

5. Question 5-19 Each flask contained 25.00 mL of serum, varying additions of 2.640 M NaCl standard and a total volume of 50.00 mL. FlaskVolume of standard (mL) Na + atomic emission signal (mV) 103.13 21.0005.40 32.0007.89 43.00010.30 54.00012.48

6. FlaskVolume of standard (mL) Conc. of standard (M) [S] * dilution factor (x-axis) Signal (y-axis) 100.003.13 21.0000.055.40 32.0000.117.89 43.0000.1610.30 54.0000.2112.48

8. Solution

9. Internal Standard - Response Factor F = response factor

10. Example: 5-22 A solution containing 3.47 mM X and 1.72 S gave peak areas of 3473 and 10222, respectively in a chromatographic analysis. Then 1.00 mL of 8.47 mM S was added to 5.00 mL of unknown X and the mixture was diluted to 10.0 mL. This solution gave peak areas of 5428 and 4431 for X and S, respectively. Find [X] in the unknown solution.

11. Solve for F in known solution

12. Unknown A X = 5428 A S = 4431

13. Solve for [S] in unknown solution

14. Solve for [X] in unknown solution Are you done?

15. Use dilution factor