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

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

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

Question 5-19 Each flask contained mL of serum, varying additions of M NaCl standard and a total volume of mL. FlaskVolume of standard (mL) Na + atomic emission signal (mV)

FlaskVolume of standard (mL) Conc. of standard (M) [S] * dilution factor (x-axis) Signal (y-axis)

Solution

Internal Standard - Response Factor F = response factor

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.

Solve for F in known solution

Unknown A X = 5428 A S = 4431

Solve for [S] in unknown solution

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

Use dilution factor