Question

In a preliminary experiment, a solution containing 0.0837 M X and 0.0666 M S gave peak areas of A _x = 423 and A _s = 347. (Areas are measured in arbitrary units by the instrument's computer.) To analyze the unknown X, 10 mL of 0.146 M S were added to 10 mL of unknown, and the mixture was diluted to 25 mL in a volumetric flask. This mixture gave the chromatogram in which A _x = 553 and A _s = 582. 1) Find the value of F determined in the preliminary experiment. Report your answer to 3 significant figures. 2) Find the value of [S]final. Report your answer in M to 3 significant figures.3)Find the value of [X]final. Report your answer in M to 3 significant figures.

Answer 1

1) First use the standard mixture to find the response factor:

AX / [x] = F x {AS / [S]}

Standard mixture:

423 / 0.0837 = F {347 / 0.0666}

=> F = 0.9700

2) In the mixture of unknown plus standard, the concentration of S is:

[S] = (0.146M)(10.0mL / 25.0mL)

[S] = 0.0584M

where: 10.0mL / 25.0mL is the dilution factor

3) Using the known response factor and S concentration of the diluted sample in the equation above:

553 / [X] = 0.9700 (582 / 0.0584)

[X] = 0.05721M