Consider a solution containing 3.79 mM of an analyte, X, and 1.41 mM of a standa
ID: 525267 • Letter: C
Question
Consider a solution containing 3.79 mM of an analyte, X, and 1.41 mM of a standard, S. Upon chromatographic separation of the solution peak areas for X and S are 3711 and 10519, respectively. Determine the response factor for X relative to S. Number To determine the concentration of X in an unknown solution, 1.00 mL of 8.55 mM S was added to 5.00 mL of the unknown X solution and the mixture was diluted to 10.0 mL. After chromatographic separation, this solution gave peak areas of 5895 and 4629 for X and S, respectively. Determine the concentration of S in the 10.0 mL solution. Number mM Determine the concentration of X in the 10.0 mL solution. Number mMExplanation / Answer
Response factor
F = Ax/[X] / As/[S]
Ax = 3711
[X] = 3.79 mM
As = 10519
[S] = 1.41mM
F = [3711 / 3.79 mM] / [10519/1.41 mM]
F = 0.131
The concentration of S in the 10mL of solution
[S]f = [S]i x ( Vi / Vf)
= 8.55 mM x (1.00 mL / 10.00 mL)
[S]f = 0.855 mM
The concentration of X in the 10mL of solution
Ax/[X]f= F (As/[S]f)
[X]f = Ax / F ([S]f/As)
[X]f = [5895 / 0.131] x [ 0.855 / 4629 ]
[X]f = 8.311 mM
The concentration of X in the unknown of solution
[X]i = [X]f x ( Vf / Vi)
= 8.311 mM x (10.00mL/5.00mL)
[X]i = 16.622mM