Question

A researcher used HPLC to examine a bicomponent mixture containing 1.22 mg/L of compound A and 1.31 mg/L of compound B, which was added as an internal standard. This mixture produced peak areas for compounds A and B of 10919 and 5379 , respectively. Using the above information, determine the response factor (F).
After establishing F, the researcher prepared a solution by combining 8.18 mg of B with 10.00 mL of an unknown solution containing only A and then diluted it to a final volume of 50.00 mL. The sample was examined using HPLC and peak areas of 6065 and 9111 were observed for A and B, respectively.

Determine the concentration of A (mg/mL) in the unknown solution.

Answer 1

Answer:

Response factor A: 8950 L/mg

Response factor B: 4106 L/mg

A = 02497 mg/mL

Explanation:

The response factor (F) is defined as the ratio between the signal produced by an analyte and its concentration.

For A:

F(A) = 10919 / 1,22mg/L = 8950 L/mg

F(B) = 5379 / 1,31mg/L = 4106 L/mg

It is possible to obtain relative response factor (RRF) that is the ratio between F(A) and F(B), thus:

RRF = 8950 L/mg / 4106 L/mg = 2,180

RRF could be:

$RRF = \frac{Area_{A}}{Area_B} \frac{Concentration_B}{Concentration_A}$

That is:

$Concentration_A = \frac{Area_{A}}{Area_B} \frac{Concentration_B}{RRF}$

The concentration of B in mg/mL is:

8,18mg / (10,00mL + 50,00mL) = 0,1363 mg/mL

Replacing:

$Concentration_A = \frac{6065}{9111} \frac{0,1363mg/mL}{2,180}$

A = 0,04162 mg/mL -In the diluted solution-

The concentration in the unknown solution is:

0,04162 mg/mL ×$\frac{60,00mL}{10,00mL}$ = 02497 mg/mL

I hope it helps!