Consider a solution containing 3.54 mm of an analyte, x, and 1.23 mm of a standard, s. upon chromatographic separation of the solution peak areas for x and s are 3275 and 10829, respectively. determine the response factor for x relative to s. f= determine the concentration of x in an unknown solution, 1.00 ml of 8.27 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 5389 and 4627 for x and s, respectively. determine the concentration of s in the 10.0 ml solution.[s]= the concentration of x in the 10.0 ml solution.[x]= the concentration of x in the unknown solution.[x]=

S.F = 0,105

[S] = 0,827mM

[X] = 9,17mM in the 10,0mL solution.

18,3mM in the unknown solution

Explanation:

The response factor of X is:

RF x = peak area / concentration = 3275 / 3,54 = 925

And for S is:

RF s = 10829 / 1,23 = 8804

The response factor for X relative to S is:

S.F = 925/8804 = 0,105

As the dilution of the measured solution regard the 8,27mM is 1/10, concentration of S in the unknown solution is:

8,27mM× 1,00mL / 10,0mL= 0,827mM

Using the RRF formula:

Conc X = Peak area X* Concentration B / peak area S * RRF

[X] = 5389*0,827mM / 4627*0,105

[X] = 9,17mM in the 10,0mL solution.

As the dilution of the measured solution regard with unkown solution is 10,0mL / 5,00 mL, in the unknown solution the concentration is:

9,17mM× 10,0mL / 5,00mL =

18,3mM in the unknown solution

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- lead + phosphoric acid = hydrogen + lead(ii) phosphate