The analysis of lead, Pb, in urine was carried out using atomic absorption spect

Question

The analysis of lead, Pb, in urine was carried out using atomic absorption spectroscopy as follows.

All of the Pb in a 500.00 mL sample of urine was extracted into 5.00 mL of toluene containing a lead complexing agent. When this toluene solution was injected into an air/acetylene flame in an atomic absorption spectrometer it gave an absorbance of 0.0520 at 283.3 nm. 500.00 mL samples of standard solutions containing 0.0100 and 0.0200 ppm by weight Pb in water were extracted and examined in exactly the same way, and these gave absorbances of 0.0410 and 0.0820 respectively.

Calculate the concentration of Pb in the urine sample as: i) ppm, ii) µg L-1, iii) molarity. And, iv) calculate the mass of Pb in the urine sample. [molar mass of Pb = 207.2 g mol-1; for the purpose of calculation you may consider urine to be a dilute aqueous solution.]

Explanation / Answer

Let the concentration of Pb in the 5.00 mL toluene extract containing the lead complexing agent be x ppm. The standard samples on lead gave absorbances of 0.0410 and 0.0820 corresponding to 0.0100 ppm and 0.0200 ppm Pb.

Beer’s law is written as

A = *C*l where A = absorbance; = molar absorptivity of Pb; C= concentration of Pb in ppm and l = path length of the solution. Note that and l are constants for a particular complex and experiment.

We have

0.0410 = *(0.0100 ppm)*l …..(1)

0.0820 = *(0.0820 ppm)*l …..(2) (these two data indicate the linear dependence of A on C)

0.0520 = *(x ppm)*l …..(3)

Dividing (3) by (1) we get

0.0520/0.0410 = (x ppm)/(0.0100 ppm)

=====> 1.268 = x/0.0100

=====> x = 0.0100*1.268 = 0.01268 0.0127

(i) 5.00 mL of toluene extract contains 0.0127 ppm Pb2+; therefore, 500.00 mL extract will contain (0.0127 ppm/5.00 mL)*(500.00 mL) = 1.27 ppm Pb (ans).

(ii) We know that 1 ppm = 1 mg/L; therefore

1.27 ppm Pb = 1.27 g/L Pb = (1.27 g/L)*(1.0*106 µg/1 g) = 1.27*106 µg/L Pb (ans).

(iii) The molar mass of Pb is 207.2 g/mol; therefore,

1.27 mg/L Pb = (1.27 mg/L)*(1 g/1000 mg)*(1 mole/207.2 g) = 6.129*10-6 mol/L Pb (ans).

(iv) Mass of Pb in the urine sample = (500.0 mL)*(1 L/1000 mL)*(1.27 mg/1 L)*(1 g/1000 mg) = 6.35*10-4 g (ans).

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