Question

In: Biology

A 5-point calibration curve was made for the determination of Pb in FAA. The regression equation...

A 5-point calibration curve was made for the determination of Pb in FAA. The

regression equation was: y . 0:155x . 0:0016, where y is the signal output as absorbance,

and x is the Pb concentration in mg/L. (a) A contaminated groundwater sample

was collected, diluted from 10 to 50 mL, and analyzed without digestion. The

absorbance reading was 0.203 for the sample. Calculate the concentration of Pb in this

groundwater sample; (b) A sediment sample suspected of Pb contamination was

collected. After decanting the overlying water, a 1-g wet sediment sample was digested

and diluted to 50 mL. FAA measurement gave an absorbance reading of 0.350. A

subsample of this wet sediment was taken to measure the moisture content with oven

drying overnight. The moisture was 35%. Report the Pb concentration in sediment

sample on a dry basis.

Solutions

Expert Solution

Answer:

a)

  • Linear regression equation is Y = 0.155*X + 0.0016
  • Where Y = Signal Ouput and X = Pb concentration in mg/L
  • Contaminated ground water was diluted from 10 to 50ml, therefore fold dilution = 50/10 = 5x (5 fold)
  • Absorbance obtained = 0.203
  • Concentration (X) can be calculated from the standard curve
  • X = (Y - 0.0016)/0.155 = (0.203 - 0.0016) / 0.155 = 1.299 mg/L
  • Actual concentration = Concentration from standard curve x Fold dilution = 1.299*5 = 6.497 mg/L
  • Therefore, concentration of Pb in this groundwater sample is 6.497 mg/L

b)

  • 1g wet sediment sample --> digested and diluted to 50ml
  • Fold dilution = 50x
  • FAA absorbance = 0.350
  • Moisture content in sample = 35%
  • Therefore dry weight of 1 g sample = 1 g - 0.350g = 0.65g
  • Concentration (X) can be calculated from the standard curve
  • X = (Y - 0.0016)/0.155 = (0.350 - 0.0016) / 0.155 = 2.248 mg/L
  • Actual concentration = Concentration from standard curve x Fold dilution = 2.248*50 = 112.38 mg/L
  • Therefore, concentration of Pb in 0.65 gram (dry weight) of sample = 112.38mg/L per 0.65g
  • Concentration of Pb per gram (Dry weight) of sample = 112.38 / 0.65 = 172.9 mg/L/g of sediment sample.

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