Question

A sample of paint chips (0.6169g) was added to 10 mL of concentrated HNO3 and heated to near dryness. Approximately 35mL of water was added and the solution was biolded for 30 mins. After cooling, the solution was filtered and the filrate was diluted to volume in a 50mL volumetric flask. A 5 mL aliquot of this solution was diluted to volume in a 25mL volumetric flask. The lead concentration of the final diluted solution (in the 25mL volumetric flask) was determined by atomic absorption spectroscopy to be 0.179 ppm Pb. what is the concentration fo the Pb in the orginial paint chips expressed as %w/w. paint is considered to be lead-based wjem ots Pb concentration exceeds 0.06 % w/w

Answer 1

% Pb = (m * D / M) * 100%

where

m = Mass of Pb = (0.179 ppm x Mass of paint chips) / (1000000) = (0.179 ppm x 0.6169 g) / 1000000 = 1.104 x 10^-7 g

D = Dilution factor = Final volume / Initial volume = 25 mL/5 mL = 5
M = Mass of paint chips = 0.6169 g

% Pb = (m * D) / M = (1.104 x 10^-7 g * 5) / 0.6169 g = 0.0001%

Concentration of Pb in the orginial paint chips expressed as %w/w = 0.0001%

Since the concentration is Pb is less than 0.06 %, the paint is not harmful or lead-based.