Find the concentration of Pb2 (aq) in ppm by mass of a 1.13 L sample of contaminated water that was assayed by adding NaI(s) and obtaining 55.1x10-3 g of PbI2(s) precipitate. Assume that Pb2 (aq) is completely precipitated as PbI2(s). The molar mass of PbI2(s) is 461.01 g/mol. Assume that the density of the solution is 1.00 kg/L. Give the answer with 3 or more significant figures.

22.1 ppm

Explanation:

The equation of the reaction is;

Pb2+ (aq)  + 2NaI(s) >PbI2(s) + 2 Na^2+

Mass of precipitate =  55.1x10-3 g

Number of moles of precipitate =  55.1x10-3 g/461.01 g/mol = 1.2 * 10^-4 moles

1 mole of  Pb2+  yields 1 mole of PbI2

Hence 1.2 * 10^-4 moles of  Pb2+  also yields  1.2 * 10^-4 moles of PbI2.

Hence mass of Pb2+ present = 1.2 * 10^-4 moles * 207 g/mol = 0.025 g or 25 mg

Mass of solution = Density of solution * volume of solution

Mass of solution = 1.00 kg/L * 1.13 L

Mass of solution = 1.13 Kg

Concentration in mg/Kg(ppm) = mass of solute/mass of solution

= 25 mg/1.13 kg

= 22.1 ppm