Question

If an aqueous solution of Pb²⁺ has a concentration of 125 ppb, and the density of the solution is 0.9997 g/mL, then what is the Molarity of the solute at 25.0 ºC?

Answer 1

ppb = 10^-6 g of solute in 1000 g of solution.

125 ppb = 125 x 10^-6 g of solute in 1000 g of solution

Given that density of  Pb2+ solution = 0.9997 g/mL

Then, volume of  Pb2+ solution = mass / density = 1000 g / 0.9997 g/mL = 1000.3 mL = 1.0003 L

Then,

mass of Pb2+ = 125 x 10^-6 g

molar mass of Pb2+ = 207 g/mol

volume of Pb2+ solution = 1.0003 L

Molarity of Pb2+ = (Mass of Pb2+ / Molar mass of Pb2+) x (1/ volume of Pb2+ solution in Litres)

= (125 x 10^-6 g/ 207 g/mol) x (1/ 1.0003 L)

= 0.604 x 10^-6 M

Molarity of Pb2+ = 0.604 x 10^-6 M

Therefore, Molarity of Pb2+ at 25.0 ºC = 0.604 x 10^-6 M