Question

To remove lead from a contaminated water solution, a chemist decides to precipitate it out by adding sodium chloride to the solution. From 108.7 g of the solution with a density of 1.03 g/mL, the chemist recovers 0.152 g of PbCl2. What is the concentration of Pb2+in the original solution in molarity?

Answer 1

The concentration of Pb²⁺ in the original solution in molarity is 0.00518 M/L

Step-by-step explanation:

Given;

mass of the solution = 108.7g

density of the solution = 1.03 g/mL

Volume of solution = mass of the solution/density of the solution

Volume of solution = 108.7/1.03 = 105.53 mL

Number of mole of PbCl₂ = Reacting mass/molar mass

molar mass of PbCl₂ = (207.2 + 71) = 278.2 g/mol

Number of mole of PbCl₂ = (0.152/278.2) = 0.000546 M

PbCl₂ ⇄ Pb²⁺ + 2Cl⁻

1 : 1 : 2

1 mole of PbCl₂ = 1 mole of Pb²⁺

0.000546 mole of PbCl₂ = 0.000546 mole of Pb²⁺

Molarity of Pb²⁺ = (number of moles of Pb²⁺)/(Liters of solution)

Molarity of Pb²⁺ = (0.000546 mole)/(105.33 X 10⁻³L)

Molarity of Pb²⁺ = 0.00518 M/L

Therefore, the concentration of Pb²⁺ in the original solution in molarity is 0.00518 M/L