Lead chloride dissolves in water according to
PbCl2(s)↔Pb2++2Cl−(aq)
The solubility in pure water has been measured to be 4.44 g⋅L−1. Calculate the solubility product of lead chloride in pure water. Your answer should be in units of molarity raised to the appropriate exponent.
convert the solubility to Moles per Litre.
To calculate the solubility product of lead chloride (PbCl2) in pure water, we need to know the molar solubility of the compound.
The given solubility of lead chloride in pure water is 4.44 g⋅L^−1. We can convert this to molarity by dividing the mass of PbCl2 by its molar mass.
The molar mass of PbCl2 can be calculated by summing the atomic masses of lead (Pb) and chlorine (Cl):
Atomic mass of Pb = 207.2 g/mol
Atomic mass of Cl = 35.45 g/mol (Cl has numerous isotopes, but the average is used for molar calculations)
Molar mass of PbCl2 = (207.2 g/mol) + 2(35.45 g/mol) = 278.1 g/mol
Now we can calculate the molar solubility of PbCl2:
Molar solubility = (4.44 g/L) / (278.1 g/mol) = 0.0159 mol/L
According to the dissociation equation of PbCl2:
PbCl2(s) ↔ Pb2+ (aq) + 2Cl− (aq)
The solubility product expression for PbCl2 is:
Ksp = [Pb2+] * [Cl−]^2
Since the stoichiometric coefficient of Pb2+ is 1 and for Cl− is 2, the expression becomes:
Ksp = [Pb2+] * [Cl−]^2 = (0.0159 mol/L) * (0.0159 mol/L)^2
= 0.0159 mol/L * (0.0159^2 mol^2/L^2)
≈ 3.20 x 10^−5 mol^3/L^3
Therefore, the solubility product of lead chloride (PbCl2) in pure water is approximately 3.20 x 10^−5 mol^3/L^3.