The solubility of lead (II) chloride (PbCl2) is 1.6 x 10^-2 M. What is the Ksp of PbCl2?
To find the Ksp (solubility product constant) of PbCl2, we need to use the given solubility and the stoichiometry of the chemical equation.
The solubility of PbCl2 is given as 1.6 x 10^-2 M, which means that the concentration of lead ions (Pb2+) and chloride ions (Cl-) in the saturated solution of PbCl2 is 1.6 x 10^-2 M.
The chemical equation for the dissociation of PbCl2 in water is:
PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq)
From this equation, we can see that one mole of PbCl2 dissociates to form one mole of Pb2+ ions and two moles of Cl- ions.
Since the stoichiometry of PbCl2 is 1:1, the concentration of Pb2+ ions in the saturated solution is also 1.6 x 10^-2 M.
Using these concentrations, we can write the expression for the solubility product constant (Ksp) as:
Ksp = [Pb2+][Cl-]^2
Substituting the given concentration of Pb2+ ions (1.6 x 10^-2 M) into the expression, we get:
Ksp = (1.6 x 10^-2)(1.6 x 10^-2)^2
Ksp = 1.6 x 10^-2 x (1.6 x 10^-2)^2
Ksp = 1.6 x 10^-2 x 2.56 x 10^-4
Ksp = 4.096 x 10^-6
Therefore, the Ksp of PbCl2 is 4.096 x 10^-6.
To find the Ksp (solubility product constant) of PbCl2, we need to use the given solubility information. The solubility of PbCl2 is given as 1.6 x 10^-2 M, which means that in a saturated solution, the concentration of lead ions (Pb2+) and chloride ions (Cl-) will be equal to this value.
The balanced chemical equation for the dissociation of PbCl2 in water is:
PbCl2(s) ⇌ Pb2+(aq) + 2 Cl-(aq)
From this equation, we can see that the stoichiometric ratio of lead ions to PbCl2 is 1:1 and the stoichiometric ratio of chloride ions to PbCl2 is 2:1.
Using the solubility information, we can write the expression for the solubility product (Ksp) as follows:
Ksp = [Pb2+][Cl-]^2
Since the concentration of lead ions and chloride ions in a saturated solution of PbCl2 is equal to the solubility (1.6 x 10^-2 M), we substitute this value into the equation:
Ksp = (1.6 x 10^-2)(1.6 x 10^-2)^2
Simplifying the expression, we get:
Ksp = 2.56 x 10^-6
Therefore, the Ksp of PbCl2 is 2.56 x 10^-6.
PbCl2 ==> Pb^+2 + 2Cl^-
Ksp = (Pb^+2)(Cl^-)
You have the solubility of PbCl2. Pb^+2 will be the same; Cl^- will be twice that. Substitute those two numbers into the Ksp expression and solve for Ksp.