The solubility product of PbCl2 at 25 degrees celcius is 1.6x10-5 mol3.dm-9. What is the solubility of PbCl2?
To find the solubility of PbCl2, we can use the solubility product constant (Ksp) equation. The Ksp expression for PbCl2 is as follows:
PbCl2 ⇌ Pb2+ + 2Cl-
The Ksp equation is expressed as:
Ksp = [Pb2+][Cl-]^2
Given that the solubility product constant (Ksp) of PbCl2 is 1.6x10^-5 mol^3.dm^-9, we can equate this value to the concentrations of the Pb2+ and Cl- ions:
1.6x10^-5 = [Pb2+][Cl-]^2
Now, since PbCl2 is a strong electrolyte that fully dissociates in water, the concentration of Pb2+ will be equal to the solubility of PbCl2 (s), and the concentration of Cl- ions will be 2s (as there are two chloride ions produced for every lead ion).
Substituting these values into the equation:
1.6x10^-5 = (s)(2s)^2
1.6x10^-5 = 4s^3
Now we can solve for s (the solubility of PbCl2) by rearranging the equation:
s^3 = (1.6x10^-5) / 4
s^3 = 4x10^-6
s = (4x10^-6)^(1/3)
Evaluating this expression:
s ≈ 2.154x10^-2 mol.dm^-3
Therefore, the solubility of PbCl2 at 25 degrees Celsius is approximately 2.154x10^-2 mol.dm^-3.