The solubility of a saturated solution of lead (ll) chloride at 25 degree celcius is 0.1681M. What is the solubility product of the salt at the same temperature?
To find the solubility product (Ksp) of lead(II) chloride (PbCl2) at 25 degrees Celsius, we need to use the given solubility information and the chemical equation for the dissolution of PbCl2.
The chemical equation that represents the dissolution of PbCl2 is:
PbCl2(s) ↔ Pb2+(aq) + 2Cl-(aq)
From the equation, we know that one mole of PbCl2 produces one mole of Pb2+ and 2 moles of Cl- ions in solution.
The solubility of PbCl2, expressed in terms of molarity (M), is given as 0.1681 M. This means that at equilibrium, there is 0.1681 moles of Pb2+ ions and 2 * 0.1681 = 0.3362 moles of Cl- ions in a liter of the saturated solution.
The solubility product expression for PbCl2 can be written as:
Ksp = [Pb2+][Cl-]^2
Substituting the values we know, we get:
Ksp = (0.1681)(0.3362)^2
Now, we can calculate Ksp:
Ksp = 0.0196
Therefore, the solubility product (Ksp) of lead(II) chloride at 25 degrees Celsius is approximately 0.0196.