What is the solubility (in g/L) of PbCl2 at 25 degree celsius? The solubility product of PbCl2 at 25 degrees celcius is 1.6x10-5.
I figured out the answer. Thanks.
To determine the solubility of PbCl2 at 25 degrees Celsius, we need to use the solubility product constant (Ksp) and the given solubility product value.
The solubility product constant (Ksp) is a measure of the maximum amount of a compound that can dissolve in a solution at a given temperature. It is the product of the molar concentrations of the dissolved ions, each raised to the power of their stoichiometric coefficients, as written in the balanced equation for the dissolution reaction.
For the dissolution of PbCl2, the balanced equation is:
PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq)
According to the given information, the solubility product (Ksp) of PbCl2 at 25 degrees Celsius is 1.6x10^-5.
To find the solubility (in g/L), we need to consider the molar concentration of the dissolved ions. Since the stoichiometric coefficients in the balanced equation are 1:2 for Pb2+ and Cl-, respectively, the molar concentration of Pb2+ is equal to the solubility, and the molar concentration of Cl- is twice the solubility.
Let's assume the solubility of PbCl2 is "x" g/L.
Since we have 1 Pb2+ ion for each PbCl2 molecule, the molar concentration of Pb2+ will also be "x" M. And since we have 2 Cl- ions for each PbCl2 molecule, the molar concentration of Cl- will be "2x" M.
Now, using the balanced equation and the expression for Ksp, we can write an expression for Ksp in terms of solubility:
Ksp = [Pb2+][Cl-]^2
1.6x10^-5 = x * (2x)^2
1.6x10^-5 = 4x^3
Solving this equation will give us the value of "x," which represents the solubility of PbCl2 in g/L at 25 degrees Celsius.