What mass of PbCl2 would be found in 0.48 L of a saturated solution of PbCl2 if the Ksp of PbCl2 is 1.6 x 10-5?
I worked this for you at your earlier post. Go back and read that. I gave detailed instructions.
To find the mass of PbCl2 in the solution, we need to use the concept of solubility product constant (Ksp) and the given volume of the solution.
The Ksp expression for PbCl2 can be written as:
Ksp = [Pb2+][Cl-]^2
Since PbCl2 dissociates into one Pb2+ ion and two Cl- ions, we can rewrite the Ksp expression as:
Ksp = [Pb2+][Cl-]^2
It is given that the Ksp of PbCl2 is 1.6 x 10^-5. Since the concentration of Cl- ions is squared, we only need to consider the concentration of Pb2+ ions.
Let's assume x represents the concentration of Pb2+ ions in the saturated solution. As PbCl2 dissociates completely, the concentration of Pb2+ ions would be equal to x. Therefore, the equilibrium concentration of Pb2+ ions can be written as [Pb2+] = x.
Now, we have:
Ksp = [Pb2+][Cl-]^2
1.6 x 10^-5 = x * (2x)^2
1.6 x 10^-5 = 4x^3
Solving this equation for x, we find:
x = (1.6 x 10^-5 / 4)^(1/3)
x ≈ 0.014
The concentration of Pb2+ ions in the saturated solution is approximately 0.014 M.
Now, we can calculate the moles of PbCl2 in the given volume of the solution.
Moles = concentration * volume
Moles = 0.014 mol/L * 0.48 L
Moles ≈ 0.00672 mol
Finally, we can calculate the mass of PbCl2 using its molar mass (molar mass of Pb = 207.2 g/mol, molar mass of Cl = 35.45 g/mol):
Mass = moles * molar mass
Mass = 0.00672 mol * (207.2 g/mol + 35.45 g/mol)
Mass ≈ 1.68 g
Therefore, the mass of PbCl2 that would be found in 0.48 L of the saturated solution is approximately 1.68 grams.