What is the Ksp of Lead II Fluoride if the lead and fluoride ions have molecular concentrations of 4.8 x 10-4 M and 8.7 x 10-3 M respectively in a saturated solution at 25oC?
Please show work. I don't even know where to start.
.............PbF2 ==> Pb^2+ + 2F^-
I..............solid.........0.............0
C............solid..........x.............2x
E............solid..........x.............2x
Ksp = (Pb^2+)(F^-)^2
The problem tells you the value of x and 2x. Substitute those into the Ksp expression and solve.
@DrBob222 Thank you Sir
To calculate the solubility product constant (Ksp) of Lead II Fluoride (PbF2), you need to use the concentrations of the lead (Pb2+) and fluoride (F-) ions in a saturated solution.
The balanced chemical equation for the dissolution of PbF2 is:
PbF2(s) ⇌ Pb2+(aq) + 2F-(aq)
Now, let's start solving the problem step by step:
Step 1: Write the expression for the Ksp:
Ksp = [Pb2+][F-]^2
Step 2: Substitute the given concentrations into the Ksp expression:
[Pb2+] = 4.8 x 10^-4 M
[F-] = 8.7 x 10^-3 M
Ksp = (4.8 x 10^-4)(8.7 x 10^-3)^2
Step 3: Calculate the value of Ksp:
Ksp = 4.8 x 10^-4 x (8.7 x 10^-3)^2
At this point, we can multiply the two concentrations, and then raise it to the power of 2 to find the value of Ksp.
Ksp = 4.8 x 10^-4 x (8.7 x 10^-3)^2 = 3.0564 x 10^-10
So, the solubility product constant (Ksp) of Lead II Fluoride (PbF2) is 3.0564 x 10^-10.