What is the solubility, in moles/liter, of PbCl2 if the Ksp =7.3 x 10^12
......PbCl2 ==> Pb^2+ + 2Cl-
I.....solid......0.......0
C.....solid......x.......x
E.....solid......x.......x
Substitute the E line into Ksp expression and solve for x = solubility PbCl2.
Thanks!
To find the solubility of PbCl2, we first need to understand what the equilibrium constant, Ksp, represents. In a saturated solution, Ksp is the product of the concentrations of the ions raised to the power of their coefficients in the balanced chemical equation.
The balanced chemical equation for the dissociation of PbCl2 is:
PbCl2 ⇌ Pb2+ + 2Cl-
From the equation, we can see that the concentration of Pb2+ is the same as the concentration of PbCl2 since the coefficient in front of PbCl2 is 1. Therefore, we can represent the solubility of PbCl2 as "x."
So, the equilibrium expression for the dissociation of PbCl2 would be:
Ksp = [Pb2+][Cl-]^2
Given that Ksp = 7.3 x 10^12, we can substitute the concentration of Pb2+ as "x" and the concentration of Cl- as "2x" into the equilibrium expression:
7.3 x 10^12 = (x)(2x)^2
Now, we can solve for "x" to find the solubility of PbCl2:
7.3 x 10^12 = 4x^3
Divide both sides by 4:
(7.3 x 10^12) / 4 = x^3
Take the cube root (or raise both sides to the power of 1/3):
(x^3)^(1/3) = (7.3 x 10^12)^(1/3)
x = (7.3 x 10^12)^(1/3)
Now, calculate the cube root of (7.3 x 10^12) to find the value of "x."