what are the equilibrium concentrations of Pb2+ and F- in a saturated solution of lead (II)flouride if the Ksp for PbF2 is 3.2*10^-8?
To find the equilibrium concentrations of Pb2+ and F- in a saturated solution of lead (II) fluoride, we need to use the solubility product constant (Ksp) for PbF2. The balanced chemical equation for the dissociation of PbF2 is as follows:
PbF2(s) ⇌ Pb2+(aq) + 2F-(aq)
According to the equation, 1 mole of PbF2 produces 1 mole of Pb2+ ions and 2 moles of F- ions.
Let's assume that 'x' is the concentration (in moles per liter) of Pb2+ ions and '2x' is the concentration of F- ions in the equilibrium state. Since 1 mole of PbF2 gives 1 mole of Pb2+ ions and 2 moles of F- ions, the expression for the solubility product constant (Ksp) is:
Ksp = [Pb2+][F-]^2
Therefore, applying the values:
3.2 * 10^-8 = x * (2x)^2
3.2 * 10^-8 = 4x^3
Solving for 'x', we get:
x = (3.2 * 10^-8 / 4)^{1/3}
x ≈ 0.002
Since the concentration of F- ions is twice that of Pb2+ ions, the equilibrium concentrations can be determined as follows:
[Pb2+] ≈ 0.002 M
[F-] ≈ 2 * 0.002 M
[Pb2+] ≈ 0.002 M
[F-] ≈ 0.004 M
Therefore, the equilibrium concentrations of Pb2+ and F- in the saturated solution of lead (II) fluoride are approximately 0.002 M and 0.004 M, respectively.
To determine the equilibrium concentrations of Pb2+ and F- in a saturated solution of lead (II) fluoride (PbF2), we need to use the solubility product constant (Ksp) and solve the equilibrium expression.
The balanced equation for the dissociation of PbF2 is:
PbF2 ⇌ Pb2+ + 2F-
The equilibrium expression is written as:
Ksp = [Pb2+][F-]^2
Given that the Ksp for PbF2 is 3.2 × 10^-8, we can assign the following variables to represent the equilibrium concentrations:
Let x represent the concentration of Pb2+ in mol/L.
Let y represent the concentration of F- in mol/L.
Now, we need to set up an equation using the Ksp expression and solve for equilibrium concentrations.
Ksp = [Pb2+][F-]^2
3.2 × 10^-8 = x * y^2
Since the stoichiometric coefficient of F- is 2, the concentration of F- will be 2y.
Substituting this into the equation, we have:
3.2 × 10^-8 = x * (2y)^2
3.2 × 10^-8 = 4x * y^2
Now, we need to make an assumption for the approximation since the Ksp value is very small compared to the initial concentrations of Pb2+ and F-, which are usually negligible. We assume that the equilibrium concentrations of Pb2+ and F- will be close to their initial concentrations.
Hence, we can say that x and y are approximately equal to their initial concentrations. Let's assign the initial concentration of Pb2+ as a and the initial concentration of F- as b.
3.2 × 10^-8 = 4(a)(b)^2
Since the solution is saturated, it means that the maximum amount of PbF2 has dissolved, and hence the concentrations of Pb2+ and F- are equal to their solubilities.
Therefore, the equilibrium concentrations of Pb2+ and F- in a saturated solution of lead (II) fluoride with a Ksp value of 3.2 × 10^-8 would be approximately equal to the square root of the Ksp value:
[Pb2+] ≈ √(3.2 × 10^-8)
[F-] ≈ 2 * √(3.2 × 10^-8)
By solving these calculations, we can find the numerical values for [Pb2+] and [F-].
..........PbF2 ==> Pb^2+ + 2F^-
I.........solid.....0.......0
C.........solid.....x.......2x
E.........solid.....x.......2x
Ksp = (Pb^2+)(F^-)^2
Substitute the E line into Ksp expression and solve for x and 2x.