If the molar solubility of lead iodide is 2.4 x 10^-3M. Calculate the Ksp.
PbI2 ==> Pb^+2 + 2I^-
Ksp = (Pb^+2)(I^-)^2
Therefore, (Pb^+2) = 2.4 x 10^-3M
and (I^-) = 4.8 x 10^-3 M.
Plug into Ksp expression and solve for Ksp.
To calculate the Ksp of lead iodide (PbI2) given the molar solubility, we need to set up the solubility equilibrium equation and determine the equilibrium concentrations of the ions in solution.
The solubility equilibrium equation for lead iodide is:
PbI2 ⟷ Pb2+ + 2I-
Let's denote 's' as the molar solubility of lead iodide, which is equal to the concentration of Pb2+ and twice the concentration of I-. Therefore, [Pb2+] = s and [I-] = 2s.
Now, substitute the equilibrium concentrations into the equilibrium expression for the solubility product constant (Ksp):
Ksp = [Pb2+][I-]^2
Ksp = (s)(2s)^2
Ksp = 4s^3
Given that the molar solubility (s) of lead iodide is 2.4 x 10^-3 M, substitute this value into the equation:
Ksp = 4(2.4 x 10^-3)^3
Ksp = 4(1.3824 x 10^-8)
Ksp = 5.5296 x 10^-8
Therefore, the Ksp of lead iodide (PbI2) is approximately 5.5296 x 10^-8.
To calculate the solubility product constant (Ksp) for lead iodide (PbI2), we need to use the given molar solubility.
The molar solubility of lead iodide (PbI2) is given as 2.4 x 10^-3 M. This means that at equilibrium, the concentration of lead ions (Pb^2+) and iodide ions (I^-) will be equal to this value.
The balanced chemical equation for the dissolution of lead iodide is:
PbI2(s) ⇌ Pb^2+(aq) + 2 I^-(aq)
From the balanced equation, we can see that the concentration of Pb^2+ ions is equal to the concentration of PbI2, while the concentration of I^- ions is equal to twice the concentration of PbI2.
Therefore, [Pb^2+] = 2.4 x 10^-3 M (concentration of lead ions)
[I^-] = 2 × (2.4 x 10^-3 M) = 4.8 x 10^-3 M (concentration of iodide ions)
Now, we can substitute these values into the expression for Ksp:
Ksp = [Pb^2+] × [I^-]^2
Ksp = (2.4 x 10^-3 M) × (4.8 x 10^-3 M)^2
Ksp = 2.4 x 10^-3 M × 2.304 x 10^-5 M
Ksp = 5.5296 x 10^-8 M^3
Therefore, the solubility product constant (Ksp) for lead iodide (PbI2) is 5.5296 x 10^-8 M^3.