The concentration of iodide ions in a saturated solution of lead(II) iodide is __________ M. The solubility product constantof PbI2 is 1.4x10-8
3E-3
To find the concentration of iodide ions in a saturated solution of lead(II) iodide, we need to use the solubility product constant (Ksp) of PbI2.
The solubility product constant (Ksp) is a measure of the degree to which a compound can dissolve in water. For a sparingly soluble salt like PbI2, it can be expressed as the product of the concentration of its constituent ions raised to the power of their respective coefficients in the balanced chemical equation.
The balanced equation for the dissolution of lead(II) iodide (PbI2) is:
PbI2(s) ⇌ Pb2+(aq) + 2I-(aq)
From the balanced equation, we can see that 1 mole of PbI2 produces 1 mole of Pb2+ ions and 2 moles of I- ions.
Since the concentration of lead(II) iodide in a saturated solution is equal to its molar solubility (s), we can write the equation for Ksp as:
Ksp = [Pb2+][I-]^2
According to the given solubility product constant, Ksp = 1.4x10^-8.
Now, let's assume the concentration of iodide ions (I-) in the saturated solution is x M. Since 1 mole of PbI2 produces 2 moles of I- ions, the concentration of iodide ions in terms of solubility (s) will be 2x.
Substituting these values into the Ksp expression:
1.4x10^-8 = (x)(2x)^2
1.4x10^-8 = 4x^3
x^3 = (1.4x10^-8) / 4
x^3 = 3.5x10^-9
Now, take the cube root of both sides to solve for x:
x = (3.5x10^-9)^(1/3)
x ≈ 6.3x10^-4 M
Therefore, the concentration of iodide ions (I-) in a saturated solution of lead(II) iodide is approximately 6.3x10^-4 M.
To find the concentration of iodide ions in a saturated solution of lead(II) iodide, we need to calculate the equilibrium concentration using the solubility product constant (Ksp) for PbI2.
The balanced equation for the dissolution of lead(II) iodide is:
PbI2 (s) ⇌ Pb2+ (aq) + 2I- (aq)
The Ksp expression for this equilibrium is:
Ksp = [Pb2+][I-]^2
Given that the solubility product constant (Ksp) for PbI2 is 1.4 x 10^-8, we can assume that at equilibrium the concentrations of Pb2+ and I- will be equal since the stoichiometric coefficient for iodide ions is 2.
Therefore, we can assume that [Pb2+] = x (M) and [I-] = 2x (M).
Substituting these concentrations into the Ksp expression, we get:
Ksp = (x)(2x)^2
1.4 x 10^-8 = 4x^3
Solving this equation for x, we find:
x = ∛(1.4 x 10^-8 / 4)
x ≈ 5.70 x 10^-3 M
Since [I-] = 2x, the concentration of iodide ions in the saturated solution of PbI2 is approximately:
[I-] ≈ 2(5.70 x 10^-3) M
[I-] ≈ 1.14 x 10^-2 M
Therefore, the concentration of iodide ions in a saturated solution of lead(II) iodide is approximately 1.14 x 10^-2 M.