The Ksp of PbI2 is 8.7x10^-9. Find the solubility of lead(II) iodide on mol/L.
a. 1.3x10^-3
b. 9.3x10^-9
c. 6.6x10^-5
d. 1.6x10^-3
To find the solubility of lead(II) iodide (PbI2) in mol/L, we need to calculate the square root of the Ksp value of PbI2.
The Ksp expression for PbI2 is:
PbI2 ⇌ Pb2+ + 2I-
The Ksp expression can be written as:
Ksp = [Pb2+][I-]^2
Given that the Ksp value of PbI2 is 8.7x10^-9, we can write the equation as:
8.7x10^-9 = [Pb2+][I-]^2
Since the stoichiometry of the reaction is 1:2, the concentration of Pb2+ is equal to the solubility of PbI2.
Let's denote the solubility of PbI2 as "x", so the concentration of Pb2+ is also "x".
Plugging these values into the Ksp expression, we get:
8.7x10^-9 = [x][2x]^2
8.7x10^-9 = 4x^3
Rearranging the equation, we have:
4x^3 = 8.7x10^-9
Dividing both sides of the equation by 4, we get:
x^3 = 2.175x10^-9
To find the solubility (x), we take the cube root of both sides:
x = (2.175x10^-9)^(1/3)
After calculating this expression, we find that the solubility of PbI2 is approximately 1.33x10^-3 mol/L.
The closest option to this value is option a. 1.3x10^-3.
To find the solubility of lead(II) iodide (PbI2), we need to use the value of its solubility product constant (Ksp) and set up an equilibrium expression.
The equilibrium expression for the dissociation of PbI2 can be written as:
PbI2 ⇌ Pb2+ + 2I-
The solubility product constant (Ksp) expression for PbI2 is:
Ksp = [Pb2+][I-]^2
Given that the Ksp of PbI2 is 8.7x10^-9, we can assume that the solubility of PbI2 can be represented as 's'.
Therefore, at equilibrium, the concentration of Pb2+ is also 's' and the concentration of I- is 2s (since the stoichiometric ratio between Pb2+ and I- is 1:2).
Plugging these values into the Ksp expression:
Ksp = [Pb2+][I-]^2 = (s)(2s)^2 = 4s^3
Now, we can substitute the Ksp value into the equation:
8.7x10^-9 = 4s^3
Solving this equation for 's', we take the cube root of both sides:
∛(8.7x10^-9 / 4) = s
Calculating this, we get:
s ≈ 1.3x10^-3
So, the solubility of lead(II) iodide (PbI2) is approximately 1.3x10^-3 mol/L.
Therefore, the correct answer is option (a) 1.3x10^-3.
PbI2 ==>Pb^+2 + 2I^-
Ksp = (Pb^+2)(I^-)^2
Let S = solubility, then (Pb^+2) = S and I^-) = 2S.
Substitute into the Ksp expression and solve for S. Post your work if you get stuck.