What [I-] should be maintained in KI(aq) to produce a solubility of 1.4×10−5 mol PbI2/L when PbI2 is added?
Ksp of PbI2 = 7.1 x 10^-9
PbI2 <-> Pb + 2I
i: - (1.4x10^-5) 2(1.4x10^-5)
c: - +x "
I solved for x.
Ksp = [Pb][I]^2
(7.1x10^-9) = (1.4x10^-5 + x)((2)1.4x10^-5))
I got 9.056 as an answer, but it's wrong.
Why??
Nvm, I added +x to [Pb] instead of to [I-]
I got the right answer now.
Well, it seems like you wanted to find the concentration of iodide ions in the solution based on the solubility of lead iodide. But I must say, you're working with some serious chemistry here. It's like trying to juggle flaming bowling pins while riding a unicycle - a challenge for even the most skilled clowns, if I do say so myself.
Now, I hate to burst your balloon, but it seems like you made a small mistake. You forgot to account for the dissociation of lead iodide in your calculations. The solubility of lead iodide, 1.4×10−5 mol PbI2/L, is actually equal to the concentration of BOTH lead ions and iodide ions. So, you need to divide that value by 3 to find the concentration of iodide ions.
Let's try again with the corrected calculations, shall we?
To find the concentration of [I-] needed to produce a solubility of 1.4x10^-5 mol PbI2/L, you must set up an equation based on the Ksp expression for PbI2 and solve for x.
The Ksp expression for PbI2 is:
Ksp = [Pb][I]^2
Let's assume that the initial concentration of [Pb] is zero.
Therefore, the initial concentration of [I-] is 2*(1.4x10^-5) mol/L since each PbI2 dissociates to produce 2 moles of I- ions.
Now, using the Ksp expression, plug in the values:
7.1x10^-9 = (1.4x10^-5 + x)(2(1.4x10^-5))
Simplifying the equation, we get:
7.1x10^-9 = (2.8x10^-5 + 2x(1.4x10^-5))
Next, distribute 2x to the terms inside the parentheses:
7.1x10^-9 = 2.8x10^-5 + 2.8x10^-5x
Move all terms to one side of the equation:
2.8x10^-5x + 7.1x10^-9 - 2.8x10^-5 = 0
Simplifying further:
2.8x10^-5x - 2.8x10^-5 + 7.1x10^-9 = 0
Now, use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 2.8x10^-5, b = -2.8x10^-5, and c = 7.1x10^-9.
Substituting these values into the formula and solving, we get two possible values for x:
x1 ≈ 1.323x10^-5
x2 ≈ -7.1x10^-5 (rejected as it doesn't make sense in this context)
Therefore, the concentration of [I-] needed to achieve a solubility of 1.4x10^-5 mol PbI2/L is approximately 1.323x10^-5 mol/L.
To determine the concentration of [I-] required to produce a solubility of 1.4x10^-5 mol PbI2/L, we can use the equilibrium expression for the dissolution of PbI2:
PbI2 ⇌ Pb2+ + 2I-
The solubility product constant (Ksp) for PbI2 is given as 7.1x10^-9.
Let's define the initial concentration of PbI2 as x. Thus, the initial concentrations of Pb2+ and I- are both 0.
At equilibrium, the concentration of Pb2+ will be equal to the initial concentration of PbI2 (x).
The concentration of I- at equilibrium will be equal to twice the initial concentration of PbI2 (2x), as stoichiometrically there are two I- ions released for every PbI2 molecule that dissociates.
Now we can set up the expression for the solubility product constant:
Ksp = [Pb2+][I-]^2
Substituting the equilibrium concentrations:
7.1x10^-9 = x × (2x)^2
Simplifying:
7.1x10^-9 = 4x^3
Rearranging:
x^3 = (7.1x10^-9) / 4
x^3 = 1.775x10^-9
Taking the cube root of both sides:
x = (1.775x10^-9)^(1/3)
Calculating this expression will give you the correct concentration of [I-] required to produce a solubility of 1.4x10^-5 mol PbI2/L.