What [I] should be maintained in KI(aq) to produce a solubility of 2.0×10−5 mol PbI2/L when PbI2 is added?
i did...x^3 = 2.0*10^-5 but that didn't work out too well. help?
PbI2 ==> Pb^+2 + 2I^-
X^3 doesn't work because the problem TELLS you the solubility of PbI2 if I^- is such and such a level. The problem wants you to calculate (I^-) to maintain that level.
Ksp = (Pb^+2)(I^-)^2
You know Ksp and you know PbI2 (which, by the way, is the same as Pb^+2), so plug those in and solve for I^-.
To determine the [I] needed to produce a solubility of 2.0×10^-5 mol PbI2/L when PbI2 is added, we need to consider the solubility product constant (Ksp) of PbI2.
The equilibrium expression for the dissolution of PbI2 in water is as follows:
PbI2(s) ↔ Pb2+(aq) + 2I-(aq)
The solubility product constant expression for PbI2 can be written as:
Ksp = [Pb2+][I-]^2
Given that the solubility of PbI2 is 2.0×10^-5 mol PbI2/L, we can assume that [Pb2+] is negligible (as it is much smaller than [I-]). Therefore, we can assume that [I-] is approximately equal to the solubility concentration.
So, we can write:
Ksp = (2.0×10^-5)([I-])^2
Simplifying the expression for Ksp, we can rearrange it to find [I-]:
[I-]^2 = Ksp / (2.0×10^-5)
Taking the square root of both sides, we get:
[I-] = √(Ksp / (2.0×10^-5))
Substituting the given value of Ksp (which you haven't mentioned), you can solve this equation using a calculator to find the [I-] concentration required to achieve a solubility of 2.0×10^-5 mol PbI2/L.
Please provide the value of Ksp, and I can assist you further in calculating the [I-] concentration.