What [I^-]should be maintained in KI(aq) to produce a solubility of 1.9×10−5mol PbI2/L when PbI2 is added?
(sorry i cant find it under the posted questions )
Write the Ksp expression. Solve for concn I^-. You know Ksp and Pb^+2 since Pb^+2 is same as concn PbI2. That leaves just the one unknown of I^-.
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To determine the concentration of iodide ion ([I^-]) necessary to produce a solubility of 1.9×10^−5 mol PbI2/L when PbI2 is added, you need to consider the solubility product constant (Ksp) of PbI2. The Ksp expression for the dissolution of PbI2 is:
PbI2 ↔ Pb2+ + 2I^-
The Ksp expression can be written as:
Ksp = [Pb2+][I^-]^2
Given that the solubility of PbI2 is 1.9×10^−5 mol PbI2/L, the concentration of Pb2+ can be considered negligible since it is much smaller than the concentration of I^- ions. Therefore, you can assume [Pb2+] ≈ 0.
Now, you can substitute the values into the Ksp expression:
Ksp = (0)([I^-])^2 = 1.9×10^−5
Simplifying the equation:
[I^-]^2 = 1.9×10^−5
Taking the square root of both sides:
[I^-] = √(1.9×10^−5)
[I^-] ≈ 0.00436 M
Therefore, to produce a solubility of 1.9×10^−5 mol PbI2/L when PbI2 is added, the concentration of iodide ion ([I^-]) should be maintained at approximately 0.00436 M.