What [I-] should be maintained in KI(aq) to produce a solubility of 2.0×10−5 mol PbI2/L when PbI2 is added?
PbI2 ==> Pb^+2 + 2I^-
Write Ksp expression.
Plug in Pb^+2, and Ksp, and solve for I^-. A funny way of stating the problem, but the concn of PbI2 equals the concn of Pb^+2.
can you explain that better? i'm confused.
To determine the [I-] concentration necessary to maintain a solubility of 2.0×10−5 mol PbI2/L when PbI2 is added, you need to use the solubility product constant (Ksp) for PbI2.
The Ksp expression for PbI2 is as follows:
Ksp = [Pb2+][I-]^2
Since 1 mol of PbI2 dissociates to produce 1 mol of Pb2+ ions and 2 mol of I- ions, we can simplify the Ksp expression to:
Ksp = [I-]^2
To find the concentration of I- necessary, take the square root of the Ksp value:
[I-] = √(Ksp)
Substituting the given Ksp value of 2.0×10−5 mol PbI2/L:
[I-] = √(2.0×10−5 mol PbI2/L)
Calculating the square root:
[I-] ≈ 0.00447 mol/L
Therefore, to maintain a solubility of 2.0×10−5 mol PbI2/L when PbI2 is added, the concentration of I- should be approximately 0.00447 mol/L.