What [I^-]should be maintained in KI(aq) to produce a solubility of 1.9×10−5mol PbI2/L when PbI2 is added?
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What I^- should be maintained in KI to produce a solubility of 1.9×10−5 PbI2/L when PbI2 is added?
To determine the necessary concentration of [I^-] in the KI(aq) solution to achieve a solubility of 1.9×10^−5 mol PbI2/L, we need to use the concept of the solubility product constant (Ksp).
The solubility product constant, Ksp, is an equilibrium constant that describes the solubility of a sparingly soluble salt in a solution. It is derived from the equilibrium expression for the dissolution of the salt.
For the dissociation of PbI2, the equation is:
PbI2(s) ⇌ Pb^2+(aq) + 2I^-(aq)
The corresponding solubility product expression is:
Ksp = [Pb^2+][I^-]^2
Given that the solubility of PbI2 is 1.9×10^−5 mol/L, we can assume that the concentration of [Pb^2+] is negligible compared to [I^-] since lead (Pb^2+) is highly insoluble. Therefore, we can approximate the expression as:
Ksp ≈ [I^-]^2
Substituting the given value for Ksp and solving for [I^-], we have:
1.9×10^−5 = [I^-]^2
Taking the square root of both sides, we find:
[I^-] ≈ √(1.9×10^−5)
[I^-] ≈ 0.00436 mol/L
Therefore, the concentration of [I^-] in the KI(aq) solution should be around 0.00436 mol/L to achieve a solubility of 1.9×10^−5 mol PbI2/L when PbI2 is added.