Consider an aqueous solution which is 0.020 M in Pb²⁺ and 0.20 M in Ag⁺. If concentrated potassium iodide solution (so that volume changes may be neglected) is added gradually with good stirring to this solution, which precipitate will form first? Ksp for PbI2 = 7.1 × 10^-9; for AgI = 8.31 × 10^-17

Question

Quizplus · Accepted Answer

To determine which precipitate will form first, we need to compare the ion product, Q, of each compound to its respective solubility product, Ksp.

For PbI2, Q = [Pb2+][2I-]^2. Initially, there is no iodide ion present in the solution, so Q = 0, which is less than Ksp, indicating that PbI2 is not yet saturated and will not precipitate.

For AgI, Q = [Ag+][I-]. Initially, there is 0.20 M of Ag+ and 0 M of iodide ion present, so Q = 0, which is less than Ksp, indicating that AgI is not yet saturated and will not precipitate.

As we gradually add KI solution, the concentration of iodide ion will increase while the concentration of Ag+ and Pb2+ will remain constant. At some point, Q for AgI will exceed Ksp and AgI will begin to precipitate. PbI2 will not precipitate until the concentration of iodide ion is high enough to exceed its Ksp. Therefore, the first precipitate to form will be AgI.

The other answer choices can be eliminated for the following reasons:
A) PbI2 will not form until later.
B) KI is not a precipitate, it is a soluble salt.
D) PbI2 will not form until later, while AgI has already precipitated.
E) This notation is incorrect, but if it were meant to represent a solid solution of AgI and PbI4, it would require PbI4 to be a stable compound, which it is not.

Consider an Aqueous Solution Which Is 0