Why does a clear, saturated solution of PbI2 form a yellow precipitate when a small amount of KI is added to the solution?
A. KI is insoluble in PbI2.
B. PbI2 is always insoluble in water.
C. Adding KI is like adding I-, which drives the reaction toward the formation of I-.
D. Adding KI is like adding I-, which drives the reaction toward the formation of PbI2.
PbI2 ==> Pb^2+ + 2I^-
If you have a saturated solution of PbI2 then Qsp = Ksp = (Pb^2+)(I^-)^2
Adding KI ==> K^+ + I^-
So you are increasing the (I^-), that is adding the common ion I^- to the saturated solution so the reaction shifts to the left forming more PbI2.
OR, you can say that adding I^- increases Qsp so it is larger than Ksp so a ppt of PbI2 forms because Qsp>Ksp.
D is the answer but you need to understand why D is the answer.
We need to understand, thank you for taking the time to explain Drbob.
Thank you for your explanation!
To determine the answer to this question, we need to understand the solubility of the compounds involved and the reaction that occurs when KI is added to the clear, saturated solution of PbI2.
First, let's consider the solubility of the compounds.
PbI2, which is lead(II) iodide, is a sparingly soluble salt. This means that it does dissolve to some extent in water, but the concentration of the dissolved ions is relatively low compared to the solid form present in the solution.
KI, which is potassium iodide, is a highly soluble salt. It completely dissociates into potassium ions (K+) and iodide ions (I-) when dissolved in water.
Now, let's analyze the reaction that takes place when KI is added to the clear, saturated solution of PbI2.
When KI is added, the iodide ions (I-) from KI react with the lead(II) ions (Pb2+) from PbI2 to form lead(II) iodide (PbI2). The reaction can be represented as follows:
Pb2+ + 2I- → PbI2
This reaction is a precipitation reaction because a solid precipitate (PbI2) is formed. The formation of the yellow precipitate occurs because PbI2 is less soluble than Pb2+ and I- ions.
Now, let's go through the answer choices:
A. KI is insoluble in PbI2. This is not true because KI is highly soluble in water.
B. PbI2 is always insoluble in water. This is not true because PbI2 is sparingly soluble in water.
C. Adding KI is like adding I-, which drives the reaction toward the formation of I-. This is not true. Adding KI actually drives the reaction towards the formation of PbI2, as explained above.
D. Adding KI is like adding I-, which drives the reaction toward the formation of PbI2. This is the correct answer. By adding KI, we are introducing iodide ions (I-), which react with lead(II) ions (Pb2+) to form the yellow precipitate of PbI2.
Therefore, the correct answer is D. Adding KI is like adding I-, which drives the reaction toward the formation of PbI2.