Asked by Laura
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.
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.
Answers
Answered by
DrBob222
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.
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.
Answered by
Drbob hater
We don’t need to understand bob. Give us the answer
Answered by
Anonymous
Thank you for your explanation!
Answered by
Drbob Stan
We need to understand, thank you for taking the time to explain Drbob.
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