Question

The solubility product of pbI2 at 298K is 1.4×10-8. Calculate it's solubility at this temperature.​

Answer 1

Answer:

Explanation:

Part- A

PbI2(soln) ionises in its solution as follows

PbI2(soln)⇌Pb2++2I−

If the concentration of PbI2 in its saturated solution is xM then concentration of Pb2+ will be xM and concemtration of I− will be 2xM.

So Ksp=[Pb2+][I-]2

⇒1.4×10−8=x⋅(2x)2

⇒x3=1.44×10−8=3.5×10−9

⇒x=1.518×10−3

So the concentrattion of I− ion in solution is 2x=3.036×10−3M

Part- B

Let the solubility of PbI2 in 0.01 M NaI solution be s M.

Then in this case

[Pb2+]=sM

And

[I-]=(2s+0.01)M

So Ksp=[Pb2+][I-]2

⇒1.4×10−8=s×(2s+0.01)2

⇒1.4×10−8=s×(4s2+4s⋅0.01+(0.01)2)

neglecting s3ands2 terms we get

⇒1.4×10−8=s×(0.01)2

⇒s=1.4×10−4M

So solubilty of PbI2 in this case is =1.4×10−4M