Question

Use the value ksp=1.4x10-8 for PbI2 to solve the following problems?

A: What is the concentration of iodide ions in a saturated solution of PbI2?

B: What is the solubility of PbI2 in a 0.010M solution of NaI?

Answer 1

Part- A

`PbI_2(soln)` ionises in its solution as follows

`PbI_2("soln")rightleftharpoonsPb^"2+" +2I^-`

If the concentration of `PbI_2` in its saturated solution is xM then concentration of `Pb^"2+"` will be xM and concemtration of `I^-` will be 2xM.

So `K_"sp"=[Pb^"2+"][I^"-"]^2`

`=>1.4xx10^-8=x*(2x)^2`

`=>x^3=1.4/4xx10^-8=3.5xx10^-9`

`=>x=1.518xx10^-3`

So the concentrattion of `I^-` ion in solution is `2x=3.036xx10^-3M`

Part- B

Let the solubility of `PbI_2` in 0.01 M NaI solution be s M.

Then in this case

`[Pb^"2+"]=sM`

And

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

So `K_"sp"=[Pb^"2+"][I^"-"]^2`

`=>1.4xx10^-8=sxx(2s+0.01)^2`

`=>1.4xx10^-8=sxx(4s^2+4s*0.01+(0.01)^2)`

neglecting `s^3 and s^2` terms we get

`=>1.4xx10^-8=sxx(0.01)^2`

`=>s=1.4xx10^-4M`

So solubilty of `PbI_2` in this case is `=1.4xx10^-4M`