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?

1 Answer
Nov 3, 2016

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#