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

Question

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?

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Answer 1 · 2016-11-03T10:08:39

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 $=1.4xx10^-4M$