Question

# Part I: Calculate the concentration of IO3– in a 9.05 mM Pb(NO3)2 solution saturated with Pb(IO3)2....

Part I: Calculate the concentration of IO3– in a 9.05 mM Pb(NO3)2 solution saturated with Pb(IO3)2. The Ksp of Pb(IO3)2 is 2.5 × 10-13. Assume that Pb(IO3)2 is a negligible source of Pb2 compared to Pb(NO3)2.

Part II: A different solution contains dissolved NaIO3. What is the concentration of NaIO3 if adding excess Pb(IO3)2(s) produces [Pb2 ] = 3.80 × 10-6 M?

Please explain thoroughly. I am SO confused!! Thank you SO much in advance!

Part I

Pb(NO3)2--->Pb+2 + 2N03- meaning that the concentration of Pb+2 is equal to the concentration of Pb+2=Pb(NO3)2=9.05*10-3M.

In the other hand Pb(IO3)2 results on

Pb(IO3)2---> Pb+2 +2IO3-

Ksp= [Pb+2]*[IO3-]2=2.5*10-13

As the problem says that the only source of Pb+2 comes from the other reaction, the concentration of [Pb+2] is the same above [Pb+2]=9.05*10-3

Clearing [IO3-]= sq(2.5*10-13/9.05*10-3)=5.26*10-6

Part II

Pb(IO3)2---> Pb+2 +2IO3- and Ksp= [Pb+2]*[IO3-]2=2.5*10-13

Being now [Pb+2]=3.80*10-6, we can clear [IO-3] in solution that is 2-56*10-4

Only comming from NaIO3 dissociation the substract of the total [IO3-] minus the amounto produced by PbIO3 dissociation that is 2*3.80*10-6

So, NaIO3---> Na+ + IO3-

[NaIO3]=2.56*10-4- 2*3.86*10-6= 2.46*10-4

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