[Sarex]

Hi guys! I desperately need your help please...I can't figure out this solubility problem:
I have 50 mL of a solution 0,05 M of Pb(NO3)2 and I mix it with 40 mL of a solution 0,20 M of NaIO3.
I'm asked to calculate [Pb 2+] and [IO3-] knowing that Ksp of Pb(IO3)2 is 2,6*10^-13.
The results given by my book are 2,3*10^-10  and 0,033 respectively. Is there a mistake in my book maybe?
Thank you in advange to anyone that will help me!  and sorry for my bad English

[billnotgatez]

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[Borek]

No mistake in the book.

Hint: this is a limiting reagent problem.

[Sarex]

sorry for not respecting the forum policy,I'm not doing this mistake again 
I managed to solve the exercise, I'm gonna write the process so maybe it will be useful for someone else too.
First, I must calculate the available concentration of the ions Pb2+ and IO₃ -
[Pb²⁺]= (MV)/(total Volume) = (0,05*0,05)/0,09 = 2,8*10^-2  mol/L
[IO₃^-] =(0,20*0,04)/0,9 = 8,9*10^-2 mol/L
 
Reaction quotient Q= 2,8*10^-2 *8,9*10^-2 =2,5*10^-3 
as you can see, Q>Ksp so the precipitation occurs.
the limiting reagent in the reaction of precipitation is Pb2+; therefore we have some IO₃^- in surplus, 3,3*10^-2.
we have to calculate the solubility:
Ksp= [Pb²⁺][IO₃-]^2 = S*(S+ 3,3*10^-2) =2,6*10^-13  in which (S+3,3*10^-2) approaches to 3,3*10^-2.
So:  S=Ksp/3,3*10^-2=(2,6*10^-13)/(3,3*10^-2)= 2,3*10^-10 mol/L

Conclusion: [Pb 2+]= 2,3*10^-10  and [IO3-]=3,3*10^-2.
those are the concentrations of the ions that do not precipitate and are found in the solution.