In the process of separating pb2+ ions from cu+2 ions as sparingly soluble iodates, what is Pb2+ concentration when Cu+2 begins to precipitate as sodium iodate is added to a solution that is initially 0.0010M Pb(NO3)2 (aq) and 0.0010M Cu(NO3)2 (aq)?
I know the answer is 1.8*10^-9 M. However I would like to know how to solve this problem.
Thank you!
3 answers
What Ksp values are you using? The answer depends upon those value you use.
Use your numbers for this.
Adding NaIO3 dropwise to the solution will continue until the most insoluble material starts pptng. That will be Pb(IO3)2. Pb(IO3)2 will continue to ppt until the IO3^- rises high enough to begin pptn of Cu(IO3)2. At that point, what is the (IO3^-)? It is
Ksp = (Cu^2+)(IO3^-)^2.
Plug in 0.001 for Cu and your value for Ksp and solve for IO3^-. Then plug this IO3^- back into the Ksp expression for Pb(IO3)2 and solve for Pb^2+. Using the values I found on the web of 3.49E-13 for Pb(IO3)2 and 1.4E-7 for Cu(IO3)2 I found (Pb^2+) to be about 2.5E-9M which is close to your answer. Using your values for Ksp should give you the answer.
Adding NaIO3 dropwise to the solution will continue until the most insoluble material starts pptng. That will be Pb(IO3)2. Pb(IO3)2 will continue to ppt until the IO3^- rises high enough to begin pptn of Cu(IO3)2. At that point, what is the (IO3^-)? It is
Ksp = (Cu^2+)(IO3^-)^2.
Plug in 0.001 for Cu and your value for Ksp and solve for IO3^-. Then plug this IO3^- back into the Ksp expression for Pb(IO3)2 and solve for Pb^2+. Using the values I found on the web of 3.49E-13 for Pb(IO3)2 and 1.4E-7 for Cu(IO3)2 I found (Pb^2+) to be about 2.5E-9M which is close to your answer. Using your values for Ksp should give you the answer.
This solution actually gets you 1.8x10^-9 M!
So Ksp of Copper Iodate is 1.4x10^-7 and Ksp = [Cu^2+][IO3^-]^2. You rearrange to solve for [IO3-] by plugging in Ksp and 0.001 M as [Cu^2+]to get 0.0118 M.
Taking 0.0118 as the [IO3^-], you then focus on the concentration of Pb2+. The Ksp of Lead Iodate is 2.6x10^-13 and the Ksp = [Pb^2+][IO3^-]^2. Rearrange to solve for [Pb^2+] and plug in Ksp and 0.0118 as [IO3^-] (don't forget to square [IO3^-]!!!). The resulting concentration of the lead ion should be 1.8x10^-9 if you did everything correctly!
So Ksp of Copper Iodate is 1.4x10^-7 and Ksp = [Cu^2+][IO3^-]^2. You rearrange to solve for [IO3-] by plugging in Ksp and 0.001 M as [Cu^2+]to get 0.0118 M.
Taking 0.0118 as the [IO3^-], you then focus on the concentration of Pb2+. The Ksp of Lead Iodate is 2.6x10^-13 and the Ksp = [Pb^2+][IO3^-]^2. Rearrange to solve for [Pb^2+] and plug in Ksp and 0.0118 as [IO3^-] (don't forget to square [IO3^-]!!!). The resulting concentration of the lead ion should be 1.8x10^-9 if you did everything correctly!