Post a New Question


posted by .

Wondering If I did this correctly so far
50.0 ml of a 0.0500 M solution of lead (II) nitrate is mixed with 40.0 ml of a 0.200 M solution of sodium iodate at 25°C. Calculate the Pb2+ and IO3- concentrations when the mixture comes to equilibrium. At this temperature the Ksp for lead (II) iodate is 2.6 x 10-13. (10 marks)

The balanced equation
Pb(NO3)2 + 2NaIO3 ==> Pb(IO3)2 + 2NaNO3
Pb(NO3)2 = (0.05L)(0.05M) = 2.5 x 10-3 mols.
NaIO3 = (0.04L)(0.2M) = 8.0 x 10-3 mols.
NaIO3 is in excess.
Now if this is correct what do I do next?

So far so good.
NaIO3 is is excess; therefore, determine mols Pb(IO3)2 pptd and mols NaIO3 remaining.(All of the Pb(NO3)2 will be consumed.) Then it becomes a common ion problem with Ksp.
You will have solid Pb(IO3)2 in a saturated solution of Pb(IO3)2 that also contains xx mols NaIO3. Calculate Pb^+2.

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question