Monday
September 1, 2014

Homework Help: CHEMISTRY

Posted by Jason on Monday, May 14, 2007 at 11:09am.

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:
Answer:

Related Questions

Chemistry - 50.00 mL samples of 0.200 mol/L potassium phosphate with an excess ...
Chemistry help!!! - The class will react 50.00 mL samples of .200 mol/L ...
chem - I'm stuck on this one aswell... 50.0 ml of a 0.0500 M solution of lead (...
Chemistry - a class reacts 50.00ml samples of 0.200 mol/L potassium phosphate ...
chemistry check 2 - I know you've read this question too many times already but ...
chemistry - When aqueous solutions of sodium sulfate and lead (II) nitrate are ...
Chemistry - If 32.0 mL lead(II) nitrate solution reacts completely with excess ...
Chemistry - Suppose we have a solution of lead nitrate, Pb(NO3). A solution of ...
chemistry - I am unsure about balancing equation I would like for someone to let...
Cheměstry - The following two solutions were combined and mixed well: 150.00 mL ...

Search
Members