J.R. S. answered 04/20/20
Ph.D. in Biochemistry--University Professor--Chemistry Tutor
First, identify the precipitate....
Na2CO3(s) + Pb(Ac)2(aq) ===> 2NaAc(aq) + PbCO3(s) So, PbCO3 is the precipitate of the reaction.
Look up the Ksp for PbCO3. I find a value of 1.5x10-13
Ksp = [Pb2+][CO32-]
We can find the [Pb2+] from the information provided: [Pb2+] = 0.0611 M
Now, we can solve for [CO32-]:
1.5x10-13 = [0.0611][CO32-]
[CO32-] = 2.45x10-12 M to just initiate precipitation