Question

A 1.0 L aqueous solution contains 0.30 mol of KBr and 0.60 mol of NaBr. What is the minimum number of moles of Pb(NO3)2 that must be added to precipitate all of the bromide ions?

Answer 1

How is it solved?

Answer 2

The same solution constains 0.90 mol of bromide ions (0.30+0.60 mol) that derive from the KBr and NaBr. In order to precipitate the 0.90 mol of Br- you will need the right amount of Pb cations which have the form Pb2+. The final precipitate should be neutral (PbBr2) so you will need 0.45 mol of Pb2+ that derive from 0.45 mol of Pb(NO3)2.

0.30 mol of KBr contains: 0.30 mol of Na+ and 0.30 mol od Br-
0.60 mol of NaBr contains: 0.60 mol of K+ and 0.60 mol of Br-
=> The total moles of bromide anion are 0.90 mol.

Combining 0.90 mol of Br- with 0.45 mol of Pb2+ you would obtain 0.45 mol of PbBr2.
I hope this will help you out.

Answer 3

Yes it has thank you!