Lead ions can be precipitated from solution with NaCl
according to the following reaction.
Pb2+(aq)+2NaCl(aq)→PbCl2(s)+2Na+(aq)
When 135.8 g
of NaCl
are added to a solution containing 195.7 g
of Pb2+
, a PbCl2
precipitate forms. The precipitate is filtered and dried and found to have a mass of 257.2 g
.