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 255.3 g .