Question #072db

1 Answer
Stefan V.
Oct 20, 2015

Here's what I got.

Explanation:

You know that your solution contains metal cations.

You also know that this solution will produce a precipitate, which is an insoluble solid, when mixed with a solution of sodium chloride, #"NaCl"#.

Sodium chloride is a soluble ionic compound, which means that it will dissociate completely to form sodium cations, #"Na"^(+)#, and chloride anions, #"Cl"^(-)#.

This means that the pricipitate will be a halide, or, more specifically, a chloride.

http://highered.mheducation.com/olcweb/cgi/pluginpop.cgi?it=jpg::::::/sites/dl/free/0023654666/650262/Solubility_Rules_4_02.jpg::Solubility%20rules

If you take a look at the solubility rules for chlorides, you'll notice that three common metal ions will produce insoluble solids with the chloride anion

  • Lead(II) cation #-># #"Pb"^(2+)#
  • Mercury(I) cation #-># #"Hg"_2^(2+)#
  • Silver(I) cation #-># #"Ag"^(+)#

So, the three ionic compounds that can precipitate out of solution are

  • Silver chloride, #"AgCl"#

http://www.bbc.co.uk/schools/gcsebitesize/science/add_edexcel/ionic_compounds/ionicanalysisrev5.shtml

  • Mercury(I) chloride, #"Hg"_2"Cl"_2#

https://en.wikipedia.org/wiki/Mercury%28I%29_chloride

  • Lead(II) chloride, #"PbCl"_2#

https://en.wikipedia.org/wiki/Lead%28II%29_chloride

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