Question #072db

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Question #072db

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Answer 1 · 2015-10-20T17:41:19

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, $NaCl$ $"NaCl"$ .

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

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

![http://highered.mheducation.com/olcweb/cgi/ pluginpop.cgi

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 $\to$ $->$ ${Pb}^{2 +}$ $"Pb"^(2+)$

Mercury(I) cation $\to$ $->$ ${Hg}_{2}^{2 +}$ $"Hg"_2^(2+)$

Silver(I) cation $\to$ $->$ ${Ag}^{+}$ $"Ag"^(+)$

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

Silver chloride, $AgCl$ $"AgCl"$

![ www.bbc.co.uk )

Mercury(I) chloride, ${Hg}_{2} {Cl}_{2}$ $"Hg"_2"Cl"_2$

![https://en.wikipedia.org/wiki/Mercury%28I%29_chloride]( useruploads.socratic.org )

Lead(II) chloride, ${PbCl}_{2}$ $"PbCl"_2$

![https://en.wikipedia.org/wiki/Lead%28II%29_chloride]( useruploads.socratic.org )

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Question #072db

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