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Finding insoluble and soluble ionic compound in precipiation reaction

Submitted by eguitar on Sun, 10/23/2011 - 14:49
I have a chemistry test in which I have to create insoluble compounds from ionic compounds given to me on the sheet, how do I create an insoluble compound? An example would be greatly appreciated.

It depends on the metal ion present. You need to look at (and learn perhaps) the solubility rules. Some are simple eg all nitrates are soluble and are often used to make insoluble salts. All group 1 metal salts and ammonium salts are soluble. So if you were asked to make the insoluble salt silver(I) chloride you would mix solutions of silver(I) nitrate and a soluble chloride such as NaCl or KCl or NH4Cl.

You always mix solutions of the soluble salts as these contains separate ions.

NaCl(aq) + AgNO3(aq) ---> AgCl(s) + NaNO3(aq). So when Ag+ ions meet Cl- ions, the join and make an insoluble salt, AgCl. The remaining ions remain in solution.

Na+Cl-(aq) + Ag+NO3-(aq) ---> Ag+Cl-(s) + Na+NO3-(aq) (sometime the insoluble salt is not shown as ionic.

As soluble salts exist as separate ions in solution, it can be written as

Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) ---> AgCl(s) + Na+(aq)  + NO3-(aq)

Then spectator ions can be eliminated

Cl-(aq) + Ag+(aq) ---> AgCl(s) and this shows than any soluble chloride solution will react with any soluble silver ion solution to give a (white in this case) precipitate of silver(I) chloride

This link, like many others is useful.

To make lead(II) iodide (an insoluble salt) use a soluble nitrate such as lead(II) nitrate and a soluble iodide such as sodium iodide.

Pb(NO3)2(aq) + 2NaI(aq) ----> PbI2(s) + 2NaNO3(aq) etc

Hope this helps