What reagent could NOT be used to separate #OH^-# from #Cl^-# when added to an aqueous solution containing both?

#a. AgNO_3(aq)#
#b. Ca(NO_3)_2(aq)#
#c. Cu(NO_3)_2(aq)#
#d. Fe(NO_3)_2(aq)#
My answer sheet says the answer is A, but I don't understand why.

2 Answers
Dec 18, 2017

It may be because BOTH silver chloride and silver hydroxide will precipitate. The other three will precipitate only the hydroxide form.

Explanation:

Solubility rules:
"4". Halides i.e. chlorides (#Cl^−# ), bromides (#Br^−#) and iodides (#I^−#) are soluble except for the +halides of lead (#Pb^(2+)#), mercury (#Hg^+# and #Hg_2^(+2)# and silver (Ag ).
e.g.
If X = Cl, Br or I, then MX, MX 2 , MX 3 , etc. are soluble unless M = Pb, Hg or Ag.

"7". Hydroxides (#OH^−# ) are insoluble or slightly soluble except for the hydroxides of the alkalis.
Note: The hydroxides of group 2 (the alkaline earth metals) are slightly soluble. Virtually all other hydroxides are insoluble.

Dec 18, 2017

Look up solubility rules, and see which metal ions form precipitated with chloride and which do so with hydroxides.

Explanation:

Roughly, most chlorides are readily soluble whereas most hydroxide are not. But two exceptions apply to the compounds listed here:

1) chlorides of certain transition metals like silver and lead are insoluble or at best, have only limited solubility.

2) hydroxides of alkali metals are soluble, other hydroxides
(apart from strontiolum and barium, maybe) have at best only limited solubility.

So, for instance, if you add copper nitrate the copper and chloride ions don't form an insoluble or sparingly soluble compound, but copper and hydroxide ions form an essentially insoluble compound, so the hydroxide ions precipitate and the chloride ions remain in solution.

Now try it with the other choices. You find that (a) fails by precipitating both silver hydroxide and silver chloride, since silver chloride being essentially insoluble is one of the exceptions noted above.