I know my question is going to be quite below the level of this site but I am 1st grade in university and I can't understand anything.

I am going to give an example from the book: distinguish between $\ce{AgI}$ and $\ce{AgCl}$. Answersheet tells me to add $\ce{NH3}$; only $\ce{AgCl}$ dissolves. I don't even know why $\ce{AgI}$ precipitate and $\ce{AgCl}$ dissolves.

Or there is another question: distinguish between $\ce{Ag+}$ and $\ce{[Ag(NH3)2]+}$. And answer sheet tells to add NaCl and only $\ce{Ag+}$ gives a precipitate. Why doesn't $\ce{[Ag(NH3)2]+}$ also give a ppt.?

How do I know which cations or anions or salts are soluble in which solvent? What should I look when I am trying to distinguish ions or salts? Or how do I know if there is no reaction? I know these questions are too general but I am having a hard time understanding this class.

You can either answer the questions from the book (which I failed to understand) or help me to understand the basis of the subject by answering others. Or both. I would really appreciate any kind of answers coming from you.

  • $\begingroup$ Hi Ceren, Welcome to chemistry. SE! Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. $\endgroup$
    – Zenix
    May 23, 2020 at 20:59
  • 2
    $\begingroup$ AgI has extremely low $K_{sp}$ value, and $K_f$ of $\ce{[Ag(NH3)]2+}$ falls too less to compensate it. However AgCl has much higher $K_{sp}$ value compared to AgI and high $K_f$ value makes it soluble. For precipitation of AgCl, presence of free $\ce{Ag+}$ and $\ce{Cl-}$ is required. Coordination complexes, like $\ce{[Ag(NH3)2]+}$ here, don't break into free ions in aqueous solution. They are stable and exist as single entity, thus, won't give precipitation with NaCl. $\endgroup$
    – Zenix
    May 23, 2020 at 21:09
  • $\begingroup$ Note to OP: Feel free to upvote, downvote or ignore my answer. For what is it worth, I do NOT think you should have to know such fine points! By the way, welcome to the chemistry stack exchange! $\endgroup$
    – Ed V
    May 24, 2020 at 9:40

1 Answer 1


Answering this question requires a preliminary discussion of the solubility product constants of three silver halides and the formation constants of three silver complexes. First, note that $\ce{AgCl}$, $\ce{AgBr}$, and $\ce{AgI}$ are all insoluble in water, but insoluble is a relative term in the end. Their respective solubility equilibria and solubility product constants, i.e., equilibrium constants for their dissolution, are:

$$\ce{AgCl(s) <=> Ag^+ (aq) + Cl^- (aq) \quad $K_\mathrm{sp(1)} = \pu{1.8E-10}$ \tag 1}$$

$$\ce{AgBr(s) <=> Ag^+ (aq) + Br^- (aq) \quad $K_\mathrm{sp(2)} = \pu{5.4E-13}$ \tag 2}$$

$$\ce{AgI(s) <=> Ag^+ (aq) + I^- (aq) \quad $K_\mathrm{sp(3)} = \pu{8.3E-17}$ \tag 3}$$

Silver ions also form complexes with ammonia, thiosulfate ion, and cyanide ion. The formation equilibria and associated equilibrium constants are as follows:

$$\ce{Ag^+ (aq) + 2 NH3 (aq) <=> [Ag(NH_3)_2]^+ (aq) \quad $K_\mathrm{f(4)} = \pu{1.6E7}$ \tag 4}$$

$$\ce{Ag^+ (aq) + 2 S2O_3^2- (aq) <=> [Ag(S2O3)2]^3- (aq) \quad $K_\mathrm{f(5)} = \pu{2.0E13}$ \tag 5}$$

$$\ce{Ag^+ (aq) + 2 CN^- (aq) <=> [Ag(CN)2]^- (aq) \quad $K_\mathrm{f(6)} = \pu{1.0E21}$ \tag 6}$$

Now consider the following sequence of aqueous solution additions:

  1. An aqueous $\ce{NaCl}$ solution is added to an aqueous solution of $\ce{AgNO3}$. Then $\ce{AgCl}$ precipitates, as per equilibrium (1).

  2. Next concentrated ammonia is added in excess, i.e., well above the 2 to 1 stoichiometry of equilibrium (4). Then the following equilibrium, obtained by adding equilibria (1) and (4), occurs:

$$\ce{AgCl (s) + 2 NH3 (aq) <=> [Ag(NH3)2]^+ (aq) + Cl^- (aq) \quad $K_\mathrm{sp(1)} K_\mathrm{f(4)} = \pu{2.9E-3}$ \tag 7}$$

Although the equilibrium constant is less than unity, adding concentrated ammonia in excess (in a hood!) results in all of the $\ce{AgCl}$ dissolving: the equilibrium is driven to the right.

  1. Next $\ce{NaBr}$ is added. This results in precipitation of $\ce{AgBr}$ via the following equilibrium:

$$\ce{[Ag(NH_3)_2]^+ (aq) + Br^- (aq) <=> AgBr (s) + 2 NH3 (aq) \quad $1/(K_\mathrm{sp(2)} K_\mathrm{f(4)}) = \pu{1.2E5}$ \tag 8}$$

This equilibrium is simply the reverse of the addition of equilibria (2) and (4).

  1. Next excess sodium thiosulfate ($\ce{Na2S2O3}$) is added. Then $\ce{AgBr}$ dissolves as per the following equilibrium, obtained by adding equilibria (2) and (5):

$$\ce{AgBr (s) + 2 S2O3^2- (aq) <=> [Ag(S2O3)2]^3- (aq) + Br^- (aq) \quad $K_\mathrm{sp(2)} K_\mathrm{f(5)} = \pu{10.8}$ \tag 9}$$

  1. Next $\ce{KI}$ is added. This results in precipitation of $\ce{AgI}$ via the following equilibrium:

$$\ce{[Ag(S2O3)2]^3- (aq) + I^- (aq) <=> AgI (s) + 2 S2O3^2- (aq) \quad $1/(K_\mathrm{sp(3)} K_\mathrm{f(5)}) = \pu{6.0E2}$ \tag{10} }$$

This equilibrium is simply the reverse of the addition of equilibria (3) and (5).

  1. Lastly, $\ce{KCN}$ is added: in a hood, with proper safety precautions! Then $\ce{AgI}$ dissolves as per the following equilibrium, obtained by adding equilibria (3) and (6):

$$\ce{AgI (s) + 2 CN^- (aq) <=> [Ag(CN)2]^- (aq) + I^- (aq) \quad $K_\mathrm{sp(3)} K_\mathrm{f(6)} = \pu{8.3E4}$ \tag{11} }$$

When I did this demo in lectures, I skipped the last step in order to avoid dealing with the hazardous waste issue that cyanide solutions present.

So now, at long last, the OP's questions:

  1. Distinguish between $\ce{AgCl}$ and $\ce{AgI}$ by adding ammonia.

From equilibrium (7), $\ce{AgCl}$ can be dissolved if an excess of concentrated ammonia is added. Trust me: this should be done in a hood! But, comparing equilibria (1) and (7), $\ce{AgI}$ is more than a million times less soluble than $\ce{AgCl}$, so no amount of concentrated ammonia will significantly dissolve $\ce{AgI}$.

  1. The OP's second question involves distinguishing between silver ion and the silver ammonia complex.

The OP's answer sheet claims that adding $\ce{NaCl}$ should result in only silver ion giving a precipitate. But equilibrium (7) shows that sufficiently high chloride ion concentrations should result in $\ce{AgCl}$ precipitation. But silver also forms $\ce{[AgCl2]^-}$, $\ce{[AgCl3]^2-}$, and $\ce{[AgCl4]^3-}$ , when the chloride concentration is high. So adding chloride will not cause AgCl to precipitate, in agreement with the OP's answer sheet. See here as well.

Last thought: I have done this demo in lecture, minus the last cyanide step, and I think the OP should not be expected to know all this stuff, especially if an exam is involved. The entire point of this kind of problem is to show that we can control equilibria to our advantage in, e.g., gold extraction via the cyanide process, and insoluble is a relative term.

Source of all solubility product constants and formation constants:

Daniel C. Harris, Appendix I In Quantitative Chemical Analysis; 7th Ed.; W. H. Freeman & Company: New York, NY, 2007 (ISBN: 0-7167-7041-5; ISBN-13: 9780716770411).

  • $\begingroup$ You got great answer. $\endgroup$ May 24, 2020 at 6:34
  • $\begingroup$ @MathewMahindaratne Thanks! But the formatting appears different on a tablet than on a computer. Always something. $\endgroup$
    – Ed V
    May 24, 2020 at 9:26
  • $\begingroup$ Hope I didn't mess up. I didn't touch the equations but the formatting on the text body. $\endgroup$ May 24, 2020 at 11:34
  • $\begingroup$ Thank you so much, it couldn't have been any better. $\endgroup$
    – O.Ceren
    May 25, 2020 at 11:38
  • $\begingroup$ @O.Ceren Thanks and happy to be of assistance! $\endgroup$
    – Ed V
    May 25, 2020 at 11:56

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