Initially, we are given a solution containing two salts, they can be either carbonate or chloride salts (we do not know their composition initially).

Generally, $\ce{AgNO3}$ is used as a confirmatory test of $\ce{Cl-}$ (a white precipitate is obtained on adding silver nitrate to solution containing the chloride salt). $$\ce{NaCl + AgNO3 -> NaNO3 + \underset{white}{AgCl} \downarrow}$$ But $$\ce{Na2CO3 + 2AgNO3 -> 2NaNO3 + \underset{brownish white}{Ag2CO3} \downarrow}$$ Therefore, can we use $\ce{AgNO3}$ to directly identify chloride in the solution or do we need to remove the possible carbonate ions to confirm the presence of chloride before performing the test as mentioned above?

(A similar question may arise in the identification of bromide and iodide ions but I have ignored it for now as they are rarely asked in exams. Still, I would appreciate it if someone could answer this as well.)

The question has been asked mainly because the silver nitrate test is generally taught to be performed without mentioning the above precautions and it is also a very famous confirmative test.

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    $\begingroup$ You should work on your English sentences a little bit anyway, If you are asking how to test for $Cl^-$ reliably then ,You can add dil. $HNO_3$ to move the equilibrium to the carbonic acid side through application of Le-Chatelier Principle so as to have minimal [${CO_3}^{2-}$]in the solution which would not bring about any unreliability in the confirmation of $Cl^-$ $\endgroup$
    – Rishi
    Commented Apr 30, 2021 at 14:34
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    $\begingroup$ Yes, Moreover I think there will be some other ways as well for this but to me this is the simplest. $\endgroup$
    – Rishi
    Commented May 1, 2021 at 2:34
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    $\begingroup$ My general understanding of your question is you are given sodium chloride and sodium carbonate and you are to react them with silver nitrate in order to distinguish them. Right? In that case, it might work. Silver carbonate becomes more and more yellow and brown when excess of silver nitrate is used. Silver chloride remains white. Also, silver carbonate is soluble in nitric acid whereas silver chloride is not. $\endgroup$ Commented May 1, 2021 at 4:24
  • $\begingroup$ The final statement is false. the silver nitrate test is performed on an acidic solution to preclude any interference from carbonates, hydroxides etc. The original question is misleading because it imposes conditions that are resolved in the standard qual scheme. $\endgroup$
    – jimchmst
    Commented Jul 12 at 17:39

1 Answer 1


The test for chloride ion is silver nitrate in an acidified solution [with nitric acid]. This removes carbonate (as carbonic acid), cyanide [careful] and sulfide (as hydrogen sulfide). (Use a well-ventilated hood as the products of acidifying both sulfide and cyanide are poisonous.) Sulfate must be absent, if present it can be removed with barium nitrate (as barium sulfate). The silver chloride formed is solubilized by addition of ammonia; insoluble precipitate remaining may be bromide or iodide. The exact procedure to follow is given in reputed in-depth qualitative analysis lab books.


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