How to distinguish silver chloride, bromide and iodide?

Question

When we test for chloride, bromide and iodide ions we use the obvious silver nitrate and dilute nitric acid test. However distinguishing between the precipitates is often quite tricky. How can we tell the difference between a white, cream or yellow precipitate respectively. I understand we could use aqueous ammonia for the white precipitate, and that if it dissolved it would indicate the presence of chloride ions. However, the next two are much more difficult given my circumstances. With my practical test only aqueous ammonia is supplied, so I do not have access to concentrated ammonia. So what should I do now to distinguish between the remaining bromide and iodide ions?

Are there any tips such as holding white paper behind it to help me distinguish the two?

Answer 1

Nilay Ghosh's method is excellent, but it does not work if the original deposit is a mixture of the three precipitates $\ce{AgCl, AgBr, AgI}$. In this case, it is first necessary to get rid of the silver from the mixture of precipitates, and look for $\ce{Cl-, Br-}$ or $\ce{I-}$ in the residual solution.

The best way to do it, is to add $3-5$ mL $1$ M $\ce{H2SO4}$, then about $0.5$g zinc powder. Wait for half an hour, filtrate the mixture $\ce{Zn + Ag}$. Save the obtained solution and test the presence of $\ce{Cl-, Br-, I-}$ ions in solution.

Anion $\ce{I-}$ can be detected by adding $1$ mL hydrocarbon (like cyclohexane) and adding $1-2$ drops $0.1$ M $\ce{NaNO2}$. The hydrocarbon turns violet or purple in the presence of iodide.

Bromide $\ce{Br-}$ can be detected by first destroying iodide ions by first neutralizing the acidic solution by $\ce{Na2CO3}$, then adding enough $\ce{KMnO4}$ solution (which produces $\ce{MnO2 + I2}$ precipitates, but $\ce{KMnO4}$ gives no reaction with bromide), filtrating, and testing the filtrate by adding two drops of $\ce{Cl2}$ in solution (chlorine water). Bromide is transformed into bromine $\ce{Br2}$. Adding now $3$ mL cyclohexane, it becomes brown if bromine is in solution.

Chloride ions are detected in the original message by Gotfried Leibnitz.

Answer 2

Aqueous ammonia should do the trick. Silver chloride is soluble in dilute ammonia solution. Silver bromide is sparingly soluble in dilute ammonia solution. Silver iodide is very slightly soluble in concentrated ammonia solution (so insoluble in dilute ammonia solution). Silver bromide/iodide is off-white as compared to silver chloride, but you can easily distinguish them if put side-by-side. Silver iodide has a slightly darker tone than silver bromide.

If you are confused with the silver nitrate-ammonia test, you can opt for sulfuric acid which gives you a more definite result. Chloride will give hydrogen chloride gas (pungent smell), bromide will give bromine gas (reddish-brown vapor), iodide will give iodine gas (purple vapor).

Answer 3

I have fond memories of doing a lab like this back in high school. Our teacher said we could bring whatever we wanted from home to help with the lab. My idea was to bring in a 9V battery and two pieces of graphite from a pencil. Stick the graphite leads in a tiny 96 well plate of your test solution and run current through them. If chlorine is present you will observe the solution turning slightly yellow-green due to the presence of chlorine formation. If there is bromine or iodine present you will see a brown color change due to the presence of bromine and tri-iodine ions. It is possible to tell the difference between the electrolyzed bromine and iodine on their own, but if you want to be extremely sure add some starch to complex with the tri-iodine and you'll see a much darker blackish blue color if there is iodine compared to bromine. Hope this helps!

Answer 4

I believe that silver bromide, chloride and iodide have significantly different colours with silver chloride is white crystalline solid , silver bromide is somewhat pale yellow and silver iodide is bright yellow 🟡 , therefore physical examination must be enough for the required distinction