A colourless solution contains a halide ion. Describe a test to show which of the halide ions, chloride, CI, bromide, Br, or iodide, IF, is present in the colourless solution.
Answers
Explanation:
rying out the test
This test has to be done in solution. If you start from a solid, it must first be dissolved in pure water.
The solution is acidified by adding dilute nitric acid. (Remember: silver nitrate + dilute nitric acid.) The nitric acid reacts with, and removes, other ions that might also give a confusing precipitate with silver nitrate.
Silver nitrate solution is then added to give:
ion present observation
F- no precipitate
Cl- white precipitate
Br- very pale cream precipitate
I- very pale yellow precipitate
The chloride, bromide and iodide precipitates are shown in the photograph:
The chloride precipitate is obviously white, but the other two aren't really very different from each other. You couldn't be sure which you had unless you compared them side-by-side.
All of the precipitates change colour if they are exposed to light - taking on grey or purplish tints.
The absence of a precipitate with fluoride ions doesn't prove anything unless you already know that you must have a halogen present and are simply trying to find out which one. All the absence of a precipitate shows is that you haven't got chloride, bromide or iodide ions present.
The chemistry of the test
The precipitates are the insoluble silver halides - silver chloride, silver bromide or silver iodide.
Silver fluoride is soluble, and so you don't get a precipitate.
Confirming the precipitate using ammonia solution
Carrying out the confirmation
Ammonia solution is added to the precipitates.
original precipitate observation
AgCl precipitate dissolves to give a colourless solution
AgBr precipitate is almost unchanged using dilute ammonia solution, but dissolves in concentrated ammonia solution to give a colourless solution
AgI precipitate is insoluble in ammonia solution of any concentration
Explaining what happens
Background
There is no such thing as an absolutely insoluble ionic compound. A precipitate will only form if the concentrations of the ions in solution in water exceed a certain value - different for every different compound.
This value can be quoted as a solubility product. For the silver halides, the solubility product is given by the expression:
Ksp = [Ag+(aq)][X-(aq)]
The square brackets have their normal meaning, showing concentrations in mol dm-3.
If the actual concentrations of the ions in solution produce a value less than the solubility product, you don't get a precipitate. If the product of the concentrations would exceed this value, you do get a precipitate.
Essentially, the product of the ionic concentrations can never be greater than the solubility product value. Enough of the solid is precipitated so that the ionic product is lowered to the value of the solubility product.
Note: If your syllabus says that you need to know about solubility product calculations, you might be interested in my chemistry calculations book where they are explained in detail.
Look at the way the solubility products vary from silver chloride to silver iodide. (You can't quote a solubility product value for silver fluoride because it is too soluble. Solubility products only work with compounds which are very, very sparingly soluble.)