Confirmatory Tests

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. You add silver nitrate to two salt solutions. One gives a pale yellow precipitate, the other a yellow one. In a real test tube, can your eyes truly tell those two colours apart? Not easily, right? And here is a second puzzle: in one test, a brown colour forms only as a thin ring, at one place in the tube. Why only there? Today we learn the confirmatory tests that remove such doubts, for chloride, bromide, iodide and nitrate. They come up often in JEE salt analysis questions, so let's begin.

0:39Halides: the silver nitrate test

So, first the halides: chloride, bromide and iodide. We make the salt solution acidic with dilute nitric acid, HNO three; the small three below the line is a subscript, counting the oxygen atoms. Then we add silver nitrate, written as A g N O three on the screen. Okay? Now let me ask you this: why do we add the nitric acid first, and not the silver nitrate straight away? Yes, the acid stops other silver salts from coming down. Without it, silver carbonate, sulphide or hydroxide could precipitate too, and in nitric acid these dissolve or are decomposed. Clear? Now the precipitates. Chloride gives silver chloride, A g C l, which is white, and it dissolves in ammonium hydroxide, N H four O H, the ammonia solution. Right? Bromide gives pale yellow silver bromide, only sparingly soluble in ammonia, and iodide gives yellow silver iodide, which does not dissolve at all. So pale yellow and yellow look alike, and the colour alone is risky. The ammonia helps, and the next tests settle it. With me?

1:48Chromyl chloride test: for chloride only

Next, a test for chloride only: the chromyl chloride test. Here we take the solid salt and heat it with solid potassium dichromate, K two C r two O seven, and concentrated sulphuric acid, H two S O four, with the word concentrated shortened to c o n c on the screen. Okay? Here is the reaction, and the arrow means gives. Four moles of sodium chloride, one mole of potassium dichromate and six moles of sulphuric acid give two moles of chromyl chloride, together with the hydrogen sulphates of sodium and potassium, and water. Got it? And what do we see? Orange-red vapours of chromyl chloride, C r O two C l two. Right? We pass these vapours into sodium hydroxide, and get a yellow solution of sodium chromate. Then acetic acid and lead acetate give a yellow precipitate of lead chromate. Clear? Now, bromide and iodide are halides too. Could they give similar coloured chromyl compounds and fool us? They cannot fool it, because dichromate oxidises them to bromine and iodine, shown as B r two and I two, and these give colourless solutions in sodium hydroxide, and no yellow precipitate with lead acetate. So this test really is for chloride only. Okay?

3:07Layer test: bromide and iodide

So how do we tell bromide from iodide for sure? With the layer test. To the salt solution we add chlorine water and carbon tetrachloride, C C l four. With a bromide, bromine is set free, and the carbon tetrachloride layer turns orange-brown. With me? Now you tell me: with an iodide, what colour will this layer be? Remember the colour of iodine vapour. Yes, exactly, it is violet, because iodine is set free. So orange-brown means bromide, and violet means iodide. Clear?

3:40Brown ring test for nitrate

Now our second puzzle: the brown ring test for nitrate. To the solution we add freshly prepared ferrous sulphate, F e S O four. Then we gently pour concentrated sulphuric acid down the side of the tilted tube, so the heavy acid forms its own layer below. Okay? And where the two layers meet, a brown ring appears. Right? So what is happening here? Nitrate, N O three minus, is reduced by ferrous ions in acid to nitric oxide, and the ferrous ions become ferric. The signs up top are charges: F e two plus is ferrous, and F e three plus is ferric. Got it? Then the nitric oxide joins a hydrated ferrous ion. The square brackets hold the complex ion, and the round brackets group the water molecules: one water leaves, nitric oxide takes its place, and this new complex is the brown ring. Clear? Now let me ask you this: why only a ring? Why is the whole tube not brown? The reason is that the junction is the only place where nitrate, ferrous ions and concentrated acid are all together. So the brown complex forms only there, as a ring, and that solves our second puzzle.

4:56The trap: nitrite looks like nitrate

Okay? Now careful, here is a JEE trap. You add ferrous sulphate and some acid, and you see a brown colour. So the salt is a nitrate, right? Not necessarily, because a nitrite gives the same brown colour with ferrous sulphate, even with dilute acetic or sulphuric acid. But with a nitrite, the brown colour is throughout the solution, not a neat ring. With me? Nitrite also turns acidified potassium iodide and starch blue; on the screen that is K I, then a dash, then starch. Right? So when nitrite may be present, we destroy it first, by boiling with urea, or with ammonium chloride, N H four C l, or with sulphamic acid. Only then do we test for nitrate. Clear?

5:42Example 1: identify the anion from AgNO3 and the CCl4 layer

Alright, here is our first example. Here, a solution gives a pale yellow precipitate with silver nitrate, only partly soluble in ammonia. And with chlorine water and carbon tetrachloride, the lower layer turns orange-brown. We have to name the anion. Pause the video and try it first. Done? Okay, let's check. Pale yellow, and only partly soluble in ammonia: that matches silver bromide, which is sparingly soluble. Okay? But wait, silver iodide is yellow too. Could this salt be an iodide? It cannot be, because silver iodide does not dissolve in ammonia at all, and iodine would turn the layer violet. Orange-brown means bromine, so the anion is bromide, B r minus, and the minus sign up top is its charge. Two tests together leave no doubt, and that solves our first puzzle.

6:39Example 2: spin-only magnetic moment of brown ring complex

Got it? Now a harder one, on the brown ring complex itself. In this complex, the iron is usually taken as plus one, with N O present as N O plus. How many unpaired electrons does it have, and what is its spin-only magnetic moment? Pause the video and try it first. Done? Okay, let's check the charge first. The whole complex carries two plus. Water is neutral, and N O plus brings plus one, so the iron, F e, must be plus one. Right? Now, how many d electrons does this iron, F e plus, have? Neutral iron has eight electrons beyond the argon core, so F e plus has seven, and as the course takes it, all seven sit in the three d level, written three d seven, with the little seven up top counting them. In high spin, the five d orbitals take one electron each, then two pair up, so three stay unpaired. With me? Now the spin-only formula: mu, the Greek letter for magnetic moment, equals the square root of n times n plus two, where n is the number of unpaired electrons. With n equal to three, we get the square root of three times five, which is root fifteen, about three point eight seven Bohr magnetons, B M; the wavy equals sign means approximately. Got it?

7:59Quick revision: halide, nitrate and nitrite tests

So, quick revision. For halides, silver nitrate in nitric acid gives white, pale yellow and yellow precipitates, and ammonia helps tell them apart. Chloride is confirmed by the chromyl chloride test, and bromide and iodide by the orange-brown or violet layer. Nitrate gives a brown ring at the junction, while nitrite gives brown throughout. Now go and try the practice problems on this page. All the best!