Anions Detected with Concentrated Acid

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Take two white salts, sodium chloride and sodium iodide. Warm each one with concentrated sulphuric acid. The chloride gives a colourless gas, but the iodide gives violet vapours! Same acid, same kind of salt, so why such different results? Today we learn the concentrated acid tests for four anions, and we will solve this puzzle too. These tests are a regular part of JEE practical chemistry, so let's begin.

0:36When we use concentrated acid

So, here is the method. We warm the salt with concentrated sulphuric acid, H two S O four; the small numbers are subscripts, okay? And notice the condition on screen: we do this only if the dilute acid test gave nothing. Now let me ask you this: why must the dilute acid test come first? Because carbonate, sulphide and nitrite also react with concentrated acid. Their gases would hide or confuse the results we are looking for today. So the dilute acid always comes first. Clear? And what does the concentrated acid test find? Four anions: chloride, bromide, iodide and nitrate. Let us take them one by one. With me?

1:20Chloride: colourless, fuming HCl

First, chloride. Sodium chloride, N a C l, reacts with sulphuric acid to give sodium hydrogen sulphate and hydrogen chloride, H C l. The arrow means gives, okay? And what is this H C l like? It is a colourless gas with a pungent smell, and it fumes in moist air. But many gases are colourless, right? So we need a confirmation. Now tell me: we bring a rod dipped in ammonia near the gas. What do you expect to see? Yes, exactly, dense white fumes. Those fumes are ammonium chloride, N H four C l. Got it? Now look at the badge: we add one more reagent, manganese dioxide, M n O two. So why would we add it? Manganese dioxide oxidises the H C l to free chlorine, C l two, which is a greenish-yellow gas. And that colour is much easier to recognise. Clear?

2:18Bromide and iodide: coloured vapours

Now bromide. Look at this equation. Two sodium bromide with three sulphuric acid give sodium hydrogen sulphate, sulphur dioxide, bromine and water; the big numbers in front tell us how many of each, okay? See the sulphur dioxide, S O two? That means the acid itself got reduced. With me? So what do we see? Reddish-brown vapours of bromine, B r two, along with some H B r, as the brackets say. And manganese dioxide intensifies these vapours, it makes them stronger. Okay? Now, iodide gives vapours of iodine. What colour are they, and what happens on starch paper? Yes, violet vapours of iodine, written I two. And when they touch starch paper, the paper turns blue. Right? And iodide reduces the sulphuric acid even further than bromide does. So along with sulphur dioxide, you may also get sulphur, the S on screen, or even hydrogen sulphide, H two S, okay?

3:17Why the halides behave so differently

Now, back to our puzzle. Chloride gave us colourless H C l, but iodide gave violet iodine, with the same acid. So why? Think about which hydrogen halide is the better reducing agent. The key is reducing power. The reducing power of the hydrogen halides rises from H C l, to H B r, to H I; the pointed signs mean less than. So H I is the strongest reducing agent of the three. Clear? Now, H C l is not a strong enough reducing agent, so sulphuric acid does not oxidise it. That is why chloride gives plain H C l gas. Right? H B r is partly oxidised, so we get bromine along with some H B r. And H I is oxidised almost completely, to iodine, as fast as it forms. So our puzzle is solved: iodide gives iodine because H I is such a good reducing agent! Now one more question. Can we make pure H B r from a bromide with concentrated sulphuric acid? No, we cannot, and this is a common trap. The acid oxidises it to bromine. So for pure H B r or H I, we use concentrated phosphoric acid, which is non-oxidising. Got it?

4:34Nitrate: light brown NO₂

And the last anion, nitrate. With concentrated sulphuric acid, it gives light brown fumes of nitrogen dioxide, N O two. Okay? Where do these fumes come from? The nitric acid formed, HNO three, breaks down. Four HNO three give four N O two, plus oxygen, O two, plus two water. Keep this equation in mind; we need it in example two. But wait. Bromide also gave brown vapours, right? So how do we tell a nitrate from a bromide? We add copper turnings. With a nitrate, the brown fumes become dense, and the solution turns blue. That is how copper turnings tell a nitrate apart from a bromide. Clear? Why does this happen? Copper, C u, reduces the nitric acid, giving much more N O two, so the fumes deepen. And copper nitrate forms; in its formula, the bracket with a two outside means two nitrate groups. It is blue, and it colours the solution. With me?

5:33Example 1: find the anion from white fumes with ammonia

Alright, example one. Read it with me: a salt warmed with concentrated sulphuric acid gives a colourless, pungent gas that forms dense white fumes with ammonia. Which anion is present? Pause the video and try it first. Done? Okay, let's check. First, the colour. The gas is colourless, so it is not bromine, not iodine and not nitrogen dioxide, because those are all coloured. So bromide, iodide and nitrate are out. Right? Next, the dense white fumes with ammonia. Those are ammonium chloride, so the gas must be H C l. Okay? And H C l comes from a chloride. So the anion present is chloride. Got it?

6:17Example 2: gas volume at STP when HNO3 decomposes

Now a slightly harder one, with numbers. Zero point one moles of nitric acid decomposes completely, as in the equation on screen. Find the total volume of gas at STP, which means standard temperature and pressure; the unit m o l is short for moles. Pause the video and try it first. Hint: use the ratios in the equation. Done? Okay, let's check. Four HNO three give four N O two, so the ratio is four to four, which is one to one. So zero point one moles of nitric acid gives zero point one moles of N O two. Okay? And oxygen? Only one O two for every four nitric acid. So zero point one over four gives zero point zero two five moles of oxygen. Right? Now, the equation also gives water. Should we add the water to the gas too? No, no, careful, because this is the trap: at STP, water is a liquid, not a gas. So we do not count it. If you add it, your answer comes out too big. Clear? So the total gas is zero point one plus zero point zero two five, which gives zero point one two five moles. With me? At STP, one mole of any gas takes up twenty two point four litres. So zero point one two five times twenty two point four gives two point eight litres; the capital L means litres. Got it?

7:42Quick revision: chloride, bromide, iodide, nitrate tests

So, quick revision. Chloride gives colourless H C l, with white fumes near ammonia, and manganese dioxide gives chlorine. Bromide gives reddish-brown bromine; iodide gives violet iodine, which turns starch paper blue. Nitrate gives light brown N O two, denser with copper turnings, and the solution turns blue. And the rising reducing power, from H C l to H I, explains why the halides behave so differently. Now go and try the practice problems on this page. You can do it. All the best!