Set 1

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. In a flask, we have thirty millilitres of oxalic acid, at zero point zero five molar. In the burette, potassium permanganate, at only zero point zero two molar, less than half as concentrated. So we will need much more permanganate, maybe seventy five millilitres, right? But the actual answer is thirty millilitres, the same volume as the acid! How is that possible? We will see by the end. Set one mixes all the chapters of this unit, in the style of recent JEE Main papers. So, let's begin.

0:40Tool 1: Lassaigne's test (question 1)

So, tool one is Lassaigne's test, with question one in the box. The fusion extracts of aniline, thiourea and chlorobenzene are made with a very small piece of sodium. Which one gives a blood red colour with ferric chloride? On the right is the course chart: the sodium fusion extract at the top, and the tests on it below. Okay? In the extract, nitrogen is sodium cyanide, N a C N, and sulphur is sodium sulphide, N a two S, with a subscript two written low. When both are present, they can stay together as N a SCN, sodium thiocyanate. Now, why does the question say a very small piece of sodium? It is because excess sodium destroys the thiocyanate. On screen, sodium thiocyanate plus two sodium gives sodium cyanide plus sodium sulphide, and the arrow means gives. So only with a little sodium do nitrogen and sulphur stay together as thiocyanate. Clear? Now acidify the extract and add ferric chloride, F e C l three; on the chart, it is the box for N and S together. The ferric ion, F e three plus, joins the thiocyanate ion, SCN minus; the raised signs are charges. The new ion in square brackets is blood red. So, which of our three compounds has both nitrogen and sulphur? Yes, exactly, it is thiourea, the only one with both nitrogen and sulphur. Aniline has nitrogen but no sulphur, and chlorobenzene has chlorine instead, so neither can form thiocyanate. Got it?

2:16Tool 2: 2,4-DNP and the iodoform test (question 2)

Now tool two, for organic tests, with question two in the box. Here, a ketone, C five H ten O, gives an orange precipitate with two four DNP, but no yellow precipitate with iodine and alkali. Identify it. First, two four DNP, short for two four dinitrophenylhydrazine, gives a yellow, orange or red precipitate with aldehydes and ketones. Okay? Next, the iodoform test: iodine with sodium hydroxide, the N a O H on screen. When it is positive, we get a yellow precipitate of iodoform, CHI three. Right? But only two groups give it: C H three C O, in ethanal and the methyl ketones, or C H three, C H, then O H in brackets, in some alcohols. The dash at the end is the bond to the rest of the molecule. Our ketone gives no iodoform. So which ketone is it? Very good if you said pentan three one. No iodoform means no methyl group next to the carbon double bond oxygen group. With five carbons, that leaves only the ketone on screen, with an ethyl group on each side of the C O group. With me?

3:31Tool 3: salt analysis (question 11)

Now tool three, salt analysis, with question eleven in the box. Here, a white salt gives with dilute hydrochloric acid a white precipitate soluble in hot water, a yellow precipitate with potassium chromate, and brown fumes with concentrated sulphuric acid and copper turnings. Identify the salt. Start with the first clue. Which cation gives that? It is lead, P b two plus, as the course chart on the right shows. Dilute H C l is the group one reagent, the Roman numeral one on screen, and lead comes down as white lead chloride, P b C l two, soluble in hot water. Okay? The second clue confirms it. With potassium chromate, K two C r O four, lead gives yellow lead chromate, P b C r O four. Right? Now the anion. When a nitrate is heated with concentrated sulphuric acid, H two S O four, it gives light brown fumes of N O two, and copper turnings, the C u on screen, make them dense. Clear? So the salt is lead nitrate: P b, then N O three in round brackets, taken two times. Got it?

4:43Example 1: 0.53 g Na2CO3 against 0.1 M HCl, methyl orange

Now Example one, question seven of the set. Zero point five three grams of anhydrous sodium carbonate is made up to one hundred millilitres. What volume of zero point one molar H C l does twenty point zero millilitres of it need, with methyl orange? On screen, g is grams, m L is millilitres, and capital M means molar. Pause the video and try it first. Done? Okay, let's check. The molar mass of sodium carbonate is one hundred and six grams per mole. So zero point five three over one hundred and six gives zero point zero zero five mole, the m o l on screen. That is five point zero millimoles, m m o l, in one hundred millilitres. Okay? But we titrate only twenty millilitres, one fifth of the solution. Five point zero times twenty over one hundred gives one point zero millimole of carbonate; the cross means times. With me? One H plus turns carbonate into bicarbonate, so we need one point zero millimole of H C l, which is ten millilitres, right? Oh no, careful, ten millilitres is only the first stage, where phenolphthalein changes. Methyl orange changes at the end of the second stage, after complete neutralisation. So carbonate, C O three two minus, takes two H plus and gives water and carbon dioxide. Clear? So we need two point zero millimoles of H C l. And the volume, V, is two point zero over zero point one, which gives twenty point zero millilitres. Got it?

6:27Example 2: 0.02 M KMnO4 needed for 30 mL oxalic acid

And now Example two, question six, which is our puzzle from the start. What volume of zero point zero two molar potassium permanganate, K Mn O four, is decolourised by thirty point zero millilitres of zero point zero five molar oxalic acid, in acid? Pause the video and try it first. Done? Okay, let's check. Millilitres times molarity gives millimoles. Thirty point zero times zero point zero five gives one point five millimoles of oxalic acid. Okay? So we need one point five millimoles of permanganate too, and one point five over zero point zero two gives seventy five millilitres, right? No, no, wait, that is the trap from our puzzle, because the ratio is not one to one. In the balanced equation, two permanganate ions react with five oxalate ions and sixteen H plus ions, giving two manganese two plus ions, ten carbon dioxide and eight water molecules. With me? So the permanganate is two fifths of the oxalic acid. One point five times two over five gives zero point six millimole. Okay? And the volume is zero point six over zero point zero two, which gives thirty point zero millilitres. So our puzzle is solved, because each permanganate ion does the work of two and a half oxalate ions. Clear? In the lab, the oxalic acid is taken with dilute sulphuric acid and warmed to sixty to seventy degrees Celsius. No indicator is needed, because permanganate is a self indicator: the end point is colourless to a permanent pale pink. Got it?

8:06Quick revision: Lassaigne, iodoform, lead, titration ratio

So, quick revision. Nitrogen and sulphur together as thiocyanate give a blood red colour with ferric chloride, but only with a little sodium. The iodoform test needs the C H three C O group or the C H three, C H, O H group; and lead gives white lead chloride and yellow lead chromate. In titrations, use the ratio from the balanced equation. Now try the rest of Set one on this page, with your timer running. All the best!