Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Let us start with a small lab puzzle. An organic compound may contain seven elements: nitrogen, sulphur, chlorine, bromine, iodine, fluorine and oxygen. You fuse it with sodium and prepare the sodium fusion extract. Now, with the usual tests on that extract, how many of these seven can you actually detect? Is it all seven? Everybody has a guess, right? Hold on to yours; we settle it in the very first problem. This is the JEE Main level part of our practical chemistry problem bank: two quick problems first, and then two full titration problems. So, let's begin.

0:41Quick problem 1: seven elements, how many?

Quick problem number one is our puzzle. On screen are the seven element symbols: N for nitrogen, S for sulphur, then C l, B r and I for chlorine, bromine and iodine, F for fluorine, and O for oxygen. We want to know how many of them the sodium fusion extract can detect. Okay? First, the reason for fusing with sodium: in the compound, these elements are held by covalent bonds, so they give no ionic tests. Fusion with sodium, N a, turns them into sodium salts: sodium cyanide, N a C N, sodium sulphide, N a two S, where the small two written low counts the sodium atoms, and the sodium halides, N a X, where X stands for chlorine, bromine or iodine. Right? Now the usual tests. Nitrogen gives Prussian blue, and sulphur gives a black precipitate of lead sulphide, P b S, with lead acetate. With me? And the halides are tested with silver nitrate, A g N O three. Silver chloride, A g C l, is white, silver bromide is pale yellow, and silver iodide is yellow. Clear? So that makes five, and fluorine is also a halogen. So fluoride should give a silver precipitate too, and the answer is six, right? No, and that is exactly the trap, because silver fluoride, A g F, is soluble in water. No precipitate forms, so fluoride is not detected by this test. Got it? And oxygen has no test in this scheme at all. So the answer is five: nitrogen, sulphur, chlorine, bromine and iodine. That settles our opening puzzle. Okay?

2:23Quick problem 2: unequal acid and base

Quick problem number two comes from the physical chemistry experiments. Solutions containing zero point five moles of hydrochloric acid, H C l, and zero point three moles of sodium hydroxide, N a O H, are mixed; m o l on screen means moles. And the question asks how much heat is evolved. With me? Recall the course value. For a strong acid with a strong base, the only reaction is one hydrogen ion, H plus, with one hydroxide ion, O H minus, giving water; a q in brackets means in water, the letter l in brackets means liquid, and the arrow means gives. The enthalpy change, delta H, written with the Greek triangle delta, is minus fifty seven point one kilojoules per mole; the minus one written high means per mole. Right? Now, we have zero point five moles of acid. So the heat is zero point five times fifty seven point one kilojoules, the cross on screen meaning times, right? No, wait, that is the trap, because heat comes only from the water that actually forms. There are only zero point three moles of hydroxide ions, so sodium hydroxide is the limiting reagent, and only zero point three moles of water form. The extra acid just stays unreacted. Clear? So the heat evolved is zero point three times fifty seven point one, which is about seventeen point one kilojoules; the wavy equals sign means about. Find the limiting reagent first, always. Got it?

3:52Example 1: NaOH volume to neutralise 20 mL of 0.05 M H2SO4

Alright, example one, an easy one. What volume of zero point zero eight molar sodium hydroxide neutralises twenty point zero millilitres of zero point zero five molar sulphuric acid, H two S O four? Here capital M means molar, that is moles per litre, and m L means millilitres. Pause the video and try it first. Done? Okay, let's check. Many students write zero point zero five times twenty point zero equals zero point zero eight times V, where V is the volume of sodium hydroxide. That gives twelve point five millilitres. Is that correct? No, and this is the classic trap, because sulphuric acid gives two H plus ions per molecule. So its n-factor is two, not one. Okay? So use the course rule: normality, capital N, equals molarity times n-factor. The acid is zero point zero five times two, which gives zero point one zero normal. Sodium hydroxide has n-factor one, so it stays zero point zero eight normal. Right? Now the titration formula, N A times V A equals N B times V B: normality times volume is the same for both solutions, and the letters A and B written low label the acid and the base. For the acid, zero point one zero times twenty point zero gives two point zero millimoles of H plus; m mol on screen means millimoles. With me? So zero point zero eight times V equals two point zero, and V is two point zero over zero point zero eight, which gives twenty five point zero millilitres. That is exactly double the wrong answer, so never forget the n-factor. Clear?

5:34Example 2: Mohr's salt oxidised by 20 mL of 0.02 M KMnO4

Now example two, a little harder. What mass of Mohr's salt is oxidised by twenty point zero millilitres of zero point zero two molar potassium permanganate, K M n O four, in acid? Pause the video and try it first. Done? Okay, let's check, starting with the set-up in the course figure. Permanganate, the titrant, is in the burette, and since it is dark, the label says read the upper meniscus, not the lower one. The stop-cock lets us add drop by drop near the end point, and the white tile shows the colour change clearly. The conical flask holds a pipetted volume of the Mohr's salt solution. Usually it also gets two to three drops of indicator, but the label says none with K M n O four. Okay? Now the reaction from the course. One permanganate ion, M n O four minus, reacts with five ferrous ions, F e two plus, and eight H plus ions, giving one manganese two plus ion, five ferric ions, F e three plus, and four water molecules; the charges are written high. Right? Millilitres times molarity gives millimoles. So twenty point zero times zero point zero two gives zero point four millimoles of permanganate. Got it? Now, zero point four millimoles of permanganate. So zero point four millimoles of Mohr's salt is oxidised, right? No, no, careful, look at the equation again, because one permanganate oxidises five ferrous ions. So five times zero point four gives two point zero millimoles of ferrous ion. Clear? Each unit of Mohr's salt holds one ferrous ion, and its molar mass is three hundred and ninety two grams per mole. So two point zero times three hundred and ninety two gives seven hundred and eighty four milligrams, m g on screen, which is zero point seven eight four grams. With me? And notice, we needed no indicator at all, because permanganate is a self-indicator. At the end point, the solution in the flask changes from colourless, or very pale green, to a permanent pale pink. Got it?

7:37Quick revision: Lassaigne's test, 57.1 kJ, n-factor, 1:5

So, quick revision. Lassaigne's test, on the sodium fusion extract, detects nitrogen, sulphur, chlorine, bromine and iodine, but not fluorine, because silver fluoride is soluble, and not oxygen. Neutralising a strong acid with a strong base gives fifty seven point one kilojoules per mole of water, and the limiting reagent sets how much water forms. In titrations, normality is molarity times n-factor, and permanganate reacts with the ferrous ion one to five, with no indicator. Now try the other problems of this part on this page, each one as a test before you read its solution. All the best!