Set 3

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Someone hands you a flask with two acids mixed in it, oxalic acid and sulphuric acid. You titrate it with sodium hydroxide, but the alkali neutralises both acids, so you only get the total. So how will you find how much of each acid is there? It looks tricky, right? By the end of this video, we will find both, with just one more titration, using potassium permanganate. This problem is from Set three of our mixed problems, and titration questions like this come up often in JEE Main. So, let's begin.

0:39How to use Set 3

First, what is Set three? It has fifteen JEE Main style problems, mixing all the chapters of this unit, and you give yourself twenty five minutes for the whole set. Okay? Time yourself first, and only then check the solutions. That way you practise against the clock. Right? Today, we solve two permanganate titrations from this set, an easier one and then a harder one. But first, the tools we need. With me?

1:07Tool 1: millimoles and the 2 : 5 ratio

So, tool one: we count in millimoles, written m mol. Capital M is the molarity, in moles per litre, and capital V is the volume in millilitres, the m L in brackets. Molarity times volume gives millimoles; the cross means times. Okay? Now the reaction. Two permanganate ions, M n O four minus, five oxalate ions, C two O four two minus, and sixteen hydrogen ions, H plus, react; the arrow points to what they give. We get two manganese ions, M n two plus, ten carbon dioxide and eight water. The raised numbers and signs are the charges, and the small numbers written low count the atoms. Right? So tell me, one millimole of permanganate oxidises how many millimoles of oxalate? Very good if you said two and a half. Two permanganate ions for five oxalate ions, so one takes five over two, which is two point five millimoles of oxalate. The small fraction on screen is five over two. Clear? The course writes it as a formula. Molarity times volume of permanganate, over two, equals molarity times volume of oxalic acid, over five; each long line is a fraction bar. The labels written low name the solution, and ox stands for oxalic acid. Notice that the two goes under permanganate, its own number in the equation. Got it?

2:33Tool 2: setting up the flask

Tool two is the set-up, in the course figure: the burette above, the conical flask below, on a white tile. Permanganate goes in the burette. The flask holds the oxalate with an equal volume of dilute sulphuric acid, H two S O four, warmed to sixty to seventy degrees Celsius; the little circle is the degree sign. Okay? And the indicator? The flask label in the figure says none with K Mn O four, because permanganate is a self-indicator. At the end point, the colourless solution turns a permanent pale pink; the arrow here means changes to. Right? Now let me ask you this. Why sulphuric acid? Why not hydrochloric acid, or nitric acid? Permanganate would oxidise hydrochloric acid, H C l, to chlorine, so hydrochloric acid is not used. Clear? And nitric acid, HNO three, is not used because it is itself an oxidising agent. So dilute sulphuric acid it is; keep this in mind for Example two. Got it?

3:37Tool 3: both acids give two H⁺ ions

Tool three is for the alkali. Oxalic acid, H two C two O four, gives two H plus ions per molecule. Okay? And sulphuric acid, H two S O four? How many H plus ions does it give? It also gives two H plus ions per molecule, just like oxalic acid. Right? So one millimole of either acid needs two millimoles of N a O H, that is sodium hydroxide. Remember this factor of two; it is the trap in Example two. With me?

4:11Example 1: molarity of KMnO4 from 0.134 g sodium oxalate

Now Example one, the easier one. Zero point one three four grams of sodium oxalate needs twenty five point zero millilitres of a permanganate solution in acid. Find its molarity. Here, g means grams, and in the bracket, capital M equals one hundred and thirty four is the molar mass of sodium oxalate. Pause the video and try it first. Done? Okay, let's check. Zero point one three four grams is one hundred and thirty four milligrams, written m g. Milligrams over molar mass give millimoles, so one hundred and thirty four over one hundred and thirty four gives one point zero millimoles of oxalate. Okay? Now the permanganate. The ratio has a two and a five, so five over two times one point zero gives two point five millimoles of permanganate, right? No, wait, that ratio is upside down, and that is the trap. Permanganate is the smaller partner, two for every five oxalate. So the permanganate needed is two over five times one point zero, which gives zero point four millimoles. Clear? Finally, molarity is millimoles over millilitres. Zero point four over twenty five point zero gives zero point zero one six molar; capital M after the number means molar. The upside-down ratio would have given zero point one molar, more than six times too much. Got it?

5:40Example 2: oxalic and sulphuric acid mixture, NaOH and KMnO4

And now Example two, our puzzle from the start. Twenty five point zero millilitres of a solution containing oxalic acid and sulphuric acid needs thirty point zero millilitres of zero point one molar sodium hydroxide for neutralisation, and twenty point zero millilitres of zero point zero two molar K Mn O four for oxidation. Find the molarity of each acid. Pause the video and try it first. Here is a hint: start with the permanganate. Done? Okay, let's check. First, which of the two acids does the permanganate react with? Only the oxalic acid. Permanganate cannot oxidise sulphuric acid, and that is exactly why sulphuric acid is the acid we add to the flask for these titrations. Permanganate used: twenty point zero times zero point zero two gives zero point four millimoles. Okay? Now the two to five ratio, the right way round. Oxalic acid is zero point four times five over two, which gives one point zero millimoles. In twenty five point zero millilitres, one point zero over twenty five point zero gives zero point zero four molar. Right? Next, the alkali. Thirty point zero times zero point one gives three point zero millimoles of sodium hydroxide, for both acids. Oxalic acid gives two H plus ions, so its one point zero millimoles use two times one point zero, which is two point zero millimoles. With me? So three point zero minus two point zero leaves one point zero millimoles of sodium hydroxide for the sulphuric acid. Clear? So the sulphuric acid is one point zero millimoles, which gives zero point zero four molar, right? No, wait, careful, because sulphuric acid also gives two H plus ions per molecule. So one point zero millimoles of sodium hydroxide neutralise half as much, zero point five millimoles of sulphuric acid; the small fraction is one half. Then zero point five over twenty five point zero gives zero point zero two molar. So oxalic acid is zero point zero four molar, and sulphuric acid is zero point zero two molar. Puzzle solved: one titration counted all the H plus ions, the other picked out the oxalic acid. Very good if you got that!

8:05Quick revision: 2 to 5 ratio, pale pink end point, two H+

So, quick revision. Millimoles equal molarity times volume, and permanganate reacts with oxalate in the ratio two to five, with the two under permanganate. Use dilute sulphuric acid, warm to sixty to seventy degrees Celsius, and no indicator: the end point is a permanent pale pink. And oxalic acid and sulphuric acid each give two H plus ions to sodium hydroxide. Now set your timer for twenty five minutes, and try the problems of Set three on this page. All the best!