Titrations and Calculations

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. In a flask, we have twenty five millilitres of an iron two plus solution, and in the burette, potassium permanganate. The iron solution is five times more concentrated than the permanganate. So you would expect to need five times the volume, one hundred and twenty five millilitres of permanganate, right? But the pink end point comes at just twenty five millilitres. So where did that factor of five go? Strange, isn't it? Today we learn the one rule behind every redox titration. These calculations come up often in JEE, so let's begin.

0:39The one working rule

So, what really happens in a redox titration? The oxidant, like permanganate, gains electrons, and the reductant, like iron two plus, loses them. So at the end point, what must be equal? The electrons, yes, exactly! Every electron lost by the reductant is gained by the oxidant. So equivalents of oxidant equal equivalents of reductant. Okay? And here is how we count equivalents. n one times M one times V one, equals n two times M two times V two. The small one written low is for the oxidant, and the small two is for the reductant. Clear? So what is this n? It is the n factor: the number of electrons gained or lost by one formula unit. Molarity alone is not enough; we always multiply by the n factor. With me? Then M is the molarity, in moles per litre, and V is the volume. Millilitres or litres both work, as long as both sides use the same unit. Got it?

1:43n-factors of the two oxidants

First, K M n O four, potassium permanganate. In acid, manganese goes from plus seven to plus two, as M n two plus in brackets; the arrow means goes to, and the raised two plus is the charge. Five electrons are gained, so the n factor is five. Okay? Now let me ask you this. Is the n factor of permanganate always five? No, no, it depends on the medium! In neutral solution, manganese goes from plus seven only to plus four, as M n O two, manganese dioxide. Three electrons, so the n factor is three. Right? And in strong alkali, it goes from plus seven just to plus six, as manganate, M n O four two minus. Only one electron, so the n factor is one. So read the medium first; this is a common trap. Clear? Next, K two C r two O seven, potassium dichromate. It has two chromium atoms, and each one goes from plus six to plus three. Two times three is six, so its n factor is six. With me?

2:53n-factors of the reductants

Now the reductants, the electron givers. Iron two plus goes to iron three plus, losing just one electron, so the n factor is one. Okay? Oxalate, C two O four two minus, goes to carbon dioxide, and each oxalate ion gives two electrons. So the n factor is two. Right? Now a tricky one: F e C two O four, iron two oxalate. What is its n factor? If you said three, very good! Both parts are oxidised: the iron gives one electron, and the oxalate gives two, so one plus two is three. Students often write one or two here, so be careful. Got it? Hydrogen peroxide, H two O two, as a reductant, has an n factor of two. Clear? And last, iodide, I minus, going to iodine, I two, has an n factor of one per ion. Thiosulphate, S two O three two minus, going to tetrathionate, S four O six two minus, also has one. With me?

4:00Mohr's salt: a primary standard for Fe²⁺

For an exact amount of iron two plus, we use Mohr's salt. Its formula is F e S O four, then N H four in brackets taken twice with S O four, then six waters, H two O, okay? The raised dots join the parts that crystallise together in one solid. Its molar mass is three hundred and ninety two grams per mole; the raised minus one means per. And it is a primary standard: we weigh it accurately, and we know exactly how much iron two plus we have. Right? Each formula unit carries one iron, so the moles of iron two plus are just the mass over three hundred and ninety two. Clear?

4:41Example 1: volume of 0.02 M KMnO4 for 25 mL of 0.1 M Fe2+

Now, Example one, and it is our puzzle from the start. What volume of zero point zero two molar K M n O four oxidises twenty five millilitres of zero point one molar F e S O four, iron two sulphate, in acid? Capital M means molar, 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 two times V equals zero point one times twenty five, and get one hundred and twenty five millilitres. Is that right? No, no, wait, they forgot the n factors. Permanganate in acid has an n factor of five, and iron two plus has one. Okay? So, five times zero point zero two times V equals one times zero point one times twenty five. Both volumes are in millilitres, so we are fine. Right? The left side becomes zero point one times V, and the right side is two point five. So V is twenty five millilitres; the double arrow means so. And there is our puzzle solved! Each permanganate ion takes the electrons of five iron two plus ions, so that factor of five cancels out. Clear?

6:00Example 2: percentage of iron in an ore by K2Cr2O7 titration

Now Example two, a harder one. Zero point five grams of an iron ore is dissolved; the small g stands for grams. All the iron is reduced to iron two plus, and it is titrated with thirty millilitres of zero point zero two molar K two C r two O seven. Find the percentage of iron in the ore. Pause the video and try it first. Done? Okay, let's check. First, the n factors: six for dichromate, and one for iron two plus. With me? Now, many students take the moles of iron two plus equal to the moles of dichromate. Is that right? No, and that is exactly the trap! Each dichromate takes six electrons, but each iron gives only one, so we need six iron two plus ions for every dichromate. Moles of iron two plus equals six times zero point zero two times zero point zero three zero, which gives three point six times ten to the power minus three. Okay? Look at the volume. Here we want moles, so it must be in litres: thirty millilitres is zero point zero three zero litres. Use thirty by mistake, and the answer is a thousand times too big. Right? Now let us find the mass. Iron, F e, has a molar mass of fifty six, so three point six times ten to the power minus three, times fifty six, gives zero point two zero one six grams. Clear? Finally, the percentage: zero point two zero one six over zero point five, times one hundred, gives forty point three two percent. So about forty percent of the ore is iron; the wavy equals sign means approximately. Got it?

7:49Quick revision: n1M1V1 = n2M2V2 and the key n-factors

So, quick revision. First, equivalents of oxidant equal equivalents of reductant: n one M one V one equals n two M two V two. Second, permanganate has an n factor of five in acid, three in neutral solution and one in strong alkali, and dichromate has six. And third, the reductants: iron two plus has an n factor of one, oxalate has two, iron two oxalate has three, and hydrogen peroxide has two. Now go and try the two practice problems on this page. You will find them easy. All the best!