Iodometry and Iodimetry

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle from the lab. Suppose we want to know how much copper is in a blue solution. In the burette we have sodium thiosulphate. But the reaction in our titration is between thiosulphate and iodine. Copper is not even part of it! Still, the burette reading will tell us the copper exactly. How is that possible? The answer is a clever go-between: potassium iodide. Iodine titrations come up often in JEE redox problems, so let's begin.

0:33Iodimetry and iodometry: one letter apart

Look at the heading: iodimetry and iodometry. Only one letter is different. So they are just two spellings of the same thing, right? No, they are two completely different jobs! First, iodimetry, the one spelt with an i. Here a reductant is titrated directly with standard iodine, I two, whose concentration is known. So the iodine itself is in the burette. Okay? Which reductants? Thiosulphate, S two O three two minus; sulphite, S O three two minus; and arsenic trioxide, A s two O three. The small numbers written below are subscripts that count atoms, and the two minus written above is the charge on the ion. With me? Now iodometry, the one spelt with an o. This one is indirect. An oxidant is added to excess potassium iodide, K I, and iodine is liberated, that is, set free, in the flask. Clear? Which oxidants? Copper two plus; dichromate, C r two O seven two minus; iodate, I O three minus; chlorine; and hydrogen peroxide. Right? Then the liberated iodine is titrated with standard thiosulphate from the burette. So, in iodimetry, the iodine comes from the burette. In iodometry, the iodine is made in the flask by the oxidant. That is the whole difference. Got it?

1:52Thiosulphate meets iodine; starch shows the end

Now the titration reaction itself. I two, plus two S two O three two minus, gives two I minus, plus S four O six two minus. The arrow means gives. So the iodine goes back to iodide, okay? Look at the numbers in front. How many thiosulphate ions does one iodine molecule need? Two, exactly! One I two needs two thiosulphate ions. Keep this two to one in mind; both examples today use it. Right? And the indicator? Starch. Starch gives a blue colour with iodine. So when the last of the iodine is used up, the blue goes away, and that is our end point. Clear? But tell me, why not add the starch right at the start, like most indicators? Here is the reason. With a lot of iodine present, starch forms a stable blue complex that gives the iodine back only slowly. So the end point would come late, and it would not be sharp. That is why starch is added only near the end point. This reason comes up often in theory questions. With me?

2:58Key reactions: KI is only a go-between

Now two key reactions. First, copper. Two copper two plus, plus four iodide, gives C u two I two, plus I two. The C u two I two comes down as a white precipitate, and the iodine is set free. Okay? So two copper ions make one iodine molecule. And that one iodine molecule needs how many thiosulphate ions? Two, yes! So we write a chain: two copper two plus is equivalent to one I two, which is equivalent to two thiosulphate. The sign with three lines means is equivalent to. Drop the middle, and copper to thiosulphate is simply one to one. Right? Next, iodate, I O three minus, with acid. One iodate, plus five iodide, plus six H plus, gives three I two, plus three H two O, okay? So one iodate makes three iodine molecules. Then how many thiosulphate ions does one iodate finally need? Six! Three iodine molecules, two thiosulphate each. One iodate, three I two, six thiosulphate. Writing three here, instead of six, is a common slip. Careful, alright? Now see the big picture. Iodide gives electrons to the oxidant and becomes iodine. Then thiosulphate gives electrons to that iodine, and it becomes iodide again. So the iodide that became iodine is given back as iodide. In the end, whose electrons were really used up? The thiosulphate's, exactly! So the equivalents of oxidant equal the equivalents of thiosulphate used. That is why potassium iodide is only a go-between. Skip it in your calculation, and link the oxidant straight to the burette. Got it?

4:41Example 1: molarity of Cu2+ from a thiosulphate titration

Now, remember our copper puzzle? Let's solve it. Twenty five point zero millilitres of a copper two solution, the Roman numeral two in brackets, is treated with excess potassium iodide. The liberated iodine needs twenty point zero millilitres of point zero five zero molar sodium thiosulphate, N a two S two O three. Find the molarity of copper two plus. Here m L means millilitres, and capital M means molar. Pause the video, and try it. Done? Okay, let's check. Millilitres times molarity gives millimoles, written m m o l. Thiosulphate used: twenty point zero times point zero five zero, which is one point zero millimoles. Okay? Now, a quick shortcut many students take: one copper gives one iodine, and one iodine takes two thiosulphate, so copper is half, point five millimoles. Correct? No, no, careful! It takes two copper ions to make one I two. So two copper two plus, one I two, two thiosulphate: copper to thiosulphate is one to one. So copper two plus is also one point zero millimoles. Right? Now divide by the volume of the copper solution. The horizontal bar means divided by. One point zero millimoles over twenty five millilitres: point zero four zero molar. And there is our puzzle, solved! Copper never appeared in the titration reaction, but through the iodine, the burette counted every copper ion. Clear?

6:14Example 2: available chlorine in bleaching powder, 28.4%

Now a harder one. Two point five zero grams of bleaching powder is made up to two hundred and fifty millilitres. Twenty five millilitres of this, with excess potassium iodide and acetic acid, liberates iodine that needs twenty point zero millilitres of point one zero zero molar thiosulphate. Find the available chlorine, that is, the mass of chlorine the sample can set free, as a percentage of the sample. Pause and try. Hint: watch the volumes. Done? Okay, let's check. Chlorine is the oxidant here. One C l two sets free one I two, which needs two thiosulphate. So the chain is C l two, I two, two thiosulphate. With me? Thiosulphate used: twenty point zero times point one zero zero, that is two point zero zero millimoles. Okay? Half of that is chlorine: one point zero zero millimoles, in the twenty five millilitres we titrated. Clear? So, one millimole of chlorine is point zero seven one grams, and point zero seven one over two point five is two point eight four percent. Done, right? Oh no, careful! We titrated only twenty five millilitres out of two hundred and fifty. That is one tenth. So the whole sample has ten point zero millimoles of chlorine. With molar mass, capital M, equal to seventy one, that is point seven one zero grams. Okay? Available chlorine: point seven one zero over two point five zero, times one hundred. Twenty eight point four percent. The answer that forgets the dilution is ten times too small. So always ask: what fraction of the solution did we titrate? Got it?

7:54Quick revision: iodimetry vs iodometry, starch, role of KI

So, quick revision. Iodimetry: a reductant is titrated directly with iodine from the burette. Iodometry: the oxidant liberates iodine from excess potassium iodide. One I two needs two thiosulphate, and starch goes in only near the end point. And the equivalents of oxidant equal the equivalents of thiosulphate; potassium iodide is only a go-between. Now go and try the practice problems on this page. You are ready for them. All the best!