Part B: Harder Problems

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Suppose a student dissolves two point eight grams of iron in dilute sulphuric acid and turns it into Mohr's salt. The crystals weigh sixteen point seven grams, almost six times the mass of the iron. So where did all that extra mass come from? Strange, isn't it? This is Part B of our problem bank: harder problems, with many steps, but each step is simple if you follow the moles. Problems like these come up often in JEE practical chemistry, so let's begin.

0:37Preparations: follow the moles

So, first, the preparations. Here is Mohr's salt, in three parts joined by raised dots: ferrous sulphate, then ammonium sulphate, with the ammonium group in brackets, then six molecules of water. The small numbers below are subscripts. And capital M is the molar mass, three hundred and ninety two. Okay? Now let me ask you this. Count the iron atoms in the formula. One mole of iron can give, at most, how many moles of Mohr's salt? Yes, exactly, one mole of iron gives at most one mole of the salt. There is only one iron, F e, in each formula unit; the arrow on screen means gives, and m o l is short for mole. With me? That at most amount has a name, the theoretical yield. And it is always calculated from the limiting reagent, the one that runs out first. Clear? But in the lab we collect less. So the percentage yield is the actual yield over the theoretical yield, times one hundred; the long bar is a fraction bar, and the cross means times. Remember this for example one. Got it?

1:46Two end points: carbonate with bicarbonate

Now the second kind of problem: carbonates titrated with hydrochloric acid. Sodium carbonate is neutralised in two stages. In stage one, the carbonate ion, C O three with a two minus charge up top, takes one H plus, and becomes the bicarbonate ion, HCO three minus. This is near p H eight point three, shown in the brackets; the wavy equals sign means approximately. Right? In stage two, the bicarbonate ion takes a second H plus, and gives water and carbon dioxide. This is complete near p H four. Okay? Phenolphthalein turns colourless at the end of stage one. And V one, a capital V with a small one below it, is the volume of acid up to that end point. With me? Then methyl orange changes at the end of stage two, the complete neutralisation. V two is the further volume: only the extra acid added between the first endpoint and the second, not the whole volume measured from the very start. Is that clear? Now take a mixture of sodium carbonate and sodium bicarbonate. In stage one, only the carbonate reacts, because the bicarbonate waits for stage two. So which volume tells you the carbonate? Yes, exactly, the carbonate corresponds to V one; that three-line equals sign means corresponds to. In stage two, the bicarbonate made from that carbonate needs V one again, and the bicarbonate that was there from the start uses the rest. So the bicarbonate corresponds to V two minus V one. Got it?

3:22Example 1: Mohr's salt from 2.8 g iron at 85% yield

Alright, example one, our puzzle from the start. Two point eight grams of iron, where the small g on screen means grams, is dissolved in dilute sulphuric acid, and the ferrous sulphate is converted into Mohr's salt. The overall yield is eighty five percent. Find the mass of Mohr's salt. Pause the video and try it first; the atomic mass of iron is fifty six. Done? Okay, let's check. Moles of iron: two point eight over fifty six gives zero point zero five zero moles. And each iron atom ends up in one formula unit of Mohr's salt. Okay? So we multiply by the molar mass, three hundred and ninety two. Zero point zero five zero times three hundred and ninety two gives nineteen point six grams. So that is our answer, right? No, careful, that is the trap, because nineteen point six grams is only the theoretical yield, the most we could ever get. The question says the overall yield is eighty five percent, so this is not the answer. Clear? We collect only eighty five hundredths of the theoretical yield. Nineteen point six times zero point eight five gives sixteen point six six, which is about sixteen point seven grams. With me? And our puzzle is solved. Iron is only fifty six out of three hundred and ninety two, one seventh of Mohr's salt; the sulphates and the six waters bring in the rest. So the mass grows seven times, and eighty five percent of seven is almost six. Got it?

4:58Example 2: Na2CO3 and NaHCO3 percentages by two indicators

Now a harder one. One point zero zero grams of a mixture of sodium carbonate and sodium bicarbonate is dissolved to make two hundred and fifty millilitres; m L on screen means millilitres. Twenty five point zero millilitres of it needs five point zero millilitres of zero point one molar hydrochloric acid with phenolphthalein; here the capital M after the number means molar. Then it needs a further ten point five millilitres with methyl orange. Find the percentage of each. Pause the video and try it first. Done? Okay, let's check. For the twenty five millilitres we titrated, V one is five point zero millilitres, and V two, the further volume, is ten point five millilitres. Okay? Carbonate corresponds to V one. Millilitres times molarity gives millimoles, the m mol on screen. Five point zero times zero point one gives zero point five zero millimoles, so there are zero point five zero millimoles of sodium carbonate. With me? Now let us find the bicarbonate. Methyl orange needed ten point five millilitres, and ten point five times zero point one gives one point zero five millimoles. So the bicarbonate is one point zero five millimoles, right? No, that is exactly the trap, so be careful here. Part of that acid went to the bicarbonate made from our carbonate, and that part is again zero point five zero millimoles. So the bicarbonate corresponds to V two minus V one, and one point zero five minus zero point five zero gives zero point five five millimoles. Clear? Now here is the step students forget. These millimoles are only in the twenty five millilitres we titrated, but the whole one gram sits in two hundred and fifty millilitres. So what should we do now? Yes, exactly, multiply by ten, because two hundred and fifty over twenty five is ten. So the whole sample has five point zero millimoles of carbonate and five point five millimoles of bicarbonate. Topper's trick: work with the portion you titrated, and scale up only once, at the end. Right? Now we change to grams. The molar mass of sodium carbonate is one hundred and six, so five point zero times one hundred and six gives five hundred and thirty milligrams, the m g on screen. That is zero point five three zero grams out of one gram, and the double arrow means so, fifty three percent. Okay? And sodium bicarbonate has molar mass eighty four. Five point five times eighty four gives four hundred and sixty two milligrams, which is zero point four six two grams, about forty six percent. So the mixture is fifty three percent sodium carbonate and forty six percent sodium bicarbonate. Got it?

7:46Quick revision: yield, V1 and V2 minus V1, then scale up

So, quick revision. For preparations, find the moles of the limiting reagent, get the theoretical yield, then multiply by the yield. For carbonate with bicarbonate, carbonate corresponds to V one, and bicarbonate to V two minus V one. And when you titrate a portion, find the millimoles, then scale up to the whole solution. Now try the other Part B problems on this page, each one as a test before you read its solution. You worked through two really hard problems today, so well done, and all the best!