Acid–Base Titrations and Back Titration

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Your teacher hands you a piece of limestone and says: find out how much calcium carbonate is inside it, by titration. But limestone is a carbonate. It does not dissolve in water, and even with acid it reacts only slowly. You cannot just put it in a flask and titrate it drop by drop. So how will you titrate a rock that will not even dissolve? Hmm, sounds impossible, right? Today we will learn a very smart trick for exactly this, called back titration. And look, in JEE, purity questions on carbonates and ammonium salts keep coming, and almost every one of them is built on this one idea. So, let's begin.

0:46Direct titration: equal equivalents

So, first, the normal titration. Acid on one side, base on the other, and we add drop by drop until the end point. Now tell me: at the end point, what exactly is equal? The volumes? The moles? Yes, the equivalents! Remember the law of equivalence: substances always react in equal numbers of equivalents. So at the end point, the milliequivalents of acid equal the milliequivalents of base. The m e q you see on screen is just short for milliequivalents. Not moles, equivalents. Clear? And how do we count milliequivalents? Normality times volume in millilitres. Capital N, the normality, times V in millilitres, gives the milliequivalents directly, okay? Now a topper's trick: stay in milliequivalents all the way. Milliequivalents times the equivalent mass, the capital E on screen, gives the mass directly in milligrams. No litres, no grams, no conversion in the middle. Really simple. And most of these questions end with purity. Purity is the mass of the pure substance, divided by the mass of the whole sample, times hundred percent. Here the impurities do not react with the reagent, so only the pure part shows up in the titration, right?

2:07Back titration: add excess, titrate the rest

Alright, now back to our limestone. Some solids simply do not cooperate. Like a carbonate, or an ammonium salt that releases ammonia. They are insoluble, or they react only slowly, so a direct titration, drop by drop, simply does not work. So here is the real question. We want to know how much acid the limestone uses up. But we cannot watch it happen, drop by drop. Then how? Think of a shop. You pay with a big note, and the shopkeeper gives you change. How much did you spend? The note minus the change, right? Back titration works exactly like this. Step one: we add a measured excess of a standard reagent, more than enough for the whole solid. That is our big note. Step two: the reaction is over, and some reagent is still left. We titrate this leftover with another standard solution. That is the change we got back, okay? Now many students do this. The second titration gives a nice clean reading, so that reading is the sample, right? No, no, careful! This is the trap. The second titration only measures what was left over. It tells you the change, not what you spent. So, milliequivalents of the sample equal milliequivalents of reagent added, minus milliequivalents of reagent left. Added minus left. That is the whole idea. Write it down in your notebook. Got it?

3:36Example 1: purity of limestone by back titration

Now, let's finish our limestone puzzle. Example one. One point two five grams of impure limestone is dissolved in fifty millilitres of one normal hydrochloric acid. The excess acid needs thirty millilitres of one normal sodium hydroxide. Find the percentage of calcium carbonate. Pause the video, and try it first. Done? Okay, let's check. First, the big note. How much acid did we add? One normal times fifty millilitres: fifty milliequivalents of hydrochloric acid. Now the change. The leftover acid needed thirty millilitres of one normal sodium hydroxide. One times thirty is thirty milliequivalents of base, so thirty milliequivalents of acid were left. Clear? So the acid used by the calcium carbonate is fifty minus thirty: twenty milliequivalents. Just like the shop: paid fifty, got thirty back, spent twenty! Now our topper's trick: milliequivalents times equivalent mass gives milligrams. The equivalent mass of calcium carbonate is fifty: molar mass one hundred, divided by two. Twenty times fifty: one thousand milligrams, which is one point zero zero gram. And the purity: one point zero zero divided by one point two five, times hundred. Eighty percent. So our limestone is eighty percent calcium carbonate. Puzzle solved!

5:06Example 2: purity of ammonium sulphate via ammonia

Now a harder one, at JEE level. Read it with me. Zero point five zero gram of ammonium sulphate is boiled with excess sodium hydroxide. The ammonia is absorbed in fifty millilitres of zero point two normal sulphuric acid. And the excess acid needs twenty five millilitres of zero point one normal sodium hydroxide. Find the purity of the ammonium sulphate. Pause and try. Hint: follow the ammonia. Done? Okay, let's check. Notice, there is sodium hydroxide in two places here. Do we need the sodium hydroxide used for boiling? No! It was added in excess, only to release the ammonia, and it never reaches the acid. Only the ammonia goes into the acid. So the acid is our big note, and the ammonia spends it. Now, the acid taken. Sulphuric acid is dibasic, so we multiply by two, right? No, no, careful! This is a classic trap. The question gives normality, not molarity, and normality already includes the n factor. So don't double it again. So, zero point two normal times fifty millilitres: ten milliequivalents of acid taken. Now the change: zero point one times twenty five, that is two point five milliequivalents of acid left. So the ammonia used ten minus two point five: seven point five milliequivalents. And all this ammonia came from the ammonium sulphate, so the sample is also seven point five milliequivalents. With me? The equivalent mass of ammonium sulphate is sixty six: molar mass one hundred thirty two, divided by two. Seven point five times sixty six: four hundred ninety five milligrams. And the sample was zero point five zero gram, that is five hundred milligrams. Four hundred ninety five out of five hundred: ninety nine percent pure. See? Taken minus left, one multiplication, and a hard looking question is done. Really easy.

7:12Quick revision: meq of sample = meq added - meq left

So, quick revision. Direct titration: milliequivalents of acid equal milliequivalents of base. Back titration: milliequivalents of the sample are milliequivalents added, minus milliequivalents left. Then milliequivalents times the equivalent mass gives milligrams, and from there, the purity. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!