Volumetric Analysis

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Imagine a bottle of vinegar. How many grams of acetic acid are there in one litre of it? You cannot pick the acid out of the water and weigh it, right? So how do we find it? With a burette, a pipette and a few drops of indicator. This is volumetric analysis, and titration numericals come up often in JEE, so let's begin.

0:27The set-up for a titration

So, here is the arrangement for a titration. The long marked tube is the burette. It holds the titrant, the solution we run in. We read it at the lower meniscus, the bottom of the curved liquid surface. And the stop-cock at its foot lets us add drop by drop near the end point. Okay? Below it sits the conical flask, with a pipetted volume of the other solution and two to three drops of indicator. The white tile underneath shows the colour change clearly. Right? And one exception on screen: K M n O four, potassium permanganate. It acts as its own indicator, so we add none. And it is so dark that we read the upper meniscus. Clear?

1:10Standard solutions and primary standards

First, the standard solution: a solution whose concentration is accurately known. Okay? But how do we make the first one? From a primary standard: a substance pure, stable and non-hygroscopic enough to be weighed directly. Non-hygroscopic means it does not pick up moisture from the air. With me? The course lists four: oxalic acid dihydrate; anhydrous sodium carbonate, N a two C O three, where anhydrous means without water; Mohr's salt; and potassium dichromate, K two C r two O seven. Got it? Now let me ask you this. Can we simply weigh out sodium hydroxide and make a standard solution? No, we cannot. Solid sodium hydroxide absorbs moisture and carbon dioxide from the air, so it cannot be weighed accurately. So N a O H, H C l and K M n O four are not primary standards; their solutions are standardised by titration. Okay?

2:15Equivalence point and end point

Next, two terms students often mix up. The equivalence point is reached when the reactants have been mixed in exactly the ratio of the balanced equation. Right? And the end point is where the indicator changes colour. So one is about amounts, and the other is the colour change we see. Clear?

2:33The titration formula

Alright, now the formula behind every titration numerical. Say small a moles of capital A react with small b moles of capital B, and the arrow means they give the products. Small a and small b are the numbers in the balanced equation. Okay? Now, many students write MAVA equals MBVB for every titration. Is that always true? It holds only when the ratio is one to one. With molarities, each side goes over its own coefficient: M A times V A, over small a, equals M B times V B, over small b. Capital M is the molarity, capital V the volume, and the little letters below them, the subscripts, name the solution. Got it? For example, H two S O four plus two N a O H, sulphuric acid with sodium hydroxide. One mole of acid needs two moles of base, so small a is one and small b is two. Forget that two, and the acid's molarity comes out double, okay? Or work in normalities, capital N, with no coefficients at all, so NAVA equals NBVB, as simple as that. And normality is molarity times the n-factor. Right? Strength in g L to the power minus one, read as grams per litre, equals molarity times molar mass. Both our examples end with it. Clear?

4:04At the bench: rinsing and concordant readings

Now the bench rules. The burette and the pipette are rinsed with the solutions they are to hold. Rinse the burette with water only, and the leftover water dilutes the titrant, so more of it is run in. Okay? So the conical flask should also be rinsed with the solution that goes into it, right? No, and that is exactly the trap. The conical flask is rinsed with distilled water only. Drops of solution left behind would add to the measured volume, so more titrant would be used than the pipetted amount requires. With me? Next, we repeat the titration until two or three readings agree within zero point one millilitre, and m L on screen means millilitres. These are concordant readings. Right? Here are four burette readings from a course example: twenty one point four, twenty point nine, twenty point eight, and twenty point nine millilitres. Which value do we use? The first one is the rough reading, so it is rejected. The other three agree within zero point one millilitre, and the value used is twenty point nine millilitres. Got it?

5:14Example 1: molarity and strength of NaOH from HCl titration

Now example one, an easy start. Twenty point zero millilitres of a sodium hydroxide solution needs eighteen point four millilitres of zero point one molar hydrochloric acid, where the capital M means molar, moles per litre. Find its molarity and its strength. Pause the video and try it first. Done? Okay, let's check. Sodium hydroxide and hydrochloric acid react one to one. So M times twenty point zero equals eighteen point four times zero point one. Clear? So M is eighteen point four times zero point one, over twenty point zero, which gives zero point zero nine two molar. Now the strength. How do we turn this molarity into grams per litre? Yes, exactly, we multiply by the molar mass, which is forty for sodium hydroxide. Zero point zero nine two times forty gives three point six eight grams per litre. Very simple, right?

6:16Example 2: acetic acid per litre of vinegar, dilution trap

Now a harder one, and it is our vinegar puzzle. Ten millilitres of vinegar is diluted to one hundred millilitres, and twenty five point zero millilitres of the diluted solution needs twenty point five millilitres of zero point one molar sodium hydroxide. Find the mass of acetic acid per litre of vinegar. Pause the video and try it first. Done? Okay, let's check. Millilitres times molarity gives millimoles, the m mol on screen. Twenty point five times zero point one gives two point zero five millimoles of base. Acetic acid reacts with it one to one, so the twenty five millilitres held two point zero five millimoles of acid. Okay? So the molarity is two point zero five over twenty five, which gives zero point zero eight two molar, and then we multiply by sixty, right? That is the trap, so be careful. Zero point zero eight two molar is the diluted solution, not the vinegar. Twenty five millilitres is a quarter of the hundred, so the whole solution holds four times as much, which is eight point two millimoles. Clear? And all of that acid came from just ten millilitres of vinegar. So the vinegar is eight point two over ten, which gives zero point eight two molar. With me? Last, the molar mass of acetic acid is sixty. Zero point eight two times sixty gives forty nine point two grams per litre. And that answers our puzzle for this vinegar. Forget the dilution, and the answer comes out ten times too small, so watch for it, okay?

7:53Quick revision: primary standards, formula, rinsing

So, quick revision. Primary standards are weighed directly, while N a O H, H C l and K M n O four are standardised. MAVA over small a equals MBVB over small b, or NAVA equals NBVB, and strength is molarity times molar mass. Rinse burette and pipette with their own solutions, the flask with water only, and use concordant readings. Now go and try the practice problems on this page. All the best!