Making or preparing a standard solution is one of the required practicals detailed in your specification.

But it really comes in two parts:

The calculation—how do you actually know what mass of solid you need to make your standard solution in the first place?

The practical details—how do you actually go about making it?

We’ll start with an example.

We need an example to work from here.

Let’s say we need to make 250 cm³ of a 0.20 mol/dm³ solution of sodium hydroxide (aqueous).

Of course, we're using solid sodium hydroxide to do this.

So, we’re trying to make up a solution. We’ve got solid sodium hydroxide and some water.

How do we actually go about making 250 cm³ of a 0.20 mol/dm³ concentration?

Step 1: The Calculation

We need to know how much sodium hydroxide we need.

First, we find out how many moles of sodium hydroxide are required to make that volume at that concentration.

The number of moles is found using:

Moles = Concentration × Volume

We know the concentration and the volume:

0.20 × 250 ÷ 1000 (because we need the volume in dm³)

That gives us 0.05 moles.

Of course, we can’t measure 0.05 moles of sodium hydroxide directly, so we need to convert it to mass.

Mass is found using:

Mass = Moles × Molar Mass

The molar mass of sodium hydroxide (NaOH) is 40 g/mol, so:

0.05 × 40 = 2.0 g

So, we need 2.00 g of sodium hydroxide.

Now, we measure that out and move on to making the solution.

Step 2: The Practical Method

These are the steps we need to follow, and they are the key points to write down in an exam when asked to describe how to do this.

Weigh out 2.00 g of solid sodium hydroxide using a top pan balance.

This is a crucial point to include in your method.

Dissolve the solid sodium hydroxide in a minimum volume of distilled water in a beaker.

A glass stirring rod is used to mix, not a metal spatula (to minimize contamination).

Why do we use a minimum volume?

We don’t want to dilute it too much.

Since we are making only 250 cm³ of solution, we should use about 50 cm³ maximum to dissolve the sodium hydroxide initially.

If it doesn’t dissolve, we add a little more water while stirring until fully dissolved.

Transfer the solution into a rinsed 250 cm³ volumetric flask using a funnel.

The funnel is important—you get marks for stating that it prevents spillages.

If any solution is spilled, you must start again because you won’t know how much has been lost.

Rinse the beaker with a small amount of distilled water and add that to the volumetric flask.

This ensures no sodium hydroxide is left behind in the beaker.

Some people rinse twice to minimize losses.

Make up to the mark with distilled water.

The meniscus must be at eye level and aligned with the mark on the volumetric flask.

This is a key mark in the exam.

Invert and shake the volumetric flask to mix.

This ensures the solution is homogeneous (evenly mixed).

If the solution isn’t mixed properly, different samples could have different concentrations, leading to inaccurate results in titrations.

Key Exam Points

Calculations should be straightforward—if you know the concentration and volume, you can find the required mass.

In the exam, if asked to outline the method, make sure to include:

Type of equipment used (top pan balance, volumetric flask, funnel, glass stirring rod).

Use of distilled water (prevents contamination from other ions).

Minimizing losses (rinsing the beaker, using a funnel).

Measuring accurately (reading the meniscus at eye level).

Ensuring homogeneity (shaking the flask).

If you follow this method and include these key details, you will get full marks when asked to describe how to make a standard solution.

This tutorial’s all about setting up a titration.

Now, I'll be honest with you.

I've never seen an exam question that asks you how do you set up a titration?

But the thing is, when setting up a titration, this can be make or break for the success of it because within the setup of a titration there's going to be a number of things that could cause errors further down the line.

So you do get quite a few questions based on titrations like, "Oh, how come, you know, what effect would this have on the mean titre? What effect would this have if the student messed this up on the mean titre?" and so on and so forth.

So really, that's what this is about.

It's about how we set it up and looking out for the things that could cause errors further down the line.

First thing you want to do before setting up the titration is rinse the burette.

Okay.

Now, your burette you're gonna take off the shelf, you don't know what's been in it before, so it's really important that you rinse it firstly with distilled water.

That gets rid of any soluble impurities in there.

Okay, so distilled water, you need to rinse that right through, open the tap, get it right through there.

Secondly, you then need to rinse it with the solution that you're going to fill it with.

So you put maybe, I don't know, like a third full, a quarter to a third full with the solution that you're putting in it, and then remove that solution as well.

We need to rinse it with the solution we're putting in there.

Why is that so important?

Well, rinsing your burette with the solution you're going to be filling it with removes any excess water because you have just rinsed it with water and when you empty it, there's going to be some water hanging around in there.

Excess water in there will go ahead and dilute your solution.

So if you fill it with your acid or alkali straight after rinsing with water, some of that water, or that water that's left in there, will dilute your solution when you put it into the burette.

And if you dilute your solution, that's going to increase your mean titre because if the concentration is less than you believe it to be—if you put, I don't know, 0.1 molar HCl in there and there's some water when you fill it in the burette—then it's going to end up maybe as 0.09 moles per decimeter cubed HCl.

Who knows, because we don't know how much water is in there.

So if you rinse with the solution, that means you avoid this error.

Okay, excess water in the burette will dilute the solution and increase the value of your mean titre.

Once we've rinsed it with water and then our solution, we can go ahead and fill the burette proper.

So if we fill a burette, we need to add the solution using a funnel while it's set in the clamp and run that solution through the tap at the bottom.

And that is to remove any air bubbles that you might have in the bottom of the burette.

And again, this is another source of error.

If you have air bubbles stuck in the bottom, if you filled your burette with your solution, let's say acid or alkali, and you don't do this, then any air bubble stuck in the bottom right here underneath the tap will cause your first titration value to be larger than expected.

So let's say it would normally require 25 cm³ of acid to neutralize the alkali in the bottom.

If you've got an air bubble in the bottom, you open the tap, well, the first two or three cm³ that is measured in your burette is just gonna be fresh air.

There's nothing actually coming out the bottom of the burette, but the value on the burette will go down, it will change.

So we need to get rid of those air bubbles.

And if you have air bubbles in there, like I said, that first titration will be larger than expected because, like I said, you'll be letting out two or three cm³ of fresh air.

So make sure you do that into a waste beaker.

Don't put it into a conical flask.

Just, you know, run it through the tap and make sure that it's full of solution into a waste beaker, and then you can go ahead and fill your burette to the 0.00 cm³ line, which is, of course, at the top.

You've got to remember with a burette, all right, we're not measuring how much is in there, but we're measuring how much is released out of the bottom.

So it's slightly inverted—zero’s at the top, 50 is at the bottom.

So we refill that to the 0.00 line.

Of course, you don't necessarily need to start at 0.00 because your titration is going to be somewhere between 0 and 50, and you always record the starting value.

But your first one, I would recommend starting on the 0.00 because you don't know how much you're going to be needing in your first titration at this point.

Once you've filled your burette, what's really important, and a lot of people forget to do this, is remove the funnel from the top.

Because using a little funnel to fill your burette—if you leave that funnel in there and there's some drops of solution in it, and you go ahead and perform your titration, as you're kind of knocking the burette around a little bit, drops of solution could decrease your reading and therefore decrease your titre volume.

So let's say you did let out 25 cm³ of acid.

Okay, and then you go ahead and you go back and read your burette.

Maybe two or three drops of that solution that were stuck in the funnel before have trickled their way down into the burette and caused the, you know, the line in the burette to go up.

Of course, that's going to affect your reading, isn't it?

Okay, so you could have let out 25 cm³, but actually, the reading is going to be something like, I don't know, 24.7 or something like that.

So we know from practical situations in titrations, one drop can make all the difference.

So we need to remove that funnel.

If that’s left in, drops would decrease your titre volume.

Okay, really important—decrease your titre volume.

I already mentioned this point in making the standard solution, but it’s worth mentioning again: take that meniscus into account.

We must read the bottom of the meniscus and the bottom of the meniscus at our value.

For example, once we fill it to 0.00 cm³, we need to make sure the bottom of the meniscus is on that line.

So we always, always read that at eye level as well.

Not always easiest when it’s on a bench in the lab, but you know what? You just, you know, stand on the little bar on your chair or something, or even, you know, just bring that burette down in the clamp so you’ve got it at eye level.

So it’s really important in terms of accuracy that you read from that meniscus.

Now, after all that, that’s the obvious sorted.

All right.

Now, what about our conical flask?

Well, we do need to rinse out that conical flask as well before we put our other solution in there.

Let's say our alkali.

So we can rinse that with distilled water.

Give it a good rinse around, right around the inside.

Make sure you get all the walls of the inside of that conical flask.

This will not affect your final results.

You do not have to remove any excess water from that, okay?

Because in that conical flask, that's where your reaction is happening.

Any excess water in there isn't going to affect the number of moles of alkali that you put in there.

It's not the concentration of your alkali that's important that you're putting in there.

It's the number of moles.

So don’t worry about leaving excess distilled water in there before you start your titration, as long as it is just water, okay?

So water removes any soluble impurities from there.

Get rid of those down the sink, give it a good shake out, and then you can go ahead and add your solution.

To add our solution, we need to use a pipette and a pipette filler.

Now, you know what, there's no way I can teach you how to use one of those in a video like this, and it's always fun and games when you're first learning how to do it.

But before we do add our solution, we need to rinse out the pipette with distilled water and then the solution that we're going to be filling it with.

So, for exactly the same reasons as our burette here, okay?

We need to rinse with water and then our solution.

Like I said, for exactly the same reasons.

However, there is a difference in the way it affects your mean titre.

If you don't rinse with the solution, any distilled water left in that pipette—that glass pipette—will dilute your solution that you fill it with.

Okay?

And if you dilute that solution and then you measure out 25 cm³ and you put that into your conical flask, if it's more dilute, then it's going to decrease your mean titre.

Okay?

Because you won’t have as many moles in that conical flask as you think you do.

So it has the opposite effect.

If you look in the top right, it has the opposite effect of the same thing with your burette.

So if you have excess water in your burette, that's going to increase your mean titre.

If you have excess water in your pipette when you first fill it after rinsing with water, then that's going to decrease your mean titre, because you've got fewer moles in the conical flask, and therefore you're going to need fewer moles out of your burette to neutralize them.

So you can see, like I said, setting up a titration—you’re not going to get a question based on that.

I've never ever seen one in any specification, for any exam board, in any kind of year that I've been doing this, okay?

But when setting up these things in red here, these are all things that will affect your result, okay?

And we do get questions on those.

Okay, so, "Steve didn’t rinse out the burette with solution. What effect will that have on his mean titre?" There’s your answer there, okay?

"Steve didn’t read the meniscus at eye level. What effect will that have on the mean titre?" Well, it could be more or less, couldn’t it?

You know, it's accuracy that's important here, okay?

"Okay, you didn’t remove the funnel. What effect may that have had?" So on and so forth.

So that's why these are really, really important.

So, know how to set up your titration accurately and avoid those errors that could creep in later on during your titration.

This tutorial's all about performing a titration, and like the previous tutorial on setting up the titration, we're focusing on some of the errors that can creep in and the usual kind of procedures that we need to follow in order to maintain accuracy and minimize those errors.

Okay, so we're going to start with general technique.

When it comes to technique, it's pretty straightforward. However, it does take a little bit of practice to get right. So first thing—you open the tap with one hand and you swirl the conical flask with the other. Now, I would always recommend swirling that conical flask with your dominant hand because if you're anything like me—of course, I'm right-handed, as you've probably noticed right throughout these videos—if I use my left hand to do that, it just doesn't know what it's doing. It's shaking around and spilling it all over the place. I'm absolutely garbage at doing it with my other hand.

So use your non-dominant hand with the tap and your dominant hand to swirl. That's my recommendation, anyway.

Swirling the conical flask—well, this is important because it ensures a quick reaction, and it makes sure that your mixture is homogeneous. Why is that important? Well, you want to make sure that, you know, when all of that alkali in the conical flask has actually reacted and your indicator changes color.

So swirling is really important—not shaking—because we don’t want it to splash up on the inside of the conical flask. You need to swirl it gently, and that’s it.

Now we know how to do it. The first titration that we do is our rough titre. It's always a rough titre that you need to do first.

Big things here: it's not accurate, so you can just open that tap—swirl, swirl, swirl, swirl, swirl—and as soon as it changes color, you stop it. So, you know, we're not doing it drop by drop or anything like that. It's just not accurate at all.

Why do we do it? Well, it gives an idea of the approximate titre value so the more accurate titres can be performed later. So it just gives us an indication—"Oh, it's gonna be about 23 cm³" or "it's gonna be about 18 cm³"—so that's why it's a rough one. It gives us an idea so later on, we can slow it down as we approach that value and add it drop by drop. Because if we start adding drop by drop from the beginning, that's gonna take us forever, isn’t it? So it just helps us speed up the process a little bit.

The last thing I'll say about this is that these rough titres—even if it ends up being concordant, which I'll talk about in a second—never use it to calculate your mean titre. Never, ever, ever. Even if, just by some miracle, it ends up just the same as the rest of your accurate titrations, you still do not use it.

Okay, so that’s your rough titre. That’s the first one that you're going to do.

So, what about recording? How do we go about recording our titrations? When it comes to recording, there's a universal method for drawing the table that we need to fill out as we perform a titration.

So this table has three rows.

Now, your first row—that's the volume at the end of your titration. Seems nonsensical, but that's the volume after you've done your titration and your indicator’s changed color. That’s the volume that you write down there.

The middle is the volume at the start, so you write that down before you actually start the titration.

And at the bottom, we've got volume used.

Now, it seems like a nonsensical way of writing it down, doesn't it? You'd think, "Oh, right, my volume at the start, then the volume at the end, and then, you know, that’s the total volume used." But actually, this way, it makes the calculation a little bit easier.

If we started with a volume of 0.00, we start that in the middle. Okay? So that's our middle row, if you like—0.00. The volume at the end, after we've done our titration, ended up at 22.25. Once that’s written, you can see how the top number minus that middle number equals the bottom number.

That’s why we do it this way around. Okay? So 22.25 minus 0 is, of course, 22.25.

If you started on 0.10 and, you know, you put that in the middle, then you put the volume at the end at the top—again, the top number minus the middle number equals that bottom number.

So just putting some more numbers in here for our first accurate titration so you can see how that system works: 22.35 minus 0.1 equals 22.25. Okay?

So the volume used equals the volume at the end minus the volume at the start, because we don’t always have to start at zero.

These, really importantly, are accurate to 0.05 cm³. We cannot be any more accurate than that. Usually, they end up just to the nearest 0.1, to be honest with you, but we can write them down to the nearest 0.05. So that’s how we go ahead and record our data.

Now, I’ve done an accurate titre there—titre number one is our first accurate one. So, how do we go about making sure that we maintain accuracy? After we've done our rough one, we know whereabouts it's going to sit. As we approach that end volume there using our rough as a guide, how do we increase the accuracy of our titration?

First thing I would say, as I just mentioned, you add drop by drop as you approach that rough titre volume.

So if your rough titre was 22.25, when you get to maybe 21, okay, I would say start adding it drop by drop and swirl.

And there's a technique involved with that when you're actually doing this. It’s quite difficult to get used to how the tap works and stuff, but in an exam situation, just say, "Add drop by drop near the endpoint." Okay?

Next point—swirl that conical flask continuously. The reason, as I mentioned in the top left-hand corner here, is that it ensures a quick reaction and a homogeneous mixture. So that’s going to increase accuracy as well.

Another tip is to rinse the inside of that conical flask as you get near your endpoint.

Now, what this does is, as you've been swirling, maybe some of that solution has splashed up towards the inside of the conical flask near the neck at the top. Of course, those molecules aren’t going to be involved in the reaction that you want to happen, and this decreases accuracy.

So get that conical flask, get your water bottle, and squirt some distilled water—not loads, but just the minimum amount to make sure you’ve rinsed down any splashes from the inside of that conical flask and make sure they go into the reaction mixture.

Read the meniscus at eye level. Once you’ve got that endpoint—so you’ve got that really pale pink with, let’s say, phenolphthalein in your titration—then read that meniscus at eye level.

So you can either move that burette, bring it down, move the conical flask from underneath it, and make sure your eye is level. However you do it, just make sure you do, because that increases accuracy.

Once you get to the end, you're going to want to calculate your mean titre.

So, how do we choose our titres? Identify, with an asterisk, which ones you'd use to calculate your mean titre. You need to be able to select the right ones, and these are the criteria you use to select your correct titres.

So, that’s how you go ahead and perform a titration.

Those are the headlines, folks.

Let's say you performed your titration, you've got a load of information, and what you need to do is actually find out what the question's asking.

Now, in a titration, you've got to remember, you've always got two things.

First thing, you've got one solution that is known. Now, what I mean by that is, you know the concentration of it, you know the identity of it. That tends to be what we put in our burette.

We also have another solution, so one solution is unknown. Maybe we don't know the concentration of that, or we don't know the identity of that, and that tends to go into our conical flask.

That rule isn't hard and fast when it comes to where those things are, but generally speaking, the known one—that we know the concentration and identity of—goes in the burette. The unknown one tends to be in our conical flask. But like I said, that's not hard and fast, it just tends to be the usual setup that we have.

So we've got a load of mean titres of our known substance, okay? And we need to find out about our unknown substance that's in our conical flask.

Now, whatever they ask you to do—because there's a number of things it can ask you to identify—it's a three-step process.

I'm doing this in red because I think it's so, so important for you guys to remember in terms of clarity when looking at exam questions.

So this is a three-step process.

Step 1: Calculate the number of moles in the mean titre.

Once we've got our mean titre—after we've calculated the average of three concordant ones—we can find the number of moles of the known substance in that mean titration.

So let's say, for example, an acid. Now, to do that, it's always number of moles = concentration × volume.

We know the concentration because it's our known solution, and we know the volume because that's what we've just got using our titrations to calculate the mean.

So find the number of moles in your mean titre.

Step 2: Stoichiometry.

Now, this is looking at the reaction itself.

So my advice—if it's not given to you already—write a balanced equation to check the ratio between your acid and your base, or acid and your alkali.
Once you know that ratio, you can find the number of moles in 25 cm³ of your unknown solution.

Why do I say 25 cm³? Because that's always the standard volume that goes into our conical flask from the pipette.

So if it's a 1:1 ratio, and we've calculated that there are 0.00125 moles in our mean titre of acid, then there are going to be 0.00125 moles of our unknown solution in the conical flask.

Second step here—use that number of moles and apply stoichiometry to find the number of moles of your unknown substance in 25 cm³.

And that 25 cm³, like I say, is the standard volume that we put into the bottom of that conical flask for the reaction.

Step 3: Convert the number of moles of your unknown substance to the desired factor.

There are a number of things they could ask you to find out in terms of this acid-base titration.

Finding Concentration

If they ask you to find the concentration of that unknown solution in your conical flask, you take that number of moles and divide it by the volume.
That volume is usually 25 cm³, as I say, that's usually the portion that goes into that conical flask at the bottom.

Even if it comes from a larger stock solution, you could scale it all up, but you’re still going to get the same concentration.

So if you're asked to find the concentration, it's simply:

Number of moles ÷ Volume (in dm³).

Finding Mass

Maybe they're asking you to find a mass.
Now, once you've got that number of moles from Step 2, you can simply multiply that by the molar mass using your periodic table.

So if you know what it is and they want you to find out what mass of substance is in that conical flask, then it’s:

Number of moles × Molar mass.

And like I said, you've always got access to your molar mass using your periodic table.

Finding Molar Mass

Lastly, they could ask you to find the molar mass of the alkali—let’s say that’s present in the conical flask.
In which case, the formula is:

Molar mass = Mass ÷ Number of moles.

Somewhere in the question, they will have given you the mass of that unknown substance that was in the solution.

Stock Solution Considerations

Be wary of the last two calculations because they might be asking:
What mass was dissolved in an original stock solution?
What’s the molar mass, given the mass dissolved in a stock solution?

You've got to remember that 25 cm³ is the volume in your conical flask, and it relates to your mean titre.

So if you're asked to find the mass dissolved in an original stock solution of 250 cm³, you can find the mass in 25 cm³, but of course, you need to multiply that by 10—because you’re only using a portion of that stock solution.

Just be wary that you’re using only a portion of an original stock solution.

Key Takeaway: Three-Step Process

Like I said, this all boils down to three steps:
Calculate the number of moles of your known substance in your mean titre.
Use stoichiometry (if you haven’t got an equation, write it out).
Use that number of moles of your unknown substance to calculate whatever the question is asking—whether it’s concentration, mass, or molar mass.
Worked Example

Let’s have a look at an example.

This is rather straightforward, okay? But you know what? If you take any complex question, you can break it down into these basic facts:
25 cm³ of KOH (aq)
Required 12.25 cm³ of 0.1 M H₂SO₄
To completely neutralise
Calculate the concentration of the KOH solution.

Now, I’m going to use this three-step process, okay?

Step 1: Find the Moles of H₂SO₄

Using the formula:

Moles = Concentration × Volume
= 0.1 × (12.25 ÷ 1000)
= 1.225 × 10⁻³ moles of H₂SO₄.

Step 2: Stoichiometry

We write the balanced equation:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O.

From this, we see the ratio is 1:2.

So, the moles of KOH = (1.225 × 10⁻³) × 2
= 2.45 × 10⁻³ moles of KOH.

Step 3: Find the Concentration of KOH

Using the formula:

Concentration = Moles ÷ Volume
= (2.45 × 10⁻³) ÷ (25 ÷ 1000)
= 0.098 M KOH.

Final Summary

So, it’s always the same three-step process:
Find the number of moles in the mean titre.
Use stoichiometry to find the moles of the unknown solution.
Convert the number of moles into whatever the question asks—concentration, mass, or molar mass.

That’s your acid-base titration calculations. Hopefully, that straightens it all out for you and simplifies the process.

One particular type of titration that really worries quite a few students is this back titration.

So this tutorial is all about what a back titration is, how you spot when you're dealing with a back titration as far as an exam question is concerned.
Like I said, hopefully, this tutorial is going to straighten all that out, and no longer are you gonna fear these questions on back titrations.

Firstly, a back titration always involves two separate reactions.

So let’s deal with this first reaction first—kind of makes sense, doesn’t it?
The first reaction is of a substance with a known concentration of excess acid or alkali.

So in the question, they’ll say: “Oh, you’ve got this stuff, and we’re gonna react it with this concentration and volume of acid,” for example.
Now, for example, it could be calcium carbonate reacting with 2HCl to give CaCl₂, H₂O, and CO₂.
So that’s our first initial reaction. That’s just done in, like, a beaker or something like that.

But like I said, the one thing they’ve all got in common is that the acid or alkali involved in this first reaction must be in excess.
Because what we’re going to do next is find out how much is actually left over in excess, and we do that through the second reaction.
Before we actually look at that second reaction, though, what tends to happen?

That resulting solution from that first reaction is made up to a known volume—usually 250 cm³—using a volumetric flask.

So reaction number one happens, you put that in a volumetric flask, and you make it up to 250 cm³.

Now, why do we do this?

Because the volume may have changed during that first reaction, especially if it’s something that fizzes, like calcium carbonate in HCl.
So we can’t be sure of the actual volume after that first reaction has taken place.
So we make it up to a known volume that we can deal with later on.

Once we’ve got that solution, we then titrate it to find the number of moles of unreacted or excess acid or alkali from that first reaction.

So, let’s say we are taking our first example—hydrochloric acid reacting with calcium carbonate.
What we do is take that solution, there’s leftover HCl, and we react that with sodium hydroxide of a known concentration and volume.
And of course, that gives us sodium chloride and water.
But we’re not bothered about what we make—what we’re bothered about is finding out how much HCl is left over from that first reaction.
So if we know the concentration of NaOH and we get a mean titre for that, we can find the number of moles of HCl that’s left over after that initial reaction.

Now, long story short:
We know how many moles of hydrochloric acid we used in the first reaction.
We know it was in excess, but we also know the concentration and volume that we put in.
If we know how much we put in, and we know how much is left over,
We can find out how much was actually used during that first reaction.

And therefore, we can find out how much calcium carbonate (or whatever substance it is) reacted using stoichiometry.

Using this method, you can deduce the following things:

The initial number of moles of acid used in reaction 1.
We call that "initial" because we know the concentration and volume that we put in there in the first place.
It’s literally: n = c × V.
The number of moles of excess acid present by titration in reaction 2.
We can use a titration to find out how many moles of that initial number of moles of acid is actually left over.
How much didn’t react in that first reaction?
The number of moles of acid used in reaction 1.
We take that initial number of moles (we know how much HCl we put in during reaction 1).
We subtract the number of moles left unreacted, which we found from the titration.
Initial moles – excess moles = moles used in reaction 1.

Once you’ve got the number of moles of acid that was used up in reaction 1, you can use stoichiometry to find the number of moles of your unknown substance.

So, going back to that initial equation in reaction 1:
If I know that 0.10 moles of HCl was used during that reaction, then I know that 0.05 moles of calcium carbonate must have reacted, because it’s a 2:1 ratio.
So, like I said, if we used 0.10 moles of HCl, then, therefore, it reacted with 0.05 moles of calcium carbonate.

From there, we can find the mass of calcium carbonate that was present, or whatever else the question is asking us to do.

Spotting back titrations—it’s an art, okay?

It’s easy for me because I’ve seen hundreds and hundreds of these questions. But what you’re looking for in a question are these key clues:
An initial reaction with a known concentration and volume of an acid or alkali reacting with another substance.
A statement saying that the solution was made up to a certain volume.
A second titration of that solution.

There are two very clear reactions here. And what we need to do is work backwards:
Find out the number of moles of HCl that was actually left over.
Find out the number of moles of HCl that was put into the first reaction.
The difference between those two tells us how much HCl was used.
Then we use that to find the number of moles of our mystery substance.
But in the next tutorial, I’m going to go through a worked example with you.

Okay, here we go. This is a classic kind of standard back titration example question. So, you know, they're not all the same but this is them probably the most common example of a back titration calculation. You're going to come across in a paper.

So a powder sample contains a mixture of sodium carbonate and sodium chloride. So it's a mixture of two different solid substances. Two grams of this mixed you was reacted with a hundred centimeter cubed of one molar HCl. So this is our first initial reaction here.

So I'm going to label that one—that's reaction number one. The results and solution was made up to 250 cm using a volumetric flask. Okay, the standard, we do that.

25 centimeter cube portions of that solution then required a mean titer of 18.6 centimeter cubed of 0.5 molar NaOH to be completely neutralized. So this is our titration reaction here—reaction number two.

What it wants us to do is calculate what percentage by mass of the original solid sample was sodium carbonate. So let's have a look at this first reaction here.

What’s actually reacting? Well, what's reacting is the sodium carbonate and the hydrochloric acid. Sodium chloride will not react with hydrochloric acid. So we know it's only the sodium carbonate that's actually reacting with it.

So Na₂CO₃ + 2 HCl gives us 2 NaCl + H₂O + CO₂. That's our balanced equation for the first reaction. Of course, what's left over of our HCl ends up reacting with sodium hydroxide in reaction number two.

Reaction number two is literally just a neutralization reaction between hydrochloric acid and sodium hydroxide. So HCl + NaOH gives you NaCl and H₂O. What we need to do here with our calculation is work from the mean titer backwards. Hence, a back titration. We're not looking to find out directly how much HCl reacted with the sodium carbonate—we're finding that out indirectly. So it's—we're working backwards. That's why it’s called a back titration.

So we're going to start with our mean titer. First things first then—the number of moles of sodium hydroxide in the mean titer equals concentration times volume.

The concentration of that sodium hydroxide was 0.5 mol per decimeter cubed multiplied by the mean titer, which is 18.6 centimeters cubed. Of course, divide that by 1,000 because we're working in decimeters, and that gives us 9.3 × 10⁻³ moles of sodium hydroxide in our mean titer.

What does that tell us? Well, that tells us that therefore the number of moles of HCl in 25 centimeters cubed—that’s the volume of that solution in our titration in the conical flask—is also 9.3 × 10⁻³. Why? Because it's a one-to-one reaction in reaction number two.

One mole of HCl reacts with one mole of sodium hydroxide. So if we know the number of moles of sodium hydroxide in a titration, we know the number of moles of HCl in the case where it’s one-to-one, so they’re equal.

But that’s just the number of moles of HCl in 25 centimeters cubed. We've got a stock solution of 250 cm³, so we need to find the number of moles of HCl in that entire solution.

The number of moles of HCl in 250 cm³ is just 10 times more than we had in 25 cm³. So that's 9.3 × 10⁻². Now, this is where I just multiplied that previous number by 10, right?

Now I can say that 9.3 × 10⁻²—that's the number of moles of HCl that was left over in excess. That was the number of moles of HCl that didn’t react with that sodium carbonate in that initial reaction.

So now we know how much HCl was left over. We need to find out how much HCl we started with in the first place. So the initial number of moles of HCl we had is literally what we started with at the beginning, before reaction one took place.

So it’s the concentration times the volume. The concentration was 1 mol per decimeter cubed, and the volume was 100 centimeters cubed. We divide that by 1,000 because we're working in decimeters—standard. So we put that in and get 0.1 moles.

That’s how many moles of HCl we had initially before reaction one. So now we know how much we had in the beginning, before reaction one, and we also know that 9.3 × 10⁻² was left over.

So all we need to do is find the difference between those two numbers. That’s the number of moles of HCl used in reaction one. The number of moles of HCl used in reaction one is initial minus excess: 0.1 - 9.3 × 10⁻² = 7 × 10⁻³ moles.

That's how many moles of HCl we used in reaction one. Now that we know how many moles of HCl there were, we can find out how many moles of Na₂CO₃ they reacted with.

That is just a case of stoichiometry. The number of moles of sodium carbonate that took part in reaction one is 7 × 10⁻³ ÷ 2 = 3.5 × 10⁻³.

Why am I doing that? Well, we know the number of moles of HCl, and if we look at the stoichiometry, 2 moles of HCl react with 1 mole of sodium carbonate. So if we have the number of moles of HCl that took part in that reaction, then we know the number of moles of Na₂CO₃.

So that’s 3.5 × 10⁻³—a bit of stoichiometry there.

Now, is that our answer? No, because the question asks us to calculate what percentage by mass of the original solid sample was sodium carbonate.

We know the entire sample weighed 2 grams. If we can find out what the sodium carbonate weighs, we can find the percentage by mass.

So the mass of sodium carbonate is the number of moles times molar mass. You’ve always got access to your periodic table, so it’s 3.5 × 10⁻³ × 106. That’s the molar mass of sodium carbonate. That gives us 0.371 grams.

How do we calculate a percentage by mass? Well, it’s just asking, “What percentage of 2 grams is 0.371 grams?”

Percentage by mass = (0.371 ÷ 2) × 100 = 18.55%.

They may specify how many significant figures or decimal places you want, so just double-check that at the end of the question. But that’s your percentage by mass.

The main standard in these back titrations is that you’ve got two reactions: an initial reaction and then a titration reaction. You have to work backwards to find out whatever it is they’re asking about—that initial substance.

In this case, a percentage by mass. You find out how many moles of your excess reactant you had using the titration, then find out how many moles of that you had initially.

So for HCl there, find the difference, and you can find out how much was used. Once you know that, apply stoichiometry in reaction one to find the number of moles of your mystery substance, in this case, Na₂CO₃.

Then tweak the answer to actually give you whatever it is the question wants. Back titrations—don’t be afraid of them. Learn how to spot them. Look for two separate reactions: an initial one and a titration one. You know, this bit here about making up to 250 cm³ or something along those lines.

That’s a back titration example question. Hopefully, we’ve straightened this out for you, and they’re not going to worry you anymore.

This practical tutorial is all about the gravimetric analysis technique. Essentially, what it is, we're looking to find the dot xH₂O. Now, this is a hydrated salt.

So even when you've got solid salts, there is some water associated within the crystals. What we're looking to do using this method is find out how many water molecules are associated with each formula unit of that salt, if you like.

Essentially, we're looking to find the formula of a hydrated salt. So, for example, MgSO₄·xH₂O. So we know it's magnesium sulfate, we just don't know what that value of x is. The plan, then, during this practical, is to find x.

So, in other words, how many moles of water are associated with each mole of magnesium sulfate. Essentially, it's a molar ratio.

So, when we get to the nitty-gritty of the calculations, it's similar to finding the empirical formula. It's basically just finding the molar ratio—if we've got one mole of MgSO₄, how many moles of water are associated with it?

Let's take a look at the method.

There's a number of really important steps in this method, which we need to go through to maintain accuracy. The first thing we need to do is heat the crucible.

The crucible is the little pottery pot, if you like, that comes with the lid that we're going to be performing the reaction in. Essentially, what we're going to be heating our salt in.

If we heat that crucible first on a hot Bunsen flame—the blue Bunsen flame—with nothing in it, we're just removing any excess water.

Because it's porcelain or pottery or something like that, essentially, water could be hanging around on the surface, and because we're looking to find a molar ratio, accuracy is key.

So if we heat that crucible first, just to dry it out and make sure there's no water hanging around in there, that's a good first step.

The next step is to find the mass of the crucible and lid and nothing else.

So you need to have the overall mass of that crucible and lid with nothing in it and make a note. While you've got that crucible on the balance, you take the lid off and add a known mass of hydrated salt. It doesn't have to be massively precise, because you're going to be taking that into account in your calculations later on.

So maybe it's something like about 2.5 grams. So it's a known mass of your salt, and you need to get the mass of the crucible with the hydrated salt in and the lid all together. So you need all of those things on your balance and get the overall mass of that.

Then we get into the practical bit. What we need to do is heat that crucible with the hydrated salt strongly for five minutes. So that is a blue Bunsen flame, and every now and again just open the lid very carefully to allow that H₂O to escape. Because what's happening as we're heating it is the H₂O that's associated with the magnesium sulfate gets driven off. We're removing all of that water—we're drying that salt out. So you need to lift the lid every now and again just to allow that water to escape.

Be careful though, because that salt sometimes will actually vaporize, and you're going to get some smoke coming out. We don't want that—we just want to let the water vapor out while we're driving off that salt.

After five minutes, then what we need to do is find the mass of that crucible, lid, and salt again. So we pop that onto a balance—we’re going to make sure it cools down a little bit first, of course—and we pop that onto the balance and find our mass.

Now write that down somewhere, because what we're going to do is go back and heat it again. Now, that process of heating it strongly for five minutes and finding the mass—you repeat and repeat until there's no further change in mass. In other words, you're going to heat it to what's known as a constant mass.

So every time you heat it, some water is going to evaporate. You weigh it. You find that some more water has evaporated, so you take it back and heat it again. Then you take it back to the balance and check again. If it's gone lighter again, you must have removed more water. So you keep repeating that until there's no further change in mass.

You could have heated it for five minutes, then taken it back to the balance, and if there's no change in mass, that means you've driven off all the water. There's no water left in there for you to evaporate or dissociate away from that salt. So by heating it to a constant mass, you are sure that you've driven off all that water. That’s the term you should use when describing it in an exam question.

So you heat it until there is no further change in mass, or you can say you heat it to a constant mass. What you’re left with, then, is the dehydrated salt. So the dehydrated salt remains in that crucible.

Now, the most important mass that you're going to record down is that final mass of the crucible with the lid and the dehydrated salt. So all those masses you wrote down while getting to a constant mass—you don’t really need those. It's this last one you need—that’s the mass of the dehydrated salt that's left over with the crucible and lid.

So we've got three really important records here of mass:
The crucible and lid before we even started.
The crucible with the lid and the hydrated salt before we've started heating.
The mass of the crucible, lid, and the dehydrated salt—the final record after driving off all the water.

You need those three bits of information if you're going to find that dot xH₂O.

This method is really, really important.

You need to know the method and understand the pitfalls with it. For example, if MgSO₄ escapes as vapor, that’s going to affect our final result because of the final mass. We need to make sure that the crucible is dry to start with, we heat it strongly, get the masses we need, and heat to a constant mass.

In any exam question, you're going to get these values. Maybe you need to do some adding and subtracting to get them, but you need these three values. We can use these values and a little bit of mole calculation to actually find the number of moles of our salt, which doesn’t have to be MgSO₄—it could be CuSO₄ or anything else—and of course, the number of moles of water as well.

Once you've found those two things, it's a simple ratio to find the molar ratio, just like with empirical formula. Gravimetric analysis is not always called that, but that's what this method is. Essentially, it's finding the x in that hydrated salt formula, dot xH₂O.

So this is your method. This is how we work it out, and it is a fairly common exam question. You will have come across some other examples of this.

So that's your gravimetric analysis.

When it comes to enthalpy-based practicals, this is an absolute classic, and it's in fact the one that's called on the most when it comes to exam questions. We're looking to measure the enthalpy change of combustion of a fuel. So what we do is burn a fuel, heat some water with it, and using the data, we can actually find out the Delta H, the enthalpy of combustion. In other words, how much energy is released per mole of fuel.

Let's look at what we've got in terms of equipment before we start talking about the method. Well, of course, first and foremost, we have our thermometer. Now, our thermometer has to be a regular mercury thermometer, not a digital one, just in case it gets damaged because we're burning a fuel right underneath it. Also, we have what's known as a calorimeter. That's basically just the vessel in which we're holding the water. And, of course, we are going to put a known volume of water in here because we need to know how much water we've got for our calculations later on. This is a known volume of water, and like I said, that's really important when we come to our calculations later on.

So we've got this calorimeter. When I was in college, we used old Coke cans to do this—an aluminum can. We can't, of course, use polystyrene because, down here, we have our spirit burner. Now, our spirit burner is basically just a wick in a glass vessel. When you light that, heat is released, and we use it to heat the water in the calorimeter. We can also put a lid on this just to make sure our data is a bit more accurate, but more on that later. This is our basic setup. The calorimeter tends to be clamped somewhere, and our spirit burner is placed underneath it.

So let's take a look at the actual method. The main steps of the method are as follows. The first thing you should do is measure a known volume of H2O in the calorimeter. That makes it sound like you are using the calorimeter to measure it, but you're not. You should use a measuring cylinder, which is perfectly accurate enough for this. So get yourself a known volume of water, like 50 cm³, and pour that into your calorimeter.

Next, before you start, you must record the starting temperature of the water. This tends to be around room temperature, about 22 degrees, but you need to get the exact starting temperature. You also need to get the starting mass of the spirit burner, which contains the fuel you're testing. Usually, it comes with a small lid to help extinguish the flame. Place the spirit burner on a balance, zero the balance, and obtain an accurate mass for the spirit burner and its fuel before you start.

Now, we begin heating the water. Place the spirit burner underneath, light it, and heat the water until it reaches a predetermined temperature increase—for example, 30 degrees. It doesn’t have to be 30 degrees, as long as you know exactly how much you are increasing the temperature by. So, if you start at 22 degrees and want to increase it by 30, you stop heating and extinguish the flame when the temperature reaches 52 degrees. You need to keep an eye on that temperature. There is no need to time it—we are heating the water to a specific increase in temperature, not for a specific time.

All that's left to do now is reweigh your spirit burner. You need to determine how much fuel was used. If you have recorded the starting mass before lighting it, once you have raised the temperature by your set amount, extinguish the flame, take it back to the same balance, and reweigh it. The difference between the initial and final mass gives you the mass of fuel that was actually used to raise the temperature of the water.

You can repeat this process for different fuels if you want to compare them. Now, in terms of analysis, we need to calculate the heat change—how much energy it took to heat the water by the set temperature increase. This brings us to the familiar equation:

Q = -m × c × ΔT

I call this "minus MCAT."

A brief explanation of why I include the minus sign: Many chemistry teachers give you Q = MCΔT without the minus. I like including the minus because it eliminates the need to remember to put the correct sign on when you calculate ΔH. If you include the minus sign, then in any exothermic reaction where the temperature increases, all your values will automatically have a negative sign. If you have an endothermic reaction where the temperature decreases, the two negatives will cancel out, and ΔH will automatically be positive. Some like to do it this way, some don’t—it’s completely up to you, as long as you ensure the correct sign at the end.

Let’s break this equation down specifically. M represents the mass of H2O. It is not the mass of fuel or anything else solid; it is always the mass of the liquid being heated. In this case, it’s water. We know the mass of the water because we know the volume, and since the density of water is 1 g/cm³, if we have 50 cm³, that means we have 50 g of water.

C is the specific heat capacity. This will be given in the question and is typically 4.18 J/g·K. Sometimes it’s rounded to 4.2—just use whatever number is given in the exam question. This value tells us that 4.18 J of energy is required to heat 1 g of water by 1 K.

ΔT is the temperature change. Since this is an exothermic reaction, the temperature increases, so ΔT will be a positive number. For example, if we set the increase to 30 degrees, ΔT is 30. If it were an endothermic reaction, the temperature would decrease, and ΔT would be negative.

Once you have calculated Q, which gives you the heat change in joules, you need to do a separate calculation to find the number of moles of fuel used. This is found using:

Moles = Mass / Molar Mass

The mass is the difference between the initial and final mass of the spirit burner. The molar mass of the fuel depends on what you are burning, and you need to know that value.

The final step is bringing these two calculations together. We know the heat change and the number of moles of fuel. So, ΔH (enthalpy of combustion) is given by:

ΔH = Q / Moles

This will give you your answer in joules per mole. If you haven't already converted your heat change into kilojoules, you must divide by 1,000 to get your final answer in kJ/mol.

If you included the negative sign in Q = -m × c × ΔT, then this value will automatically be negative for an exothermic reaction. If you didn’t include the negative, you must manually add the negative sign at the end, as combustion is exothermic.

Now, let’s talk about sources of error. This is one of the most common enthalpy questions in exams, and sources of error are always asked about. Here are four key ones:

Heat loss to surroundings – Not all the energy released by the fuel is transferred into the water. Some is lost to the air.
Heat loss from the water – As soon as water is heated, it begins losing heat, which affects accuracy.
Incomplete combustion – If there is not enough oxygen, the fuel does not fully combust, forming carbon monoxide instead of carbon dioxide, releasing less energy.
Heat capacity of the calorimeter – The calorimeter itself absorbs some heat, which is not accounted for in our calculations.

These factors cause our experimental ΔH to be less negative than more reliable data from a bomb calorimeter, which minimizes heat loss and maximizes accuracy.

Final word: Be prepared to recall this method in exams. The calculation is simple—find heat change, find moles, and divide. Be wary of units and always include the correct sign. This practical is very common in exams, so make sure to practice related questions.

That is how we measure the enthalpy of combustion.

I'm calling the method. I'm about to go through measuring the enthalpy of neutralization because that's the one it tends to be used for the most, but actually, this method can be used for most reactions.
I say between a solid and liquid—even like magnesium and HCl or calcium carbonate and H₂SO₄. So any other kind of reaction that you can think of, maybe involving two liquids or a solid in a liquid, you can use this method as well. But I'm focusing on the enthalpy of neutralization where you've got an acid and an alkali. For example, this is the one I'm running with: the reaction between HCl and NaOH, which of course gives us sodium chloride and water—a very straightforward neutralization reaction.

What have we got over here? Well, of course, we've got our main player in this because we're measuring enthalpy. We need a thermometer. Now, I've just drawn a standard thermometer here. You can use a digital one here because, you know, some temperature changes aren't very big with different reactions, and they can record to the nearest 0.1 of a degree. Whereas, obviously, these are less accurate, so you can use a digital thermometer here.

This container here, of course, is known as a calorimeter. Our calorimeter, in this case, because we're not heating it directly underneath, is just measuring the enthalpy change of the reaction going on inside. This can be a polystyrene cup—that tends to be what we use here. Underneath, what I've got here is a magnetic stirrer. I love these; it's my favorite bit of kit in the lab.

Okay, so we've got what's called—I’m not joking—a "flea" in the bottom of the calorimeter here. It's literally just a magnet, like a pill-shaped magnet, that spins around with a motorized magnet that's underneath the polystyrene cup. So it stirs it and makes the solution homogeneous as the reaction is occurring. We get a fast reaction, and we get a homogeneous mixture, so all the reactants mix very, very quickly, and we get that reaction happening really quickly.

Okay, so that's our setup. Like I said, not just for neutralization, but for all the reactions as well between, you know, two solutions or a solid in a solution.

So what's the method?

Only four steps to this method. It's nice and straightforward. The first thing you should do—as I'm talking about HCl and NaOH—is add 25 cm³ of 2 molar sodium hydroxide to the calorimeter using a pipette. We're being very accurate with our volumes here, hence why we're using a glass pipette to measure out that 25 cm³ of sodium hydroxide.

The next thing we're going to do doesn't even involve the HCl yet. We're going to start our stirrer and get that turning around at a nice gentle pace and get a little whirlpool in the bottom. We don't want it going crazy, but we’ve got to start a timer at the same time. Now, what we're going to do is record the temperature every 30 seconds for three minutes. This is before we've even added the HCl.

What's the point of that? We need to get a constant base temperature to start with. We're getting our starting temperature here because you don't know where that sodium hydroxide—or whatever liquid you're using—has been stored. So if we put it into the calorimeter and get it swirling around for three minutes, it's going to come to the ambient temperature and get a nice, level base temperature for our delta T that we need to calculate later on. So that's why we do this to start with: to get a level starting temperature or base temperature.

Once we've done that and recorded it up to three minutes, at three minutes and 30 seconds—if I can spell "min"—we add 25 cm³ of 2 molar HCl.

I've kept the concentrations and the volumes the same here just for simplicity's sake. So at three minutes 30, you don't take a temperature. You're just adding the acid to the alkali, okay? And like I said, if it was another reaction, you'd just be adding your second reagent at that point. At three minutes 30 seconds, you do not take a temperature because you can't do two things at the same time.

Now comes what's considered the boring bit. All we're doing here is continuing to record the temperature as it's swirling around and the reaction is proceeding—every 30 seconds, up to 10 minutes at least.

Ten minutes because what we want is to know that the reaction is finished, and what we're going to notice is that after we get an initial increase in temperature—well, because this particular reaction is exothermic—the temperature is going to start to decrease. It's going to start cooling down because, of course, once you heat something up and you're no longer heating it, then it just starts to cool down.

So we need to record that temperature every 30 seconds, up to 10 minutes at the very least—maybe go a couple of minutes further if you need to. And the reason for that is we need that data to get an accurate change in temperature, an accurate delta T, which I'm going to show you now.

When it comes to our analysis, it's not just calculations. It's a bit more involved to find our delta T here.
What we do is we've got our results—from zero to 10 minutes, or whatever it is. Every 30 seconds, we've got a temperature. So we plot a graph. On the y-axis, we've got temperature in degrees Celsius; on the x-axis, we've got our time in minutes, or 1, 1.5, 2.5—or it could be in seconds, whatever floats your boat.

What we need to do then is actually plot our data. So what I'm doing here—obviously making this data up—what we tend to find is for our first six data points (okay, three, four, five, six), these are the records of the temperature before the reactions actually happened. So when we've got NaOH, and it's swirling around, hopefully, we don't see any change in temperature there. But this, like I said, is to make sure we've got a steady starting temperature.

Then at three minutes 30, what actually happens is we add our HCl. What we do is we put a line on the graph like this, and I tend to call it "t-mix." Okay, so that's the time at which we mix the two reagents together.

So no temperature is recorded at that point. That's just the point at which we added the acid into the sodium hydroxide—or if we're using different reagents, we mix them together.

Now, for an exothermic reaction, of course, the next temperature we get will be much higher. And what we'll find is that even though we're stirring it, sometimes, just sometimes, we find the temperature is still increasing, you know, 30 seconds or a minute after we've done the mixing. But then what will happen is it starts to decrease over time.

Now, I'm not going up to 10 minutes here, but you get the idea. After the initial reaction, when there's no more reaction happening, it's just going to start cooling down, and that's where we get this decrease in temperature—just a typical exothermic reaction in terms of data we get from this practical.

If it was an endothermic reaction, what's going to happen is—we're going to have a different axis and stuff—but our starting temperature is going to be here. It's going to go down in temperature and then rise back up again. So it's going to be the inverse of this, if you like. Again, we treat that in the same way, but this is an exothermic reaction.

What we need to do here is extrapolate. What I'm doing here with the initial temperature is extrapolating the line of best fit through that point of mixing and seeing where that crosses the line there. That's what we want.

Of course, what we need to do here is extrapolate back. Now, extrapolating back—what we always do is go through the highest point. This here, the reaction is still happening, of course, and it's getting hotter.

So if you do get one like this, you ignore it. Sometimes, it's up here, the highest one, but if it's not, then ignore it. But what we do is we always make sure we get a line of best fit through that highest temperature again, right through this time of mixing.

Okay. Now what we've got here is through these two extrapolations, the difference between this point and this point—this, ladies and gentlemen, is our delta T. We can take one from the other and find out what the increase in temperature—or indeed decrease in temperature—was if it was an endothermic reaction.

So this is your delta T. You need to do it this way for accuracy. The reason why we do this is to take into account heat loss. I'm going to write that down in a second because as soon as you start heating something, it starts cooling down again. It starts releasing heat energy.

So we extrapolate to account for heat loss. And why? Well, this is basically what we’d expect if we didn’t get any heat loss. This is the increase in temperature that we'd expect to get. So this is just a graphical method we use to account for heat loss.

So let me put all these points I made in bullet points over here for you in summary:

We plot time versus temperature.
After we've plotted, we extrapolate on both sides to find our delta T.
We extrapolate on both sides back to the t-mix line, and it's the difference between where those lines intersect—that's your delta T.

Once we've got our delta T, we can go ahead and apply -MCΔT (minus M-cat) to find our heat change.

The M being the total volume of liquid here. Okay, so it's 50 cm³ or 50 grams. This will be because we had 25 of NaOH and 25 of HCl, because when the reaction is happening, it's heating 50 cm³ of water, or 50 grams of water.

The C—you can take as 4.18 or whatever value is given in the exam question. And of course, we've just found our delta T.

Next step: because we're looking at ΔH neutralization specifically in this example, you need to calculate the number of moles of water produced because that is your definition of ΔH neutralization.

Okay, it's the enthalpy change when one mole of water is produced during a neutralization reaction. So that's what we do. We find the number of moles of water produced here.

If you're looking at another reaction, let's say between a metal and acid, or I don't know, like a calcium carbonate and acid or something like that, you find the number of moles of the limiting reagent in another reaction.

Okay. So if your acid is in excess and you're reacting it with magnesium—if the acid's in excess, use the number of moles of magnesium to find your ΔH for the reaction, because, you know, that's what limits it. That's the number of moles of reactant that's going to limit how much energy gets released.

Okay, you're not going to want to include any excess acid in that because that wouldn't give a, you know, a good picture of how much energy is released per mole, would it now? Okay. So in this case, use the number of moles of H₂O produced.

In any other case, find which reagent is limiting the reaction and use the number of moles of that—really, really important.

Lastly, then ΔH—well, that's your energy change, Q, divided by the number of moles of H₂O.

In this case, of course, like I said, in another reaction just use the number of moles of your limiting reagent, and that will give you your ΔH for the reaction.

Again, I'll remind you here two things:
Make sure you get the correct sign on your ΔH. Always put a sign on it—positive or negative. Never leave it blank.
Make sure it's in kilojoules per mole, because when you first get Q, it's going to be in joules. So you need to divide by a thousand to get it into kilojoules per mole.

So this experiment here, this practical setup, is mostly used for measuring the enthalpy of neutralization.

But like I said, this method can be used to find the enthalpy of any other reaction you want, as long as you've got two aqueous solutions or a solid in an aqueous solution, and this will work absolutely perfectly for you.

Okay. So your method here—this is probably the most important bit, okay? Finding your delta T, the reason for extrapolation, taking into account heat loss.

All right, and then it's the same as enthalpy of combustion:
You find Q with -MCΔT.
You find the number of moles of your limiting reagent—or indeed the number of moles of water in this case.
Then find ΔH using those two things.

In terms of sources of error, I'm not going to write these down here, but they're pretty much the same as enthalpy of combustion.

Heat loss to the surroundings: At the top of the calorimeter, you can put a lid on it to make it more accurate.
Using a digital thermometer: This can make it more accurate—accurate to 0.1 of a degree Celsius.

We are taking into account some heat loss by doing this, though.

Okay, so just don't worry about those too much, but it's the process here that's very important.

So that is how we measure the enthalpy of neutralization.