1.7 ACID - BASE TITRATIONS AND CALCULATIONS
- Chemistry at glance
- Jul 6, 2020
- 7 min read
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Hello everyone! It's me again with a new topic. This is the last post in AS-level chemistry and then we're going to start with A-level. You can still visit my youtube channel for past paper questions. I think, now that we're done with individual topics, I might make some videos with random past paper questions.
Today, we're going to talk about titrations and more specifically acid-base titrations. How do you perform them? What happens during titration? And at the end, we'll look at some example calculations involving titration.
Acid-base titration - what is it?
During a titration, we react a volume of acid with a known volume of a base ( or vice versa; a volume of a base with a known volume of acid) until we reach a neutralisation point which is measure by an indicator.
Neutralisation point is reached when salt is formed.
An acid-base titration is readily used to form salts but it's also practised to find unknown concentration of one of the reactants. And this is something we're going to focus on now.
To obtain the unknown concentration, we need to know the exact concentration of the other reactant ( for example if we want to find the concentration of acid, we need to know the exact concentration of the base), therefore, we use a standard solution of that reactant.
Standard solution - solutions whose concentration is accurately known.
You may get asked about the process of making standard solutions and that's what we're going to go through now.
One little important point: for this process, we need a primary standard which is a very pure reagent (give us a true representative of moles) and one that can be easily weighed.
1. Accurately weigh the amount of primary standard you require (you need different amounts depending on the concentration you are looking to get) in a weighing bottle.
2. Put all of the solid into a beaker, remembering to wash the weighing bottle with distilled water and run everything to the beaker ( to get all of the weighings). Then, add just enough water to dissolve the solid and stir.
3. Pour the solutions into a volumetric flask using a funnel, remembering to wash all the solution out of the beaker. Add water just under the graduation mark.
4. Finally, add water drop by drop until you reach the graduation mark (we do it drop by drop at the end to not overflow the graduation mark).
This is the whole process! It is not complicated and I think you can manage to remember it with no problem.
Ok, so we know what titration is, and what reactants we need, and what we're going to get out of titration. Now, it's time for us to get to know how to perform a titration. There is a process you always follow and I am now going to present it to you:
1. Put one of the reactants into a burette (e.g. the acid) and read the burette - this is our INITIAL VOLUME.
2. Put the other reactant into a conical flask (e.g. the base). It doesn't matter if you put them the other way, just know the volume that's in the burette and in the conical flask.
3. Add a few drops of indicator into the conical flask - to see the endpoint of the reaction.
4. Run the solution from the burette (in this instance the acid) to the flask, swirling the flask at the same time.
5. When the indicator just changes colour, stop the solution from running anymore- this is your ENDPOINT.
Always do a rough titration first- it will show you when approximately the solution will change colour and next time you will be prepared to decrease the running about that time. This ensures the endpoint is not overshot.
6. Read the burette again - FINAL VOLUME
7. Calculate the titre (INITIAL VOLUME - FINAL VOLUME). This is the amount of acid that was needed to neutralise the base.
This is how you perform a standard acid-base titration. Although many different reactants can be used, most titrations follow the same rules.
Let's try an example together:
e.g. 1
On the photo above I included a WJEC GCE Chemistry past paper question. When I get questions where there is a lot of text, I always try to break it down a little. First of all, you are told some numbers. In titration calculations, you're always going to use the equation n=cv, therefore, I look for solutions which have both c and v allocated to them (this is the solution with known concentration and known volume because it is measured).
In this example, that solution (with both concentration and volume known) is sulfuric acid, so you find the number of moles of it;
Then, scan the question again, to determine what you're looking for.
In our example, we are trying to find the concentration of sodium hydroxide. We already know the volume and we want to find the concentration, so let's figure out the number of moles first;
To do this, we use the mole ratio. As we already found the number of moles of sulfuric acid and the mole ratio is 2:1 (look at the coefficients), we just need to multiply the moles of sulfuric acid by 2 (to get the number of moles of NaOH);
Lastly, we use n=cv but rearranged to find the concentration;
e.g. 2
In the second example, we're just asked to find the mean titre. So, this shouldn't be much of a trouble. However, there is one mistake that everyone seems to make. When you calculate the mean of all the titres, you only include those that are within 0.20cm^3. Hence, 25 is the titre we do not use as it is not within 0.20cm^3 with the other titres.
Then, it is just a normal 'mean finding' calculation.
Double titration:
Double titration- it is slightly different from the standard titration and I couldn't get around it during my as-level. As the name suggests, here we have to titrate something twice. We use it if we have 2 bases in a solution which can be neutralised by one acid. Both of the bases change colour at different pH, so we use two different indicators which will change colour at two different times.
To be honest, I have never seen double titration calculation question in the as-level exam when I was revising for it, but it is in the book, so I guess you never know.
Here is an example:
As this is a more difficult example, I prepared steps for you to follow.
1. This is the first stage; I included equations of reactions that happen in the first stage above in blue colour. Both of these reactions happen at this stage. We can use the volume of HCl used and the known concentration to find the number of moles of HCl used in the 1st stage (when the first indicator changes colour).
2. The moles of OH^- and CO3^2- will also be the same as the number of moles of HCl in this stage.
3. Now, we find the number of moles of HCl during the 2nd stage (when the second indicator changes colour). We only have one equation in this stage and we are told, 8.25cm^3 were used with the same concentration as before. So, we again use n=cv to find n.
4. The moles of HCO3^- is the same as for HCl at this stage.
5. In the first stage (in the equation) we have CO3^2- in 1:1 ratio with HCO3^-.
6. Therefore, CO3^2- must have 8.25x10-4mol just like HCO3^-.
7. Therefore, the OH- in the first stage must be 2.2x10-3 - 8.25x10-4. All of this happens because we have both carbonates and hydroxides in the first stage but only carbonates in the second stage.
8. Now that we know the number of moles of carbonates and hydroxides we can calculate their concentrations using c=n/v (we're told the volume at the very beginning of the question).
If you're not with me, try to work through it again and again until it clicks. It worked for me! Good luck. :)
Back titration:
Here is the last titration for today, a back titration. Again, the overall performance of this titration is similar to the standard one. However, in back titration, we have an excess of one of the reagents (let's say an acid) which reacts with, for example, some sort of a solid. After the reaction, the excess of the 'acid' is reacted by titration with a base (a neutralisation reaction).
We usually use back titration when it is not as easy to determine the endpoint or the reaction would be too slow.
Again, let's have a look at an example together.
It is called back titration because we're kind of working backwards with our calculations.
Here are the steps:
Here we have (in the photo above) two equations. The first one is of the acid with solid and the second one is the titration reaction.
1. In step one we use the 2nd equation (going backwards). We find the number of moles of NaOH (number of moles that reacted with excess HCl).
2. The mole ratio between NaOH and HCl is 1:1.
3. Therefore, the number of moles of HCl is 2.28x10^-3 and this is the excess acid.
4. Next, you find the number of moles of HCl that reacted with the solid (limestone). We have concentration and volume, so n=cv. This is the number of moles of HCl at the beginning.
5. Now, you want to find the number of moles that reacted with limestone-> beginning moles- excess moles.
6. As the aim of this question is to find the amount of calcium carbonate in limestone, we need to know the moles of calcium carbonate (you always need the moles of everything). Here comes in the 1st equation. The mole ratio is 2:1, so divide the moles of HCl by 2.
7. Now, let's find the mass of calcium carbonate in the limestone using m=nxMr.
8. Lastly, you take the mass of calcium carbonate and divide over the mass of the limestone and multiply by a 100. This will give you the per cent of calcium carbonate in the limestone.
I think back titration is much nicer than the double titration.
If you're interested in some past paper questions, visit my youtube channel, I left the link at the top of this page.
PS. Please remember, I am only a student, and as anyone, I can make mistakes. If you think you can see one, don't hesitate and comment (either here on on my youtube channel) Thank you!
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