I Don't Understand Titration

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M.A.A
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#1
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#1
I seriously don't understand titration, and my chem teacher isn't making it any easier. She rushes us through lessons because we're Triple students, even though that doesn't make sense: admitted, I see what she's saying, but the idea is to, you know, TEACH us.
I'm struggling to remember general formulas too, like calculating the moles, finding the missing mass etc.
Titration though. Oh. My. God. I don't understand a thing. I just don't get it. I have a chem test coming up, and the teacher's told us that we're gonna have a few Qs on titration, maybe a 6-marker. I understand the method of titration, like the practical element to it, but I don't understand calculating the missing acid/alkali using the numbers given to you regarding the other. I need it really 'baby-fied' to understand, lol.
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hunchoM
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#2
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#2
Tried reading a text book? CGP books explain it well
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M.A.A
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#3
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#3
(Original post by hunchoM)
Tried reading a text book? CGP books explain it well
I just don't understand it in general. The only part I understand is that you find a ratio for the balanced chemical equation and, depending on which is acid or alkali, you either half or double or triple etc. the number you've already found via the actual titration you physically do. Calculating the mass of it and the moles is beyond me lol. I don't see why a lot of people think chem is easy - imo, it's on par with physics' difficulty. IMHO, bio is pretty simple.
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hunchoM
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#4
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#4
(Original post by M.A.A)
I just don't understand it in general. The only part I understand is that you find a ratio for the balanced chemical equation and, depending on which is acid or alkali, you either half or double or triple etc. the number you've already found via the actual titration you physically do. Calculating the mass of it and the moles is beyond me lol. I don't see why a lot of people think chem is easy - imo, it's on par with physics' difficulty. IMHO, bio is pretty simple.
Ahahah, I’m the opposite, I find chem easier than bio. What I found useful was going through each step of a worked example and making sure to understand each step and where they got the numbers from.
Then finding similar questions and doing them by myself.
If I’m honest, sometimes I don’t even understand what’s going on. I just memorise the steps and what numbers to use from the question and which equations. It usually works lol
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Kian Stevens
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#5
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#5
Fundamentally, titrations are used to determine the concentration of a known volume of an analyte, by titrating it against a known concentration of titrant—this is in the burette—and the volume required to do this is the titre

There are many different types of titration, but in principle they all exploit one simple thing: stoichiometry
Generally speaking, when the titrant has completely reacted with the analyte—based on the reaction’s stoichiometry—this is known as the equivalence point, and is where the moles of titrant are equal to the moles of analyte
The volume of titrant required to reach the equivalence point is, as I mentioned before, called the titre; and this can be used in conjunction with the titrant’s concentration, and the reaction’s stoichiometry, to determine the unknown concentration of the analyte

Of course, you need some way of indicating when the titrant has completely reacted with the analyte, and there are quite a few; here are just a couple of examples:
  • pH indicator, such as phenolphthalein: used in acid-base titrations, and the colour change of the indicator—known as the endpoint—tells you when the equivalence point has been reached, i.e. when the acid has been neutralised by the base, or vice versa
  • Potassium permanganate (KMnO4) solution: used to determine the concentration of some aqueous metals, here the titrant itself is used as an indicator, in which the permanganate will instantly turn colourless as soon as it dissolves in the analyte—characteristic of the reduction of MnO4- to colourless Mn2+—and thus the equivalence point is indicated as soon as the analyte starts to turn a purple colour, as this is when there’s no more analyte to reduce the permanganate

Since you mentioned it in your comment, let’s consider a simple acid-base titration between 50 cm3 hydrochloric acid (in this case, the analyte) and 1 M sodium hydroxide sodium (in this case, the titrant)
Of course, the reaction between these two species is HCl + NaOH \rightarrow NaCl + H2O, and so based on the reaction’s stoichiometry, the moles of NaOH used to reach the equivalence point should be equal to the moles of HCl present in the analyte solution
So, let’s assume that you did a rough titration and found that the indicator’s endpoint was reached at 50 cm3; you would then repeat the titration until you acquired at least three concordant titres, i.e. they’re within 0.01 cm3 of each other
After further titrations, you find that the average titre—i.e. the average of all concordant titres—is 50 cm3; this is what you use in your calculations

If the average titre was 50 cm3, and you used a 1 M solution of NaOH, then the moles of NaOH used to reach the endpoint was conc. = \frac{mol}{vol} \therefore mol = conc. \cdot vol = 1 \cdot \frac{50}{1000} = 0.05 mol (divided by 1000 to convert cm3 to dm3)
According to the reaction’s stoichiometry, and since the moles of NaOH used are equal to the moles of HCl present, the moles of HCl are also 0.05 mol
Since you used 50 cm3 of analyte, this means that the concentration of it was: conc. = \frac{0.05}{50} \cdot 1000 = 1 M (multiplied by 1000 to convert cm3 to dm3)

Also, here’s a tip: since mol = conc. \cdot vol, and the moles of both species must be equal at the equivalence point, it is true that CAVA = CTVT, where C and V are concentration and volume, respectively; and A and T are analyte and titrant, respectively
Hence, you can rearrange this equation to make CA the subject, in order to easily calculate the concentration of the analyte if you know the other three terms; you don’t have to convert between cm3 and dm3 using this equation either, which makes life a bit simpler

And so there’s an example of doing a titration to determine the unknown concentration of an analyte, by using a known concentration of a titrant: all titrations boil down to this, as that’s the entire point of them, so once you learn the basics of titrations you can essentially tackle any titrations problem

I hope I helped clear things up slightly...!
Last edited by Kian Stevens; 2 years ago
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M.A.A
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#6
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#6
(Original post by Kian Stevens)
Fundamentally, titrations are used to determine the concentration of a known volume of an analyte, by titrating it against a known concentration of titrant—this is in the burette—and the volume required to do this is the titre

There are many different types of titration, but in principle they all exploit one simple thing: stoichiometry
Generally speaking, when the titrant has completely reacted with the analyte—based on the reaction’s stoichiometry—this is known as the equivalence point, and is where the moles of titrant are equal to the moles of analyte
The volume of titrant required to reach the equivalence point is, as I mentioned before, called the titre; and this can be used in conjunction with the titrant’s concentration, and the reaction’s stoichiometry, to determine the unknown concentration of the analyte

Of course, you need some way of indicating when the titrant has completely reacted with the analyte, and there are quite a few; here are just a couple of examples:
  • pH indicator, such as phenolphthalein: used in acid-base titrations, and the colour change of the indicator—known as the endpoint—tells you when the equivalence point has been reached, i.e. when the acid has been neutralised by the base, or vice versa
  • Potassium permanganate (KMnO4) solution: used to determine the concentration of some aqueous metals, here the titrant itself is used as an indicator, in which the permanganate will instantly turn colourless as soon as it dissolves in the analyte—characteristic of the reduction of MnO4- to colourless Mn2+—and thus the equivalence point is indicated as soon as the analyte starts to turn a purple colour, as this is when there’s no more analyte to reduce the permanganate

Since you mentioned it in your comment, let’s consider a simple acid-base titration between 50 cm3 hydrochloric acid (in this case, the analyte) and 1 M sodium hydroxide sodium (in this case, the titrant)
Of course, the reaction between these two species is HCl + NaOH \rightarrow NaCl + H2O, and so based on the reaction’s stoichiometry, the moles of NaOH used to reach the equivalence point should be equal to the moles of HCl present in the analyte solution
So, let’s assume that you did a rough titration and found that the indicator’s endpoint was reached at 50 cm3; you would then repeat the titration until you acquired at least three concordant titres, i.e. they’re within 0.01 cm3 of each other
After further titrations, you find that the average titre—i.e. the average of all concordant titres—is 50 cm3; this is what you use in your calculations

If the average titre was 50 cm3, and you used a 1 M solution of NaOH, then the moles of NaOH used to reach the endpoint was conc. = \frac{mol}{vol} \therefore mol = conc. \cdot vol = 1 \cdot \frac{50}{1000} = 0.05 mol (divided by 1000 to convert cm3 to dm3)
According to the reaction’s stoichiometry, and since the moles of NaOH used are equal to the moles of HCl present, the moles of HCl are also 0.05 mol
Since you used 50 cm3 of analyte, this means that the concentration of it was: conc. = \frac{0.05}{50} \cdot 1000 = 1 M (multiplied by 1000 to convert cm3 to dm3)

Also, here’s a tip: since mol = conc. \cdot vol, and the moles of both species must be equal at the equivalence point, it is true that CAVA = CTVT, where C and V are concentration and volume, respectively; and A and T are analyte and titrant, respectively
Hence, you can rearrange this equation to make CA the subject, in order to easily calculate the concentration of the analyte if you know the other three terms; you don’t have to convert between cm3 and dm3 using this equation either, which makes life a bit simpler

And so there’s an example of doing a titration to determine the unknown concentration of an analyte, by using a known concentration of a titrant: all titrations boil down to this, as that’s the entire point of them, so once you learn the basics of titrations you can essentially tackle any titrations problem

I hope I helped clear things up slightly...!
Thank you!! This helped so, so, so much. Like, seriously, thanks for all the time and effort you put into this to try and help me!
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young.one
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#7
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#7
what even is the general purpose of determining the concentration of a known volume?

sorry to jump in, i am completely at the beginning of taking on titration and am phased by it somewhat also.

the explanation has helped a lot but i feel like i don't even know the actual purpose of titration - in what way does it help?
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