/sci/ - Science & Math » Thread #13005173

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Fucking Titrations

Anonymous Tue 20 Apr 14:36:11 2021 No.13005173 View Reply Original Report

Quoted By: >>13005275 >>13005294 >>13005307 >>13005435 >>13005613 >>13005732

Does anybody have good sources on titration calculations with a CONSISTENT FORMAT?
I would like to read the steps in a manner that isn't written like a schizo post.
Maybe I'm just a brainlet but I can't stand this textbook where the calculations are juxtaposed 60 different ways before it even reaches an explanation.

Anonymous

Anonymous Tue 20 Apr 2021 14:48:32 No.13005216 Report

Quoted By: >>13005224

analitical chemists are all midwits that should promtly kill themselves

Anonymous

Anonymous Tue 20 Apr 2021 14:50:39 No.13005224 Report

Quoted By: >>13005268 >>13005310

>>13005216
I agree but do you have an actual contribution? I'm not majoring in analytical chemistry but I still have to do this shit.

Anonymous

Anonymous Tue 20 Apr 2021 15:01:29 No.13005268 Report

Quoted By: >>13005341

>>13005224
no

Anonymous

Anonymous Tue 20 Apr 2021 15:02:57 No.13005275 Report

Quoted By: >>13005284 >>13005495

>>13005173
he he stopcock

Anonymous

Anonymous Tue 20 Apr 2021 15:05:38 No.13005284 Report

Quoted By:

>>13005275
lol

Dolphin with superior cetacean intelligence

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Dolphin with superior cetacean intelligence Tue 20 Apr 2021 15:09:00 No.13005294 Report

Quoted By:

>>13005173
>op doesn't have the Midas touch

Anonymous

Anonymous Tue 20 Apr 2021 15:12:45 No.13005307 Report

Quoted By: >>13005341

>>13005173
fucking titrations make me seethe so hard. For some reason every chemist who writes a textbook is an absolute faggot who can't explain anything in an intuitive way. I have not come across a single physics book that's anywhere near as bad as the average chemistry book.
Even fucking David Morin's book is intuitive and easy to understand if you're comfortable with the math. In fact, I would say David Morin's book is far easier to understand than Taylor's since he doesn't waste time on bloated, unintuitive explanations that take several paragraphs to explain things.

Anonymous

Anonymous Tue 20 Apr 2021 15:14:34 No.13005310 Report

Quoted By: >>13005347

>>13005224
Are you from India, what the fuck is the problem. Just Titrate the damn thing and calculate how much solution is left after the color-shift. These are among the easiest calculations in chemistry.

Anonymous

Anonymous Tue 20 Apr 2021 15:24:16 No.13005341 Report

Quoted By:

>>13005268
Then please go back to your mental masturbation containment threads where they're discussing 5th dimensional faggot mechanics.

>>13005307
Hey, somebody who understands. I'll at least look into those authors and see if I can find any improvement. I'm reading Daniel Harris & every time math comes in he has to write like 3 different instances of a reciprocated fraction. And then add ten million schizophrenic footnotes that reference steps of the equation that you're not even currently working on.
It's just a clusterfuck and superfluous shit like that gets me fucked up regardless of how simple the underlying concept is.

Anonymous

Anonymous Tue 20 Apr 2021 15:25:24 No.13005347 Report

Quoted By: >>13005351

>>13005310
That's not what I mean, dickhead. I mean the superfluous titration homework questions that arealiciously worded to get you fucked up. I WISH I was doing physical titrations.

Anonymous

Anonymous Tue 20 Apr 2021 15:26:29 No.13005351 Report

Quoted By:

>>13005347
Fuck
>Maliciously worded

Anonymous

Anonymous Tue 20 Apr 2021 15:39:16 No.13005398 Report

Quoted By: >>13005553

The absolute state of chemistry threads on this board. You fucking mental degenerate. How do you not comprehend this basic calculation. Go shit in a hat you moron. Write a chemical equation of the neutralization reaction like the witless bastard that you are.
H2SO4 + 2 NaOH -> Na2SO4 + your moms pussy juice
2n(acid) = n(base) in this case you bitchass brainlet
2 [c(acid) x V(acid)] = c(base) x V(base) you spineless slimy fucker
2 [m(acid) / M(acid)] = m(base) / M(base)
Go fuck yourself.

Anonymous

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Anonymous Tue 20 Apr 2021 15:45:46 No.13005419 Report

Quoted By:

>strong-strong
equilibrium when H3O moles are equal to OH moles
>strong-weak
same thing but equilibrium will not be at seven because the weak acid/base will create a base/acid. use -log[pKa]
>i keep overshooting
do a fast analysis first to see at approximately which point it will tip over then do a second analysis and slow down 5-10mL before that point, keep going slower and slower as you get closer

Anonymous

Anonymous Tue 20 Apr 2021 15:50:15 No.13005435 Report

Quoted By: >>13005560

>>13005173
Are all the other chemists on this board midwits? This is a basic experiment. Literally just titrate in a strong base into a strong acid. When the pH is 7, your moles of base added and acid are equal. If its a weak acid and strong base, its when the pH is equal to the pKa.

The state of chemistry on this board is depressing.

Anonymous

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Anonymous Tue 20 Apr 2021 16:00:36 No.13005495 Report

Quoted By:

>>13005275
>stopcock
>he doesn't stop
Sasuga OP

Anonymous

Anonymous Tue 20 Apr 2021 16:08:46 No.13005553 Report

Quoted By: >>13005580 >>13005769

>>13005398
Alright, well you got a laugh out of me but this still wasn't what I was asking for. I understand a regular titration, I was not asking for basic ass concentration from a titration. I don't know why I didn't expect /sci/ to be filled with presumptuous fucks.
I'm doing titration problems that require you to keep track of multiple reactions, multiple volumes & concentrations both L/mL and Mol/mMol.

I understand how to find concentrations from the neutralization. What I was asking for is a concise reference to all the algebraic manipulations that might occur in a titration. Because regardless of how simple the basis of the math is, I get fucked up when there are multiple reactions being referred to within the same problem & when said problem is specifically worded to get you confused about which reaction happened last.

Holy fuck, id rather just trudge my way through it at a snail's pace than deal with this retarded board.

Anonymous

Anonymous Tue 20 Apr 2021 16:10:23 No.13005560 Report

Quoted By: >>13005600

>>13005435
Dumbass read my posts. I'm not asking about some basic ass organic-1 experimental titration.

Anonymous

Anonymous Tue 20 Apr 2021 16:14:15 No.13005580 Report

Quoted By: >>13005641

>>13005553
Can you give an example of a titration that was difficult for you to understand? I'm having a hard time understanding what is tripping you up so badly.

Anonymous

Anonymous Tue 20 Apr 2021 16:17:46 No.13005600 Report

Quoted By:

>>13005560
>Basic ass organic 1 experimental titration
They're all the same anon. If you understand one you understand them all. Its moles of whatever you add in equals moles of whatever you're adding to at a certain point. What kind of titration you're doing determines the point(s) at which this occurs.

You didn't say what kind of titration you're asking for. There is no general format other than what I told you. It all depends on what kind you're doing. There's no general math if its that complex, you're just going to have to use your big boy brain and figure out how your system will act as you add your titrant.

Just purify your system if there's so much shit going on you can't keep track of it.

Anonymous

Anonymous Tue 20 Apr 2021 16:21:10 No.13005613 Report

Quoted By:

>>13005173
Dude if I could do titrations in high school while half asleep then you at (I assume) university age should be able to do it no problem. It really isn't that difficult

Anonymous

Anonymous Tue 20 Apr 2021 16:30:30 No.13005641 Report

Quoted By: >>13005695 >>13005696

>>13005580

Zn + 2HCl --> ZnCl2 + H2

Zinc content of 1.88g ore sample was determined by dissolving it in HCl. Excess HCl is neutralized with NaOH.

Ore was dissolved in 150mL of 0.600M HCl
Solution was diluted to 300mL

20.0mL aliquot of the Final Solution required 9.33mL of 0.543M NaOH to neutralize excess HCl

Find the mass % of zinc in the ore sample.

I feel like I know the steps to this problem, it's just the burden of perceiving all those fucking numbers at one time.
The moles of zinc will be found by relating them to [reacted HCl], which is calculated from the [excess HCl], right? Or is there something I'm not getting? I was just looking for better references for these multiple step questions so I can keep all this shit straight in my head. The textbook does a shit job at keeping anything concise.

Anonymous

Anonymous Tue 20 Apr 2021 16:40:47 No.13005695 Report

Quoted By: >>13005733

>>13005641
Just work backwards.

Find moles of excess HCl from NaOH added.

With that, you can find the concentration of excess HCl in your 20 mL aliquot and from there you can go backwards with your dilutions to get the excess HCl concentration in your 150 mL sample.

Knowing how much HCl was added and how much remained, you can subtract and find how much reacted.

Remembering that you need 2 HCls per Zn, you can then find how many moles of zinc were present.

Anonymous

Anonymous Tue 20 Apr 2021 16:41:03 No.13005696 Report

Quoted By: >>13005720 >>13005733

>>13005641
This is what's called a backtitration (you add a reagent in known amount which is in excess, then you titrate the excess back to find the difference).

Let's walk through it then.

Every mole of zinc consumes two moles of hydrochloric acid. Keep this in mind.

If you dilute 150 mL of 0.6 M HCl to 300 mL, the concentration will half, so we are dealing with a 0.3 M HCl here.
If you take 20 mL of that this means 20 mL x 0.3 M = 6 mmol of HCl.
Now we actually used up 9.33 mL of 0.543 M NaOH, and they react 1:1.
9.33 mL x 0.543 M = 5.06619 mmol of NaOH that was consumed.

But we should have had 6 mmol of HCl. Some got consumed by the zinc, 6 - 5.06619 = 0.93381 mmol did.

Now every mmol of zinc consumes two mmol of HCl, so for this 20 mL aliquot 0.93381/2 = 0.466905 mmol of zinc must have been there.
The full volume was 300 mL, so 300/20 * 0.466905 = 7.003575 mmol of zinc have been dissolved originally, which are by m = n*M approx. 457.9 mg.

The zinc content was therefore 0.4759/1.88 = 25.3%.

It's always best to walk through problems like these step by step. None of the steps are hard, but trying to do them all at once will not work.

Anonymous

Anonymous Tue 20 Apr 2021 16:44:24 No.13005720 Report

Quoted By: >>13005746

>>13005696
Your math is wrong.
You had 5.06619 mmol of HCl in the 20 mL aliquot, not the whole system.

Idk where you got 6 mmol from. You initially added 90 mmols.

Anonymous

Anonymous Tue 20 Apr 2021 16:46:26 No.13005732 Report

Quoted By: >>13005775

>>13005173
sure, why not.
https://apps.dtic.mil/dtic/tr/fulltext/u2/a296181.pdf
enjoy

Anonymous

Anonymous Tue 20 Apr 2021 16:46:42 No.13005733 Report

Quoted By:

>>13005695
>>13005696
Thank you. I'm sorry if I wasn't clear enough at the beginning. But honestly thanks guys

Anonymous

Anonymous Tue 20 Apr 2021 16:48:35 No.13005746 Report

Quoted By: >>13005759

>>13005720
>You had 5.06619 mmol of HCl in the 20 mL aliquot, not the whole system
Yes, precisely what I calculated. That's why I then calculated the amount of zinc in this 20 mL sample and then extrapolated to the full volume.

90 mmols is indeed the amount of HCl in the full 300 mL. 6 mmol is the amount that would be in the 20 mL aliquot.

Anonymous

Anonymous Tue 20 Apr 2021 16:50:46 No.13005759 Report

Quoted By:

>>13005746
Yeah I see now, you did the dilution math implicitly there.

Anonymous

Anonymous Tue 20 Apr 2021 16:52:45 No.13005769 Report

Quoted By:

>>13005553
maybe you should post the question then

Anonymous

Anonymous Tue 20 Apr 2021 16:53:55 No.13005775 Report

Quoted By:

>>13005732
The only non-brainlet answer so far.

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