17.2 Titrations and Titration Curves
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- เผยแพร่เมื่อ 16 มิ.ย. 2024
- Chad breaks down Titrations and Titration Curves for Strong Acid-Strong Base, Weak Acid-Strong Base, and Weak Base-Strong Acid Titrations.
00:00 - Three Common Types of Titrations
00:55 - How do you Recognize a Titration?
01:04 - Examples
03:05 - pH @ Equivalence Point
05:29 - Strong Titration Example (pH 7 at Equivalence Point)
09:04 - Titration Curves
15:20 - Titration of Polyprotic Acid
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Here we are in 2024, best explanation I have heard! I especially love the passion you carry and ability to teach bro! Appreciate you.
Thanks for your kind words!
I spent my first couple minutes of 2021 watching this great explanation. Thanks!
Glad it was helpful, Muhammad. Happy 2021.
Thank u professor for ur marvellous way of representing titration.
pro tip : watch series on flixzone. Me and my gf have been using them for watching a lot of movies lately.
@Armani Oakley yup, I have been watching on Flixzone for months myself :D
Excellent explanations!! Thank you so much for making this video!
You're welcome Hadeer. Glad you found it helpful!
These chapters series are great, thank you so much
You're welcome - thank you.
I've been confused this entire unit and my professor just never explained it in simple terms life this. THANK YOU
Glad it finally made sense, Veronica - persevere & succeed!
This helped me sooo much!!!! I was so confused!!! Thank you for this video!!
You are welcome, nraishanzr - glad it helped!
This channel is GREAT!
Your comment is great, Shean - thanks!
This helped me so much for AP Chemistry! Thanks!
Glad to hear it, Spacix - You're welcome!
Thanks sir i was having hard time trying to understand buffer and titration, for my fortunate i came across to your videos and it's pretty much clear now.
Glad the video helped you out!
Seeing in2019! The best video on titrations! Awesome sir!
Glad you found it helpful Satya!
thank u a lot I have been trying to understand it for two days that's
amazing honestly
Glad to hear it, Just August - glad you found the channel!
Thank you so much...this was so concise!
Glad it was helpful, Daniel, and thanks for your comment!
Hi Chad! Thanks for the video. I have a question at 13:10, if you add the weak base(NH3) instead of strong acid (HCl), would the pH graph of the HCl solution present a flat shape initially instead of “curving upwards”?
You're a god among men, titrations have been puzzling me and you just made it all click in a single minute!
Glad to hear the video helped you - thanks for commenting!
Amazing explanation, WOW
Glad you found it helpful Madalena!
Chad, you did it again. thank you thank you thank you
Glad to hear it, Oriana Reyes - You're welcome!
Love the expression..."plug and chug"!
Really captures the process doesn't it?!
WOW. I just wasted all this time reading Kaplan books. I understand this so much better
Love the explanation
Glad it was helpful!
Cool!
I wish my Chemistry teacher was like you😪
It seems we all want a short, bald chemistry teacher! 😜
@@ChadsPrepHello professor, I also taught chemistry at school (in Uzbekistan). I'm not bald but too short with being just 1.63 m)))
Hi Chad, you mentioned at 12:40 that it would take the exact same number of moles of NaOH to neutralize HCl or HF. Could you please clarify why that is? In the buffer video, thought we needed 2:1 ratio for weak acid to strong base, so I would've assumed that with 2 moles of HF, NaOH would have neutralized 1 mole and left over 1 for the buffer? Sorry if this sounds confusing and thank you for your help!
At 12:40 I am not referring to how to make a buffer, but simply to the amount of NaOH required to completely neutralize either HCl or HF. If you completely neutralize HF with NaOH you definitely would not have a buffer. A buffer would be created when you neutralize roughly half of your HF which explains why using a 2:1 ratio creates a buffer. The idea is that when you mix 2 moles of HF with 1 mole of NaOH, 1 mole of the HF is converted to F-, the conjugate base, and the other mole of HF still remains. So you'll still have 1 mole of HF left and now 1 mole of F- formed. This is a weak acid and its conjugate base in a 1:1 ratio, a buffer. Hope this helps!
Chad's Prep thank you so much for the clarification!
I believe it's around 10:40. I understand that 2:1 is the half equivalence point. But if we plug that into the HH equation wouldn't it be log(1/2)? i thought we needed a log(1/1) to get the pH=pKa.
Edit: I just read a previous comment on how the NaF has 5 mols and can further react with the remaining HF has 5 mols, so there is a 1:1 ratio. However, is HF and NaOH, combined, not a buffer on its own?
Hello Jess! The key is that if you mix an acid and a base, if either is strong (like NaOH in this example) then a neutralization reaction will occur to completion. You will then calculate the resulting pH in the solution based not on what you mixed, but on what is left over after this neutralization reaction. So if you mix 10 moles HF with 5 moles NaOH (a 2:1 ratio) they will react completely as NaOH is a strong base according to the following neutralization reaction:
HF + NaOH --> NaF + H2O
When this reaction is complete you will still have 5 moles of HF left and will have formed 5 moles of NaF. It is this 1:1 ratio that is plugged into the Henderson-Hasselbalch equation.
Hope this helps!
Like your shirt. You are a great communicator too.
Thank you, Sean - appreciate the comment!
in buffer HF+NaOH Strong base and weak acid but in titration also there is a type of SB/WA. How can be decide whether it is buffer or titration? Thank you for your all videos!
I know you explained this in video but I couldn't understand unfortunatelly :(
It could be both Cagla. If you look at a weak/strong titration curve we refer to the region around halfway to the equivalence point as the buffer region on the curve. So if you mix a weak acid and strong base in a roughly 2:1 ratio you could be doing that in the course of a titration or to create a buffer. But whether you were trying to or not the solution would be a buffer at that point and we would calculate the pH as a buffer solution (like using Henderson-Hasselbalch). Hope this helps!
Hey Chad, I'm a little confused about the "half equivalence point". You mention it at 10:50 but I'm not sure what it means or what its significance is.
Hi Genevieve! The half-equivalence point is the point at which half of your weak acid has been neutralized and coverted into its conjugate base. The result is that the solution will contain equal amounts of weak acid and conjugate base. This makes it a special point in the titration for two reasons.
1. Having equal amounts of weak acid and conjugate base the solution will be a buffer with maximum buffering capacity.
2. It is a point at which it is rather easy to calculate the pH of the solution as it is the point where pH = pKa. With the weak acid and conjugate base concentrations equal the 2nd term (the log term) in the Henderson-Hasselbalch equation cancels leaving pH = pKa.
That is pretty much the two-fold significance of the point. Hope this helps!
Chad's Prep thank you so much!
does it matter if we watch this 17.2 vid before the 17.1? Only cause my professor is doing this titrations stuff before buffers for some reason
It does. A little ways after the initial point in a titration and a little ways before the equivalence point you will actually have a buffer during a titration and you will end up calculating the pH just as you would for a buffer solution. So definitely watch pH calculations for buffers in 17.1 first. Best!
thanks sir
You're welcome!
Why is the ratio 2:1 for half-equivalence and 1:1 for equivalence? I had trouble understanding that part
1:1 for the Equivalence point is simply the definition; we've added an equivalent amount of acid to base or base to acid which for monoprotic acids/bases will be a 1:1 ratio. So say you started with 10 moles of HF, it would take 10 moles of NaOH to reach the equivalence point (as that is an equivalent amount.
The half-equivalence point follows. If it takes 10 moles of NaOH to reach the equivalence point then it would only take 5 moles to reach the half-equivalence point. And at this point you would still have 5 moles of HF remaining whereas the other 5 will have been neutralized being converted into F-. This is why you'd have an equal number of moles of both the acid (HF) and its conjugate base (F-) at the half-equivalence point.
Hope this helps!
@@ChadsPrep Thank you!
10:34 is it ok to explain again why is it 2:1and for 11:50 at 10:34 you mentioned the weak: strong ratio must be 2 to 1 but when you asked whether it is true or false, you said wrong, they have the exact same mole ratio. Your thoughtful response will be greatly appreciated.
2:1 ratio is only when you wanna create a buffer solution, not when you wanna neutralize it completely
That help me alot thanx
JOB THAT IS AWESOME
Glad you enjoyed the video, Mekdes!
Thanks
You're welcome!
Thank u
U are welcome!
Consistent DAT question
Please help. How to know when a salt is negligible?
Got a lesson on that here: th-cam.com/video/KW7c3m8q0Qc/w-d-xo.html
Hope this helps and Happy Studying!
@@ChadsPrep thanks...a lot
Seeing in 2K20.Awesome
why use kaplan when chad can easily explain these concepts in under 20 minutes (10 minutes if you double speed like i do LOL)
You know how to use your time wisely, luis - Good job!