Titration of a weak acid with a strong base | Chemistry | Khan Academy
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- Опубліковано 5 сер 2024
- Calculating the pH for titration of acetic acid with strong base NaOH before adding any base and at half-equivalence point. Created by Jay.
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2:16 = How I look after studying all night for this exam.
Just wanted to say thank you to Khan Academy, you guys are always really helpful when I don't understand something!
WHY DONT THEY EVER JUST GIVE YOU Ka IN THE QUESTION. WHY DO THEY ALWAYS MAKE YOU GO LOOK IT UP. AHHHHHHHHHH I AM ENRAGED
+Michael Buss I feel you bro
Michael Buss EXACTLY
After you do them enough, you begin to just know them by heart.....until after the test that is then....they're gone again.
Normally I do very well in my AP Chem class but titration made me struggle a lot and it killed me. Thank you so much!
thank you khan academy, they explained it much better than my chemistry teacher. He is explaining it in the most difficult way possible. No wonder noone in my class rlly understand his lessons
i have a question.... why did u do the ice table in part a using concentration (M) but in part b u used mol? how do u know when to use what?
I think as long as you're not adding any new reagent to the solution (and hence changing the volume) you can do your ice tables using concentration terms itself. That's why in part (a) he uses concentration terms for the ice tables, because we're currently not adding any NaOH to the solution. But in part (b), we've just added some NaOH, and the volume of the solution is constantly changing during the reaction and so I don't think using concentration terms will be helpful. You can see He does this in the next video also.
Thanks a lot! You're the real MVP
thank you. but suggestion, you should increase the level volume of your voice. I can't hear you in my laptop =.= . good for for people with pc's
+Alyanna Darza buy new laptop bro, I'm okay with it
you must be listening on a quiet area, i'm listening in a classroom
and if you compare this from other videos, his' do has lower volume
Same here. Just use earphones.
thanks a lot... it's really help me to finish my home work
Very helpful. Thanks!
I have a problem with what happens at 9:00. The hydroxide takes half of the acetic acid and turns it into acetate and water. But it seems to me that once that has happened you still need to do an equalibrium reaction between the remaining acetic acid and the remaining acetate.
Great Video
You save lives!
how do I know when to use molarity or moles in my table?? This video uses moles but a previous Khan Academy video I watched, seemingly about the same thing, used molarity/concentration.
There's an error in the calculation of OH-. You have excess OH- so because of this your acid is the limiting reagent and the OH- is the excess. Don't you have to subtract the acid from the OH- to get 0.0400 mol of OH- while the acid is neutralized completely. Or am I wrong?
Thank u😍
So that people might practice, it might be good to include everything that is needed for the problem in the typed out version as if it is a test (i.e. the Ka...)
Best titration vids hands down
Good
Is there any way to know the end point of titration apart from based on the changes in indicator?
if you know the exact amount it takes to reach the endpoint, or if you have a pH reader, stick it in the solution, and slowly drop your solution, once you see a drastic change in pH, thats your endpoint.
how about acids with 2 or more hydrogen(as H2SO4 , H3PO4 ). are their titration goes with this way as well ?, someone said me no, there is some differences..please someone explains. i have exam after 2 days
All the Khan Organic videos are too low in volume.
what if I had less than 0.005 moles of OH-? it would still be a 1-1 reaction and all of the OH would be gone and according to your reaction I would have the same A-/HA thus log would be again one , thus pKa=pH again? , I dont get why especially 0.005 moles of OH bring us to the half equivalence while even if we had less than that it would have been the exact same thing
Good video though
At 13.13 pH should be greater than 7 because salt dissociation generates NaOH which creates oh - ions ..
What happens if the concentration isn't the same. Can you use the H-H equation then?
Ben Toby yeah dude
How do we get the Ka (1.8x 10^-5)
you could get the Ka if you already knew the pH and the conc. but then again, this is a question about finding pH, so Ka is always given, or to trick you, they'd give you Kb, or pKa to see if you know the conversions...lol
Hi
In the first part we calculated the pH, as well as the amount of CH3COO- at equilibrium. But then in the second part we're saying that we start with 0mol CH3COO-???? I am so confused about this process.... we found that at equilibrium because it's a weak acid there's already a certain amount of CH3COO-. why does it go away for the start of the second part?
Yes, indeed. I think this is a mistake. The amount of CH3COO- should be taken into consideration as well imo!
اشكر امك على هالانتاج
Khant acaemy :D
im not sure if part B is correct. all of the acid gets used up.
Lifehack: put it at the speed 1.5 and Khan Academy will speak like normal human 😄
This shit wild
You are supposed to check because you ignored the x in the denominator. Ignore the x, but check the answer by doing (x/0.2) x 100. If the answer is
You don't need to check when the Ka is 3 orders of magnitude larger than the concentration.
Martial Really? I always was taught to always go back and check the "5% rule"
its good to check, nothing wrong it.. plus it literally takes 2 seconds once you fine the [H+]
Where did he get the Ka 1.8x10(-5) from ??
it is one of those things that you should memorize or if your school is nice then they will give you a tabell with Ka and Kb of those common solutions
@@user-ri8ps6cl4w Good to know..Thank you I thought I missed something in one of your steps.
@@ondreiat6674 you are welcome but i am not the guy in the video xD
@@user-ri8ps6cl4w Thank you
Is it really necessary to put 0.200? That's just a waste of space, just write 0.2.
good practice to ALWAYS write in 3sf. In an exam you may get penalised for not writing numbers to 3sf. You need to be clear.
0.2 could be 0.228584023 rounded, so writing 0.200 makes it clearer that it is exactly 0.2
I don't write 0.200 out when I'm doing my work... but nevertheless I probably should. Sig Figs aren't really much of an issue as the AP exam as well as my teacher go by +/- 1 on your answers so yeah... Sig Figs don't really matter. Outside of this I'm sure it would be an issue, but 2.74 isn't that much different than 2.7 in a highschool classroom setting.
CH3COOH(C-CA)=CH3COO-(CA)+H+(CA) HERE IF I M NT WRONG THEN THIS H+ IN (CA) AMOUNT GONNA BE REACTING WITH THE OH_ THAT IS COMING FROM THE FULL DISSOCIATION OF NAOH i.e the whole (ca) amount acetic acid is going to finish but here in this video there is no calculation for (ca) for acetic acid when it is reacting with NAOH
sound is terrible...can you please speak up...thanks
I CANT HEAR A THING
THE AUDIO IS TOO LOW
I can barely hear him and the volume is maxed out.
Neither can I. I use earphones. They work great.