Buffers

Buffers and the Henderson-Hasselbalch equation

[@RImAPhan]

You always save our asses. Thank you.

[@EvelcyclopS]

why is it, that youtube is now better than any of my chemistry teachers in my entire education history (1st school through to undergraduate)? Great videos mate...

[@Brendanfarr29]

+EvelcyclopS education system is trying their hardest, sorry it's not good enough for you.

[@rae3775]

GOOD LUCK ON THE EXAM EVERYONE

[@maisoonpharmad3320]

Rae Amiiiin

[@shoaeb19]

I never knew buffers can be learnt without wanting to stop. I thank you Sal, for helping us helpless students.

[@MarkIsJustKidding]

Thank you so much! Good luck to everyone who is having exam tomorrow

[@kuraudo9700]

All i need is a text book and this mans tutorials, then i'll be an A+ student.

[@haakonjohansen]

What you're doing for the education of the world is so great. You deserve a universal medal!

[@denzelnolet]

It's worthy to note that for the buffer to actual "work," you need significant amounts of the weak acid/base and its conjugate. So if you're adding a large amount of HCl to a buffer with small amounts of the buffering components (ie a weak base/acid with it's conj salt), the pH will change significantly.

[@RaoufHasan]

love your videos, i always search khan academy for chemistry and biology questions i have. Thank you so much

[@awaizy]

Awesome. Your explanation at the end was done well, too. I was wondering about that in class today and you cleared my doubts.

[@sonyasammarco5672]

This has been breathtakingly helpful, thank you for uploading!

[@ryttut]

You make pure gold, easy-digestible gold. Really, my chemistry book makes my head-ache while this easy to digest even while tired

[@SRGjager]

Water will add H+ and OH- ions, which will use up part of the buffering capacity. This means that when an acid or base is added, there is less capability for the buffer to act.

[@XfactorfreakX12]

you have NOOOO idea how much i love you.

[@Skateboardfreakist]

Suddenly Chemics is getting interesting to me

[@elijahbarela5137]

I was actually looking for a solution for internet buffer lol

[@CharTiger]

Same😂😂

[@mpendulojayiya6203]

Thumbs up to your videos! Not even my chemistry lecturers explain the way you do. :)

[@Caradepato]

KHAAAAAN! gread videos. Very usefull, and they are a big help in my studying for my may IB exams..

[@jimmibooyah]

i cant thank you enough! literally saved my academic life.

[@blubberinghumdinger35]

Ikr, his videos make me do that. And that too chemistry! can't stop!

[@gnosiophobe]

i don't really remember any of the log laws, but there's a negative sign in front of the log (xy). so if log (xy) = log x + log y then -log (xy) = log x - log y

[@josvrijdag5089]

Personally, I think his voice is quite enjoyable.

[@seafoamren]

this is helping me pass chemistry. THANK YOU!!!

[@QNeville]

5 stars are getting played out so I feel like coming over there and high fiveing you. THANK YOU!

[@NaturallyHigh]

@stretchyrubberbands It won't go down because water will dissociate to form H+.. and it won't go up because it's also forming OH-... I think...

[@shellshockcake]

thank you soo much! great overview, just what I needed!

[@Hadoyful]

Thank you very much you really saved my life ..

[@zainabsyedahmed5194]

thanksz for the video...it relived my concept of buffer :)

[@mohamedownalla5809]

you don't know how much I love you

[@Bugz904]

OK as useful as that was for understanding, some of the words used wouldn't be awarded marks , especially with my exam board. They want you to say a buffer is a solution that minimizes or resists changes in pH. The mark scheme says you need a series of equations which you demonstrated well. And also to say what happens when you add an acid ( [H+] increases reacts with negatively charged base from acid dissociation. Etc. After doing a few questions its more clear what they are after.

[@minerman6767]

@imalazykid the OH- is a strong base meaning it is far more likely to react with that instead of the A-

[@gekkaa]

this movie saved me 4 lessons chemistry _o_

[@athena7417]

Its really helpful.. thanks ^.^

[@josephbenobasa]

Thx helps me for my upcoming physical chemistry lab

[@ginahuh8538]

you saved my life.. thank you

[@ilikebananasorange]

Oh, my God. This guy is SO funny!! :')

[@Abhii0]

"Straight up water" awesome :)

[@jpizz7]

Chem final today... Wish me luck!

[@StarShooting03]

+James Ward good luck!

[@hi-z]

khan academy, you might just save my life

[@deveshrawat1334]

it is getting helpful for me

[@melrakan]

React with the hydrogen and eat it all up! React with the hydrogen and eat it all up!

[@artiomgretsky6951]

NaOH isn't a molecule Khan. It's an ionic compound. Molecules are typically held together by covalent bonds.

[@claytonchristudass161]

Artiom Gretsky face palm

[@oridniv]

you have a sexy voice...it's nice to have it in memory when not stalling on my finals..thanx so much

[@imalazykid]

@7:30, sal mentions that the proton from HCl would react with OH- and neutralizes the reaction. And then, the reaction shifts to the right to make up for lost Hydroxide (OH-) ions. I get that. But isn't there a chance that same proton can also react with the A- too instead of OH-? And lets say, it did react with the A-, then according to Le chatelier's principle, wouldn't the reaction shift to the right, creating more OH- and thus increasing the pH?

[@PallavShinghal]

17:21 pH goes UP not DOWN

[@spycowboi2190]

Thanks!

[@mathieuvpone]

thank you very much

[@8ballsix27]

this could have helped as review before my final exam...

[@isabellaalvarez2548]

i love you thank you.!!!

[@purplepick1]

Thank you for explaining to me WHYYYYYY!!!!!!!!! ahhhhhhhhh SO FRICKEN HAPPY!!!! HHAHAHAHHA

[@Be1smaht]

thank you soooooooo much

[@sciencenerd7639]

thank you

[@GrayGhosting]

Love your videos, I am learning much in my retirement years. But simply dropping the units m/L from Ka at time 10:35 bothered me and since I don't believe log of a dimensional quantity is mathematically defined, I attempted to see if I could derive the H-H equation without violating mathematics or at least within some more explicit definitions. Here is the math: for compactness of notation: - anything inside [ ] is in m/L (moles/liter). - if a variable or constant is not inside [ ] then it is dimensionless, unless otherwise explicitly noted. - I drop the ionic charges. Your initial equation for equilibrium constant at 10:35 should have been written [Ka] * [HA]/[A] = [H] Note units on left and right side of my equation match, where yours do not at time 10:35 Now to get rid of the dimensions m/L from Ka so that we can take the log of it: multiply both sides by L/m Ka * [HA]/[A] = H take the -log of both sides: pKa - log([HA]/[A]) = -log H To get pH as defined within the Chem community we need to get back the dimensions for H by multiplying by m/L. This is equivalent to adding the log of m/L, then of course doing it on both sides of the eq. pKa - log([HA]/[A]) + log(m/L) = -log(H) + log(m/L) = -log[H] ...... -log[H]=pH by definition only. Then since pH is dimensionless and pH = -log(H * 1m/L) = -log(H)+log(1m/L), by definition log(1m/L) must=0 so... pKa - log([HA]/[A]) = pH or pKa + log( [A]/[HA] ) = pH since [HA]/[A] is dimensionless we get the H-H equation balanced with all terms dimensionless So only by defining log(m/L)=0 the H-H equation can be derived without erroneous math.... at least to my satisfaction.

[@GrayGhosting]

The mathematical reason your "amazing" conclusion at the end of the video was wrong is that a log of a dimensional quantity is undefined. So the log(moles/moles) is undefined and therefore you can not interpret that equation to say that the ph is independent of volume and only dependent on the ratio of moles of constituents. To be defined you need to remove the dimensions, i.e. divide m/L by m/L. This leaves no opportunity for the faulty interpretation. Or if one followed my math above one would see this equation.. pKa + log( [A]/[HA] ) = pH ... is only correct if Log(m/L) = 0, and this can only be correct if m/L is dimensionless. After consideration one can see it is in fact dimensionless when we realize that Liters are just acting as a scale factor for moles. That is to say, for example, we could write the very first equation with the ratio of atoms of H to atoms of H2O, which give a dimensionless ratio instead of the dimension of moles to liters. Following that approach throughout there would have been no opportunity in the last equation to remove the dimension of volume.

[@davidkim0678]

i guess Sal's good at explaining concepts but i wish he would solve some more complicated questions...maybe University style questions

[@matthewchoi1092]

now i understand what buffer is...... one day before ap exam..... haha........

[@dorkychamp3269]

Wait so when equilibrium is reastiblished H+ will increase because of water right?

[@jerryc2691]

As true as that may be, you may want to watch more of "there" videos on grammar.

[@angelicas9914]

This is the 2022 comment you’ve been looking for 😅🥰

[@BereketAbHailemariam-wq8jp]

I wish to meet u one day

[@Zzz230210]

thanks a lot u explain it very well (Y)

[@theyeow92]

why does the buffer capability decreases as the buffer solution is diluted with water

[@ThePeterDislikeShow]

Really? You can make an acid with ammonium ion?

[@overtonwindowsill5218]

FortNikitaBullion yah

[@l0veforpi3]

dude totally !

[@TehJayPee]

AP Chem Test tomorrow. Lets Go!!!

[@MissAshleyWen]

Me too! Haha good luck

[@akshaykamath7719]

hey buddy u told that because of NaoH the H+ get disapeared and the HA wl dissociate to undo the loss of hydrogen ,but my question is, what about the A ? IT GOES ON INCREASING and it should under go backward rxn to balance the A .then y doesnt it do a bkward rxn?

[@hamidnia101]

go to 13:04 -log(xy) = - log(x) - log(y)

[@MRFARMS2]

I got a question!! Aint the conjugate base of a weak acid strong??? not weak??

[@FarrahGreye]

ur awesome

[@yoochanhhong7223]

thanks KHAN

[@hamidnia101]

makes sense to me

[@MrBlack69Ops]

Thats why youre watching the vid

[@badshabz1]

you know when you have the log (ha/a) is that not just log(ha) - log(a), could you write it that way?

[@mkruthik6316]

Actually the dilution won't affect the ''log'' term (a part which you got right but again you got wrong in the mid video comment) but instead it will affect the pka of the solution.

[@lenilaud4214]

Hi everyone! I have been looking around the internet for articles discussing how much buffer to add into a solution but all I've found are calculations on buffer preparation. I am preparing an antibiotic solution (synthetic antibiotic-contaminated wastewater) which will be treated afterwards at constant pH. I will be adding NaOH and H2SO4 for the initial pH adjustments, then phosphate buffers to maintain the set pH. My problem is, I have no idea how much of the NaOH, H2SO4, and buffer will I add to my solution. Could you please help me with this? I'd really, really appreciate it. Thank you very much!

[@Troynjk]

thx

[@f.mmaisha1607]

thanks

[@jaikohaal]

i have it in 1.5 hours....cramming

[@sneakysnicker]

my teacher told me to take the day off and relax for the ap test tomorrow. well I guess that's not happening.

[@bhrzali]

Sal the god!

[@weihaihe9839]

MMmmmmm..so khan is 30 years old.....

[@shubhbisht9363]

42 in real life

[@TheChuggtom88]

you too?!?! haha khanacademy all nightt!!

[@rajatdixit007]

Khan-jugates !

[@mustanglp50]

ok WTF i go to the website and dont see this video under his playlist for chemistry

[@snatch147at85]

@africanchicaful WORD! Dude I read the chapter 3 times and thought a buffer was a vitamin supplement. After this video? I tell the exam......come at me bro! but please no hard questions

[@ddaannssttaarr]

I get it now

[@overtonwindowsill5218]

That statement at end is only true if the chemical is in excess.

[@overtonwindowsill5218]

Oh right he addressed it

[@oshklhgaewrf3309]

Yes I KHAN! :)

[@jlin9268]

there no such thing as a ph 0 btw

[@krisnika1123]

Comment on the edit at the end: wrong again. Initially, the math was correct (getting rid of the volume), but its interpretation was wrong... and the whole math thing was useless/needless anyway. The edit (the comments) were wrong.... when trying to discredit the 'correct' math with the example of diluting the solution with H2O - of course the pH will change due to Chatelier's principle. Your initial equation relating pH, pKa, and number of moles, is correct and APPLIES only at a particular instant/concentrations.

[@Usman9217119]

kris nika he is not gonna reply because the video post on 2009

[@thelearner3962]

The case in which adding infinitely excess water violates the assumptions made during derivation of HH equation, that's why we won't use HH for that case. That's all :p

[@dtoxwOw]

Didn't he make a mistake by using log instead of lg. There's a big difference ...

[@melrakan]

+dtoxwOw It doesn't matter. Since this form of calculations only uses the base logarithm, lg and log are interchangeable.

[@Jarrodmontelius]

my final is thursday!! hahahahha!!

[@TheNilan92]

Déjà vu ! o.O :D

[@Be1smaht]

oh you kids...lol

[@skateandtennis]

Vitsmin supplement??? lmfao!! :P

[@nurhdyh]

UT CHEMISTRY TODAY. #RPLIFE HAHAHAH

[@backhandable]

3 teachers lost there jobs in the making of this video.

[@abbyhooker]

ACKWEEUS

[@maisoonpharmad3320]

Can U put Arabic translation pls 😊..

[@habeebhabib3160]

ههههههههههههه

[@thepokigur]

Are you single?!?