Advice: make sure you have a Ka value. My textbook gave me an intial Kb value and I spent 45 minutes trying to figure out what was wrong. Remember, you can convert Ka to Kb with the expression Ka * Kb =1.0E-14 then simply take the negative log to get Pka! Cheers!
Hi I know its been a year ago. But I just wanted to know that in order to get the Ka, do we must use this expression as CONSTANT? and is it applicable to all related problems of missing the pKa value?
OMG this is the first time doing biochemistry and people who did it before me kept on saying the calculations part, specifically buffers are hard. Clearly they haven't seen your videos. I love how you work out the questions, the way you explain.. THANK YOU FOR HELPING US.
I'm graduate from Bachelor of Art by now I'm preparing for an MCAT test . These video series help me get a good score. Thank for Khan academy and AAMC.
Much appreciated. It was a very comprehensive tutorial. In my class i don't have lectures.. it's all from the textbook so i rely on vids like this to get me through
I get why you divided the 0.05 molNaOH/0.50L to find the molarity, but why is the same not done for the buffer componenets? we have only 0.50 L of solution, so shouldn't the concentration of NH3 be 0.50 L(0.24mol)/1L ? and then add to it the 0.01M of NaOH?
+Sebastian Gulbransen just think about it, the hydrogen binds to NH3 to form NH4+, and the Cloride will dissolve easily with the H2O in the solution. It becomes a spectator ion. this is why we wanna stay with K, Na, and other metals in easily dissolved salts to add to buffers, they dissolve easily.
Im having a hard time between knowing when to do it this way vs using the common ion and ICE method as both problems seem pretty similar at first in terms of approach. Any advice?
they are essentially the same thing. If you notice, in this video, he is creating an ice table. With initial concentration, then the change, and the result
U need ICE for Common ion, so to look for concentration, it is easier to go with common ion. Other than that, u can opt for Henderson-Hasselbalch or buffers equation. Pretty convenient:))
The chemical equation is pH = pKa + log([conjugate base]/[weak acid]). The best time to use this is when you're dealing with a buffer problem involving an acid with its conjugate base or a base with its conjugate acid.
for this question 2:54 why all of this just go with moles and add them to the base and remove from the acid, more easier! pH = 9.25 + lg (0.12+0.005 / 0.1 - 0.005 ) = 9.369
ik i probably wont get a reply for like 5 years but for problems where a strong acid/base is added to a buffer (2:51 and 7:19), shouldn't the concentrations change after the strong base reacts with some of the weak acid (or vice versa) because of the equilibrium? do we just not consider this because the Ka value is usually very low? if it's not very low, should we draw an ICE table after calculating the effect of the strong reactants?
Hi sir how do u get the the ka for the solution ? Is it a constant value. I thought the salt was suppose to be divided by the base? Am not quite sure any one who could pls explain it better pls it's urgent
how do you know which species in solution will be used up? for example in one of the problems you say all of the acid will react so we subtract that concentration in the "change" row of the ice table.
to solve these questions what I think is that we need to be very thorough with all the assumptions... okay so like HCl donates proton to ammonia via irreverservible reactions , sorry if I am being too much but its important as far as I can see :-/
I appreciate the explanations for the chemistry, but there is no calculator allowed on the MCAT, which makes the calculations very difficult. Is there a way to learn it without a calculator?
For the situation where we added acid to our buffer, why is the procedure the same as when we added base to our buffer since in the case of adding acid, the chemical equation on screen is a base dissociation equation which would use Kb, while the Henderson-Hasselbach equation is derived from acid dissociation?
Not sure if its correct but here in Singapore they taught us that the NH3/NH4Cl pair is a congugate base pair, as NH3 acts as a weak base when it ionises in water to give NH4+ and OH-. So our subsequent calculations would include the PKb instead of Pka and finding pOH first then pH, but i still end up with the same values as you do for my final answer. So does NH3/NH4Cl pair act as a congugate base pair or as a congugate acid pair? Thanks
Its still a base buffer the fact that in relative terms its conjugate is termed as acid n it can be put in the same equation using ka still gives the same ans
+Danica Bernas notice that we added an extra OH- which reacts with NH4+ and thus NH4+ is losing moles and forming NH3 and water in the process, Le Chatelier's principle basically
For the base question, how did we know NH3 was the base and NH4 was the acid when doing the Henderson equation? silly question but aside from knowing ammonium is the acid, and seeing how a hydrogen is included in both, how would you differentiate them then.
4:49 All of the hydroxide reacts with ammonium because the kb for nh3 is a really small number meaning that nh4 adheres weakly to its extra proton, right? that's why we assume the hydroxide will rip off all the protons it can from it.
It gets always talked about the addition of a strong acid or strong base... my question would be how to calculate the ph with the addition of a weak acid? Lets say the addition of acetic acid to a phosphate buffer
Petition for Khan academy to receive an honorary Nobel prize
Advice: make sure you have a Ka value. My textbook gave me an intial Kb value and I spent 45 minutes trying to figure out what was wrong. Remember, you can convert Ka to Kb with the expression Ka * Kb =1.0E-14
then simply take the negative log to get Pka!
Cheers!
what is E here?
actually you can still find pH value by using Kb. if you use Kb value, what you find is pOH. so pH=14-pOH
Oh man you are a saviour. ThAnks a bunch bro😭😭😭
@@rojansayami177 its just the simplified version of 1.0 * 10^-14
Hi I know its been a year ago. But I just wanted to know that in order to get the Ka, do we must use this expression as CONSTANT? and is it applicable to all related problems of missing the pKa value?
I had SO many question about Henderson-Hasselbalch equation before I watch this vedio.
Thanks a lot. This vedio is really helpful.
TH-cam is my new teacher, thanks a lot!!!!
OMG this is the first time doing biochemistry and people who did it before me kept on saying the calculations part, specifically buffers are hard. Clearly they haven't seen your videos. I love how you work out the questions, the way you explain.. THANK YOU FOR HELPING US.
I'm graduate from Bachelor of Art by now I'm preparing for an MCAT test . These video series help me get a good score. Thank for Khan academy and AAMC.
So so helpful! thanks, now I don't need to wait for an office appointement with my Biochem professor. Gosh I love our the 21century!
I know, and this crams our 2 weeks of stuff into 10 mins of video!
Much appreciated. It was a very comprehensive tutorial. In my class i don't have lectures.. it's all from the textbook so i rely on vids like this to get me through
I get the equation and sometimes get the right answer but I really never understood what is actually happening until I watched your video. Thank you
Thanks a lot.You Will be always in my heart
Do you play rust?
thank you so much it's so helpful I wish u r my teacher
Khan academy is the best of chemistry in my heart. 👏🏼
For the first time in forever my concepts become crystal clear!
I get why you divided the 0.05 molNaOH/0.50L to find the molarity, but why is the same not done for the buffer componenets? we have only 0.50 L of solution, so shouldn't the concentration of NH3 be 0.50 L(0.24mol)/1L ? and then add to it the 0.01M of NaOH?
Thanks! Think it's starting to make some more sense, although i don't understand why the hydronium gets canceled out/"used up"
+Sebastian Gulbransen just think about it, the hydrogen binds to NH3 to form NH4+, and the Cloride will dissolve easily with the H2O in the solution. It becomes a spectator ion.
this is why we wanna stay with K, Na, and other metals in easily dissolved salts to add to buffers, they dissolve easily.
+Elektro Techniek Haha wow, i do NOT know what i was thinking when i wrote my answer xD
haha i am out, i wrote about spectator ions, whaat the hell xD
+lasse skarpengland lol this cracked me up
Behold: The Preventer of Mental Breakdowns
Thank you
Im having a hard time between knowing when to do it this way vs using the common ion and ICE method as both problems seem pretty similar at first in terms of approach. Any advice?
they are essentially the same thing. If you notice, in this video, he is creating an ice table. With initial concentration, then the change, and the result
after the first problem, of course
U need ICE for Common ion, so to look for concentration, it is easier to go with common ion. Other than that, u can opt for Henderson-Hasselbalch or buffers equation. Pretty convenient:))
Good explanation but how can you come up with the chemical equation? I'm having a hard time creating the equation hELP
The chemical equation is pH = pKa + log([conjugate base]/[weak acid]). The best time to use this is when you're dealing with a buffer problem involving an acid with its conjugate base or a base with its conjugate acid.
I think that they're talking about like the ACTUAL equation. Like the __ + __ --> __ +__. If it isn't, can you help me with finding that equation?
That is the one
@@esa2236 Great side note about when is the best time to use Hendnerson Hasselbach equation. ;)
Sameee that’s the part that’s messing me up
for this question 2:54
why all of this
just go with moles and add them to the base and remove from the acid, more easier!
pH = 9.25 + lg (0.12+0.005 / 0.1 - 0.005 ) = 9.369
Thank you so much! But what happens when you add 0,24M H3O+ or higher? Because then the Henderson-Hasselbalch equation won't work??
4:25 why does it go to completion? not equilibrium
BIG THANKS TO KHAN ACADEMY
how would you find the ka if you were using something like sodium benzoate or benzoic acid
Thank you for saving a live today.
Thanks a lot sir u saved my life
This video is a lifesaver
That's very helpful, thanks for uploading this video!
"Let's say we already know the Ka..."
*My professor laughs maniacally*
lol
But it's so easy to calculate Ka if you know the concentrations..
ik i probably wont get a reply for like 5 years but for problems where a strong acid/base is added to a buffer (2:51 and 7:19), shouldn't the concentrations change after the strong base reacts with some of the weak acid (or vice versa) because of the equilibrium? do we just not consider this because the Ka value is usually very low? if it's not very low, should we draw an ICE table after calculating the effect of the strong reactants?
brilliant...thnk u so much😇
Hi sir how do u get the the ka for the solution ?
Is it a constant value.
I thought the salt was suppose to be divided by the base?
Am not quite sure any one who could pls explain it better pls it's urgent
Thank you for basic concept 👍👍👍🧾
how do you know which species in solution will be used up? for example in one of the problems you say all of the acid will react so we subtract that concentration in the "change" row of the ice table.
BUT WHY
But why you did not use the following equation and you had weak base with it is salt??
PoH=Pkb+log (salt)/ (base)
And thank you alot.
right????
Ikr.
Thanks for the video. Really helpful but how did u get the Ka value tho?
great explanation!
I am from India we study this in our final sem of school thank you for the best education you provide to us ❤️
thank you very much !! it's very clear now !!
Extremely helpful, thank you!
Thanks a lot sir all the from Uganda
It helped me a lot. Thank you.
This is really helpful... Thanku Sir
That's why I always rely on @ Khan Academy!! ☝️☝️😎😁
Thank u so much
it was.......so grate full nd awesome
Thank you so much sir!
how do you find pKa or its constant as 5.6 * 10 to power -10
constant is usually given
Excellent explanation!!!
thank u very much sir
n khann academy
My god. I finally get this. THANK YOU!
thanku sir 🙏🏻
Where did u get the 0.16
pH = 9.25 - 0.16?
11:14
This is very helpful.
Thanks.It's really helpful!
holy hell... i actually understand chemistry now; thank you!
Thank you! very useful
to solve these questions what I think is that we need to be very thorough with all the assumptions... okay so like HCl donates proton to ammonia via irreverservible reactions , sorry if I am being too much but its important as far as I can see :-/
I appreciate the explanations for the chemistry, but there is no calculator allowed on the MCAT, which makes the calculations very difficult. Is there a way to learn it without a calculator?
Great Help!!
I don't need to go to school anymore.
For the situation where we added acid to our buffer, why is the procedure the same as when we added base to our buffer since in the case of adding acid, the chemical equation on screen is a base dissociation equation which would use Kb, while the Henderson-Hasselbach equation is derived from acid dissociation?
How to know when NH4+ donate or accept proton? Also why does 4:24 NH3 doesn't have a +?
Why did it react 100% of the added OH- with NH4+ and not according to NH4+ Ka ?
Ka is for NH4--NH3+H+ for this equilibrium reaction and not for NH4OH
Tell us that... how to calculate ph
Of buffer with out Hasselbalch equation
extremely helpful, thank you
LOVED IT
Not sure if its correct but here in Singapore they taught us that the NH3/NH4Cl pair is a congugate base pair, as NH3 acts as a weak base when it ionises in water to give NH4+ and OH-. So our subsequent calculations would include the PKb instead of Pka and finding pOH first then pH, but i still end up with the same values as you do for my final answer. So does NH3/NH4Cl pair act as a congugate base pair or as a congugate acid pair? Thanks
Its still a base buffer the fact that in relative terms its conjugate is termed as acid n it can be put in the same equation using ka still gives the same ans
thank u so much, dude
Ph = pka + log [HA]\A...... I think. Numerator or denominator messed up. Kindly correct me if I am wrong.
BioMania it’s A-/HA for the regular equation so pH = pKa + log [A-/HA]
A- = base, HA = acid
Thanks my G
What if we dont already know the Ka value and it is not given in the question as well
Are we supposed to learn the dissociation constant Ka value?
and what about the additional concentration of NH4+ that is produced when NH3 is dissolved?
Thanksssss😍😍😍😍😍
What is this program you are using in your videos
thanks
Sir How you put the value of salt on the place of base in numerical 1 . Did not get that please clarify that sir.
why did we add 0.01 in NH3?
and why we subtracted 0.01 from NH4 and OH^- ?
thank you!
+Danica Bernas notice that we added an extra OH- which reacts with NH4+ and thus NH4+ is losing moles and forming NH3 and water in the process, Le Chatelier's principle basically
Thanks! :)
Why did it react 100% of the added OH- with NH4+ and not according to NH4+ Ka ?
Reactions of strong acids and strong bases go to completion.
thanks teach
Lets see who's here in 2021 👍👍😊
For the base question, how did we know NH3 was the base and NH4 was the acid when doing the Henderson equation? silly question but aside from knowing ammonium is the acid, and seeing how a hydrogen is included in both, how would you differentiate them then.
NH3 is a weak base. NH4 is the conjugate of NH3 because it gain a hydrogen.
thank you :)
4:49 All of the hydroxide reacts with ammonium because the kb for nh3 is a really small number meaning that nh4 adheres weakly to its extra proton, right? that's why we assume the hydroxide will rip off all the protons it can from it.
can this be formulae replace ice table in finding the ph of a buffer solution?
May I ask, will it be always H30 whenever we compute for addition of base or acid?
Please calculate the ph of Tris buffer solution
Why did you plus the second time you use the HH equation? Since it is pka-log(HA/A^-) ? I do not understand
What would the pH without the buffer be?
So you’re telling me, we have to look up the pka in order to do this problem?
how do you get the equation? why would it not be NH4+ + H20 ---> NH3 + H30+?
Hey anyone notice ammonia is a base so that the formula will be P^OH=14-Pka+[salt]/[base]
Is the value of ka common in every solution
how do you know which one would act as the acid and which one the base
i have the same confusion. if you get to know, please remember to tell me too!
Basically the acid contains the most number of hydrogen while a base contains fewer number of hydrogen
It gets always talked about the addition of a strong acid or strong base... my question would be how to calculate the ph with the addition of a weak acid? Lets say the addition of acetic acid to a phosphate buffer
How to find Ka?
where? is it constant?
How did you realize acid and base in this question ?
NH4+--->NH3+H+ NH4+ is determined as a wake acid and we can use buffer formula for this system
the acid is the one with more hydrogens typically and the base will have fewer hydrogens pr an OH
Sir, ph=pka+log(salt/acid) but why you used it in different
Exactly
How do you determine the PKA if it's not given in the problem? I'm given the two concentrations like you had just not the Ka or PKA
+Tyler Crandall Ka should be given to do these types of problems.
Yeah or basically the ka is constant therefore pKa should be - log Ka
Why did you put the strong acid on the top? I thought weak acid/base always goes to the top and that conjugate acid/base goes to the bottom?