Buffer solution pH calculations | Chemistry | Khan Academy

[@MynameisJapan]

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!

[@rojansayami177]

what is E here?

[@zijian8147]

actually you can still find pH value by using Kb. if you use Kb value, what you find is pOH. so pH=14-pOH

[@demodema5192]

Oh man you are a saviour. ThAnks a bunch bro😭😭😭

[@seungjunbaek3410]

@@rojansayami177 its just the simplified version of 1.0 * 10^-14

[@AlgSub22]

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?

[@noahsalomons9687]

Petition for Khan academy to receive an honorary Nobel prize

[@yx_png]

I had SO many question about Henderson-Hasselbalch equation before I watch this vedio. Thanks a lot. This vedio is really helpful.

[@rofhiwalivingstone1552]

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.

[@nicaalyssashanearauag8029]

KZbin is my new teacher, thanks a lot!!!!

[@psyp2013]

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.

[@alessandrac1940]

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

[@farhansadikabtahifh-6158]

Thanks a lot.You Will be always in my heart

[@UTFT27]

Do you play rust?

[@graceloussakou7345]

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!

[@HoshinoMirai]

I know, and this crams our 2 weeks of stuff into 10 mins of video!

[@johnguillen68]

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

[@wafasaliha1552]

thank you so much it's so helpful I wish u r my teacher

[@puretwr899]

Khan academy is the best of chemistry in my heart. 👏🏼

[@mahmoudemadeddin]

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

[@Ulumanyano07]

Thank you

[@aishan.7703]

For the first time in forever my concepts become crystal clear!

[@sara-kf7vh]

4:25 why does it go to completion? not equilibrium

[@saucemonster7137]

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?

[@wildzach]

"Let's say we already know the Ka..." *My professor laughs maniacally*

[@NomanKhan-bd4st]

lol

[@georgemoustakas4030]

But it's so easy to calculate Ka if you know the concentrations..

[@victoroduntan]

Thank you for saving a live today.

[@mpilow9463]

BIG THANKS TO KHAN ACADEMY

[@DrCrochetMadeWithLove]

Thank u so much

[@brittanykosterlytzky7248]

how would you find the ka if you were using something like sodium benzoate or benzoic acid

[@NCBrianS]

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?

[@owensmith3940]

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

[@owensmith3940]

after the first problem, of course

[@kochosks]

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:))

[@quantum_psi]

Thanks! Think it's starting to make some more sense, although i don't understand why the hydronium gets canceled out/"used up"

[@skarpengland]

+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.

[@skarpengland]

+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

[@athena7417]

+lasse skarpengland lol this cracked me up

[@sarahfaress5957]

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?

[@nnennaagidi]

Thanks a lot sir u saved my life

[@bethanyalvarez521]

This video is a lifesaver

[@phumudzomamatsiari787]

brilliant...thnk u so much😇

[@nancyliu5099]

That's very helpful, thanks for uploading this video!

[@jeroenfranzel3994]

Thank you so much! But what happens when you add 0,24M H3O+ or higher? Because then the Henderson-Hasselbalch equation won't work??

[@keeravyas9849]

I am from India we study this in our final sem of school thank you for the best education you provide to us ❤️

[@halidsufiyan3663]

Thank you for basic concept 👍👍👍🧾

[@sylviakazi5629]

That's why I always rely on @ Khan Academy!! ☝️☝️😎😁

[@ShallomJoseph]

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

[@Hoxgene]

My god. I finally get this. THANK YOU!

[@RickRudy]

Behold: The Preventer of Mental Breakdowns

[@torchlight3145]

Good explanation but how can you come up with the chemical equation? I'm having a hard time creating the equation hELP

[@esa2236]

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.

[@judypark8977]

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?

[@n7pako367]

That is the one

[@comoelitamelendez8467]

@@esa2236 Great side note about when is the best time to use Hendnerson Hasselbach equation. ;)

[@laxsplash78]

Sameee that’s the part that’s messing me up

[@nakatorose7795]

Thanks a lot sir all the from Uganda

[@nurhusna1407]

How to know when NH4+ donate or accept proton? Also why does 4:24 NH3 doesn't have a +?

[@noemiez2076]

thank you very much !! it's very clear now !!

[@mak1x3]

and what about the additional concentration of NH4+ that is produced when NH3 is dissolved?

[@divya-c6v]

thanku sir 🙏🏻

[@oznuryldz8662]

It helped me a lot. Thank you.

[@Joydeddeh]

Thanks for the video. Really helpful but how did u get the Ka value tho?

[@Manikaim]

This is really helpful... Thanku Sir

[@ariaarmstrong2487]

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.

[@ariaarmstrong2487]

BUT WHY

[@jennifermadrid7221]

Extremely helpful, thank you!

[@fenn8723]

Where did u get the 0.16 pH = 9.25 - 0.16? 11:14

[@deflationsmc3301]

great explanation!

[@codeonline9577]

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?

[@nataliaurrutia3882]

holy hell... i actually understand chemistry now; thank you!

[@Biraj456]

thank u very much sir n khann academy

[@ay_yowth]

thanks

[@millionmelesse1518]

Thank you so much sir!

[@moristhetiger]

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 :-/

[@darlingdarling1688]

it was.......so grate full nd awesome

[@abdulfattahmuhammad3923]

Tell us that... how to calculate ph Of buffer with out Hasselbalch equation

[@EddieOverwatch]

Thanks my G

[@christinemukuni5608]

What if we dont already know the Ka value and it is not given in the question as well

[@MuhammadShafiq-jc7rx]

Sir How you put the value of salt on the place of base in numerical 1 . Did not get that please clarify that sir.

[@mimi-jt3yg]

Thanks.It's really helpful!

[@lazolampapela6190]

Excellent explanation!!!

[@aisyahAAZ]

LOVED IT

[@AVADHESHCHAUHAN01091995]

Please calculate the ph of Tris buffer solution

[@sdghalehbandi16]

Thanksssss😍😍😍😍😍

[@haddadmj96]

Thank you! very useful

[@sarasherman2662]

This is very helpful.

[@MohamedKhaniHassan]

Is the value of ka common in every solution

[@aestheticxgirll]

Are we supposed to learn the dissociation constant Ka value?

[@AnimeloveNaruto-i3c]

Hey anyone notice ammonia is a base so that the formula will be P^OH=14-Pka+[salt]/[base]

[@KenyanLondonor]

how do you find pKa or its constant as 5.6 * 10 to power -10

[@Hellamoody]

constant is usually given

[@hannahhomerski586]

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?

[@xenawu1588]

Great Help!!

[@leoningman62]

Why did you plus the second time you use the HH equation? Since it is pka-log(HA/A^-) ? I do not understand

[@isihummetli4671]

thank u so much, dude

[@EMNE-zi3]

wow!

[@marcosmoran8835]

extremely helpful, thank you

[@GEGE77R6]

thanks teach

[@fadhladel247]

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.

[@AmeliaKhairi]

right????

[@hriturajdas224]

Ikr.

[@sccm100]

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.

[@rish_official1934]

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

[@mariya1941]

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

[@Music-zx4po]

What would the pH without the buffer be?

[@wryder156]

thank you :)

[@successwokili]

can this be formulae replace ice table in finding the ph of a buffer solution?

[@paulrodriguez2901]

May I ask, will it be always H30 whenever we compute for addition of base or acid?

[@nabilhaqim7757]

How to find Ka? where? is it constant?

[@awaisabdulrehman]

How can I solve questions like 100.0 mL of 0.025 M formic acid and 0.015 M sodium formate, %0.0 mL of 0.12 M NH3 and 3.50 mL of 1.0 M HCl, and 5.00 g of Na2CO3 and 5.00g of NaHCO3 diluted to 100 L?

[@martinfederico7269]

Why did it react 100% of the added OH- with NH4+ and not according to NH4+ Ka ?

[@pink_floyd1]

Ka is for NH4--NH3+H+ for this equilibrium reaction and not for NH4OH

[@belle8606]

Thaaaaaaaaaaank youuuuuuuu

[@florf619]

Guys I need help, please: Acetic acid is added to water until the ph value reaches 4. What is the total concentration of the added acetic acid?

[@yejichaeryeong5545]

What happen if HCl. Is added to NaCl? Does it affect the NaCl's PH

[@Dmplivemail]

I don't need to go to school anymore.

[@patrickstar7693]

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

[@mayaaljundi5279]

how did we know the Ka value from the begining??

[@mhmdageeli9974]

What is this program you are using in your videos

[@muddassirrehman4623]

How we can directly find the pOH of basic buffer?

[@tyronwood2231]

how do you get the equation? why would it not be NH4+ + H20 ---> NH3 + H30+?

[@bradley7118]

So you’re telling me, we have to look up the pka in order to do this problem?

[@alokneet]

Ph = pka + log [HA]\A...... I think. Numerator or denominator messed up. Kindly correct me if I am wrong.

[@jaimearredondo787]

BioMania it’s A-/HA for the regular equation so pH = pKa + log [A-/HA] A- = base, HA = acid

[@ا.حيدرالقيسي]

we have weak base and salt why you use ka we shoud uses kb