Buffer solution pH calculations | Chemistry | Khan Academy

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  • Опубліковано 1 січ 2025

КОМЕНТАРІ • 220

  • @MynameisJapan
    @MynameisJapan 7 років тому +218

    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
      @rojansayami177 6 років тому +2

      what is E here?

    • @zijian8147
      @zijian8147 6 років тому +22

      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
      @demodema5192 4 роки тому

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

    • @seungjunbaek3410
      @seungjunbaek3410 3 роки тому +6

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

    • @AlgSub22
      @AlgSub22 2 роки тому

      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
    @noahsalomons9687 Рік тому +78

    Petition for Khan academy to receive an honorary Nobel prize

  • @yx_png
    @yx_png 8 років тому +36

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

  • @nicaalyssashanearauag8029
    @nicaalyssashanearauag8029 2 роки тому +4

    UA-cam is my new teacher, thanks a lot!!!!

  • @rofhiwalivingstone1552
    @rofhiwalivingstone1552 8 років тому +32

    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.

  • @farhansadikabtahifh-6158
    @farhansadikabtahifh-6158 9 років тому +82

    Thanks a lot.You Will be always in my heart

    • @UTFT27
      @UTFT27 8 років тому +3

      Do you play rust?

  • @psyp2013
    @psyp2013 2 роки тому +15

    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
    @alessandrac1940 9 місяців тому +1

    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

  • @graceloussakou7345
    @graceloussakou7345 8 років тому +9

    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
      @HoshinoMirai 6 років тому +3

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

  • @johnguillen68
    @johnguillen68 4 роки тому +5

    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
    @wafasaliha1552 7 років тому +6

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

  • @puretwr899
    @puretwr899 4 роки тому +2

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

  • @ShallomJoseph
    @ShallomJoseph Рік тому +1

    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

  • @torchlight3145
    @torchlight3145 8 років тому +44

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

    • @esa2236
      @esa2236 8 років тому +5

      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
      @judypark8977 8 років тому +25

      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
      @n7pako367 7 років тому +1

      That is the one

    • @comoelitamelendez8467
      @comoelitamelendez8467 5 років тому

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

    • @laxsplash78
      @laxsplash78 Рік тому

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

  • @sarahfaress5957
    @sarahfaress5957 6 років тому +4

    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?

  • @NCBrianS
    @NCBrianS 7 років тому +19

    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
      @owensmith3940 3 роки тому +4

      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
      @owensmith3940 3 роки тому +1

      after the first problem, of course

    • @kochosks
      @kochosks Рік тому

      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:))

  • @phumudzomamatsiari787
    @phumudzomamatsiari787 2 роки тому

    brilliant...thnk u so much😇

  • @Ulumanyano07
    @Ulumanyano07 3 місяці тому +1

    Thank you

  • @nancyliu5099
    @nancyliu5099 7 років тому +1

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

  • @brittanykosterlytzky7248
    @brittanykosterlytzky7248 4 роки тому +1

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

  • @victoroduntan
    @victoroduntan 11 місяців тому

    Thank you for saving a live today.

  • @halidsufiyan3663
    @halidsufiyan3663 2 роки тому

    Thank you for basic concept 👍👍👍🧾

  • @aishan.7703
    @aishan.7703 5 років тому +2

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

  • @wildzach
    @wildzach 4 роки тому +19

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

  • @nnennaagidi
    @nnennaagidi 3 роки тому

    Thanks a lot sir u saved my life

  • @saucemonster7137
    @saucemonster7137 Рік тому

    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?

  • @bethanyalvarez521
    @bethanyalvarez521 9 років тому

    This video is a lifesaver

  • @darlingdarling1688
    @darlingdarling1688 8 років тому

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

  • @deflationsmc3301
    @deflationsmc3301 2 роки тому

    great explanation!

  • @sara-kf7vh
    @sara-kf7vh 2 роки тому +1

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

  • @quantum_psi
    @quantum_psi 9 років тому +1

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

    • @skarpengland
      @skarpengland 9 років тому

      +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
      @skarpengland 9 років тому +1

      +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
      @athena7417 8 років тому

      +lasse skarpengland lol this cracked me up

  • @Manikaim
    @Manikaim 7 років тому

    This is really helpful... Thanku Sir

  • @nakatorose7795
    @nakatorose7795 4 роки тому

    Thanks a lot sir all the from Uganda

  • @noemiez2076
    @noemiez2076 9 років тому

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

  • @jeroenfranzel3994
    @jeroenfranzel3994 7 років тому +2

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

  • @mpilow9463
    @mpilow9463 8 років тому

    BIG THANKS TO KHAN ACADEMY

  • @jennifermadrid7221
    @jennifermadrid7221 7 років тому +1

    Extremely helpful, thank you!

  • @oznuryldz8662
    @oznuryldz8662 4 роки тому

    It helped me a lot. Thank you.

  • @sylviakazi5629
    @sylviakazi5629 3 роки тому

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

  • @millionmelesse1518
    @millionmelesse1518 3 роки тому

    Thank you so much sir!

  • @RickRudy
    @RickRudy Рік тому +2

    Behold: The Preventer of Mental Breakdowns

  • @KenyanLondonor
    @KenyanLondonor 8 років тому +6

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

    • @Hellamoody
      @Hellamoody 7 років тому

      constant is usually given

  • @Biraj456
    @Biraj456 7 років тому

    thank u very much sir
    n khann academy

  • @sarasherman2662
    @sarasherman2662 8 років тому

    This is very helpful.

  • @lazolampapela6190
    @lazolampapela6190 7 років тому

    Excellent explanation!!!

  • @divya-c6v
    @divya-c6v Рік тому

    thanku sir 🙏🏻

  • @nataliaurrutia3882
    @nataliaurrutia3882 5 років тому +2

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

  • @aisyahAAZ
    @aisyahAAZ 8 років тому

    LOVED IT

  • @ariaarmstrong2487
    @ariaarmstrong2487 5 років тому +1

    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.

  • @Joydeddeh
    @Joydeddeh Рік тому

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

  • @mimi-jt3yg
    @mimi-jt3yg 6 років тому

    Thanks.It's really helpful!

  • @keeravyas9849
    @keeravyas9849 4 роки тому +2

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

  • @haddadmj96
    @haddadmj96 8 років тому

    Thank you! very useful

  • @DrCrochetMadeWithLove
    @DrCrochetMadeWithLove 3 роки тому

    Thank u so much

  • @sdghalehbandi16
    @sdghalehbandi16 6 років тому

    Thanksssss😍😍😍😍😍

  • @mizramontana8000
    @mizramontana8000 3 роки тому +1

    Lets see who's here in 2021 👍👍😊

  • @ay_yowth
    @ay_yowth 2 роки тому

    thanks

  • @EddieOverwatch
    @EddieOverwatch 2 роки тому

    Thanks my G

  • @xenawu1588
    @xenawu1588 9 років тому

    Great Help!!

  • @marcosmoran8835
    @marcosmoran8835 8 років тому

    extremely helpful, thank you

  • @Hoxgene
    @Hoxgene 4 роки тому

    My god. I finally get this. THANK YOU!

  • @isihummetli4671
    @isihummetli4671 8 років тому

    thank u so much, dude

  • @Dmplivemail
    @Dmplivemail 3 роки тому +2

    I don't need to go to school anymore.

  • @mahmoudemadeddin
    @mahmoudemadeddin Рік тому

    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

  • @GEGE77R6
    @GEGE77R6 7 років тому

    thanks teach

  • @mak1x3
    @mak1x3 10 місяців тому

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

  • @fadhladel247
    @fadhladel247 7 років тому +3

    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.

  • @wryder156
    @wryder156 3 роки тому

    thank you :)

  • @MohamedKhaniHassan
    @MohamedKhaniHassan 2 місяці тому

    Is the value of ka common in every solution

  • @christinemukuni5608
    @christinemukuni5608 8 місяців тому

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

  • @abdulfattahmuhammad3923
    @abdulfattahmuhammad3923 3 роки тому

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

  • @MuhammadShafiq-jc7rx
    @MuhammadShafiq-jc7rx Рік тому

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

  • @nurhusna1407
    @nurhusna1407 4 роки тому

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

  • @aestheticxgirll
    @aestheticxgirll 2 роки тому

    Are we supposed to learn the dissociation constant Ka value?

  • @leoningman62
    @leoningman62 3 роки тому

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

  • @codeonline9577
    @codeonline9577 9 років тому

    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?

  • @martinfederico7269
    @martinfederico7269 8 років тому +3

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

    • @pink_floyd1
      @pink_floyd1 6 років тому

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

  • @fenn8723
    @fenn8723 9 місяців тому

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

  • @rish_official1934
    @rish_official1934 7 років тому +1

    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
      @mariya1941 Рік тому

      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

  • @successwokili
    @successwokili 8 років тому

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

  • @EMNE-zi3
    @EMNE-zi3 Місяць тому

    wow!

  • @hannahhomerski586
    @hannahhomerski586 6 місяців тому

    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?

  • @nabilhaqim7757
    @nabilhaqim7757 3 роки тому

    How to find Ka?
    where? is it constant?

  • @belle8606
    @belle8606 3 роки тому

    Thaaaaaaaaaaank youuuuuuuu

  • @fehimaa1609
    @fehimaa1609 3 роки тому

    10Q teacher

  • @chaks94
    @chaks94 8 років тому

    how do you know which one would act as the acid and which one the base

    • @fatimakhan409
      @fatimakhan409 8 років тому

      i have the same confusion. if you get to know, please remember to tell me too!

    • @ivygithinji7080
      @ivygithinji7080 4 роки тому

      Basically the acid contains the most number of hydrogen while a base contains fewer number of hydrogen

  • @AVADHESHCHAUHAN01091995
    @AVADHESHCHAUHAN01091995 4 роки тому

    Please calculate the ph of Tris buffer solution

  • @ا.حيدرالقيسي
    @ا.حيدرالقيسي 2 роки тому

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

  • @alokneet
    @alokneet 7 років тому +1

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

    • @jaimearredondo787
      @jaimearredondo787 6 років тому

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

  • @paulrodriguez2901
    @paulrodriguez2901 2 роки тому

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

  • @mayaaljundi5279
    @mayaaljundi5279 3 роки тому

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

  • @tylercrandall5909
    @tylercrandall5909 8 років тому

    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

    • @turhanpathan4646
      @turhanpathan4646 8 років тому +1

      +Tyler Crandall Ka should be given to do these types of problems.

    • @ivygithinji7080
      @ivygithinji7080 4 роки тому

      Yeah or basically the ka is constant therefore pKa should be - log Ka

  • @moristhetiger
    @moristhetiger 8 років тому

    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 :-/

  • @mhmdageeli9974
    @mhmdageeli9974 6 років тому

    What is this program you are using in your videos

  • @yejichaeryeong5545
    @yejichaeryeong5545 2 роки тому

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

  • @bradley7118
    @bradley7118 6 років тому

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

  • @yumenoritsu
    @yumenoritsu 9 років тому

    Thank you!! :DD

  • @Userraw674
    @Userraw674 Місяць тому

    why was the ammonia not - if the equation is a-/ha?

  • @awaisabdulrehman
    @awaisabdulrehman 2 роки тому

    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?

  • @patiencesakala8020
    @patiencesakala8020 2 місяці тому +2

    Who is here in 2024 preparing for a test😭