Buffer solution pH calculations | Chemistry | Khan Academy

Поделиться
HTML-код
  • Опубликовано: 30 сен 2024
  • Example of calculating the pH of solution that is 1.00 M acetic acid and 1.00 M sodium acetate using ICE table. Another example of calculating pH of a solution that is 0.15 M ammonia and 0.35 M ammonium nitrate.
    Watch the next lesson: www.khanacadem...
    Missed the previous lesson? www.khanacadem...
    Chemistry on Khan Academy: Did you know that everything is made out of chemicals? Chemistry is the study of matter: its composition, properties, and reactivity. This material roughly covers a first-year high school or college course, and a good understanding of algebra is helpful.
    About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content.
    For free. For everyone. Forever. #YouCanLearnAnything
    Subscribe to Khan Academy’s Chemistry channel: / channel
    Subscribe to Khan Academy: www.youtube.co...

Комментарии • 212

  • @noahsalomons9687
    @noahsalomons9687 Год назад +61

    Petition for Khan academy to receive an honorary Nobel prize

  • @MynameisJapan
    @MynameisJapan 7 лет назад +202

    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 лет назад +2

      what is E here?

    • @zijian8147
      @zijian8147 6 лет назад +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 года назад

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

    • @seungjunbaek3410
      @seungjunbaek3410 3 года назад +6

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

    • @AlgSub22
      @AlgSub22 2 года назад

      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?

  • @RickRudy
    @RickRudy Год назад +2

    Behold: The Preventer of Mental Breakdowns

  • @yx_png
    @yx_png 8 лет назад +35

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

  • @sarahfaress5957
    @sarahfaress5957 6 лет назад +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?

  • @ShallomJoseph
    @ShallomJoseph 9 месяцев назад +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

  • @farhansadikabtahifh-6158
    @farhansadikabtahifh-6158 8 лет назад +82

    Thanks a lot.You Will be always in my heart

    • @UTFT27
      @UTFT27 8 лет назад +3

      Do you play rust?

  • @torchlight3145
    @torchlight3145 8 лет назад +43

    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 лет назад +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 7 лет назад +24

      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 лет назад +1

      That is the one

    • @comoelitamelendez8467
      @comoelitamelendez8467 5 лет назад

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

    • @laxsplash78
      @laxsplash78 Год назад

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

  • @psyp2013
    @psyp2013 2 года назад +14

    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.

  • @cherryredinguito7277
    @cherryredinguito7277 11 месяцев назад +1

    Hello sir can you help me understand this problem. The answer was 1.6 x 10^-2. And I have solve this a hundred times but I still don't arrive on the answer
    This is the problem
    the solubility product constant for ce(io3)3 is 3.2 x 10^-10. what is the ce ^3+ concentration in a solution prepared by mixing 50.00 ml of 0.0450 M ce ^3+ with 50.00 ml of 0.045 M io3^-

  • @graceloussakou7345
    @graceloussakou7345 8 лет назад +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 лет назад +3

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

  • @johnguillen68
    @johnguillen68 3 года назад +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

  • @fadhladel247
    @fadhladel247 6 лет назад +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.

  • @rofhiwalivingstone1552
    @rofhiwalivingstone1552 8 лет назад +30

    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
    @nicaalyssashanearauag8029 2 года назад +4

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

  • @wildzach
    @wildzach 4 года назад +19

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

  • @Dmplivemail
    @Dmplivemail 3 года назад +1

    I don't need to go to school anymore.

  • @wafasaliha1552
    @wafasaliha1552 7 лет назад +6

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

  • @NCBrianS
    @NCBrianS 7 лет назад +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 года назад +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 года назад +1

      after the first problem, of course

    • @kochosks
      @kochosks Год назад

      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:))

  • @KenyanLondonor
    @KenyanLondonor 7 лет назад +6

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

    • @Hellamoody
      @Hellamoody 7 лет назад

      constant is usually given

  • @sara-kf7vh
    @sara-kf7vh Год назад +1

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

  • @mizramontana8000
    @mizramontana8000 3 года назад +1

    Lets see who's here in 2021 👍👍😊

  • @brittanykosterlytzky7248
    @brittanykosterlytzky7248 4 года назад +1

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

  • @jeroenfranzel3994
    @jeroenfranzel3994 6 лет назад +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??

  • @billipbillip8411
    @billipbillip8411 5 лет назад +7

    Perfect explanation, thank you!

  • @martinfederico7269
    @martinfederico7269 7 лет назад +3

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

    • @pink_floyd1
      @pink_floyd1 6 лет назад

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

  • @alokneet
    @alokneet 6 лет назад +1

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

    • @jaimearredondo787
      @jaimearredondo787 6 лет назад

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

  • @cherryredinguito7277
    @cherryredinguito7277 11 месяцев назад

    Hello sir can you help me understand this problem. The answer was 1.6 x 10^-2. And I have solve this a hundred times but I still don't arrive on the answer
    This is the problem
    the solubility product constant for ce(io3)3 is 3.2 x 10^-10. what is the ce ^3+ concentration in a solution prepared by mixing 50.00 ml of 0.0450 M ce ^3+ with 50.00 ml of 0.045 M io3^-

  • @hannahhomerski586
    @hannahhomerski586 2 месяца назад

    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?

  • @mahmoudemadeddin
    @mahmoudemadeddin Год назад

    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

  • @puretwr899
    @puretwr899 4 года назад +2

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

  • @rish_official1934
    @rish_official1934 7 лет назад +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 Год назад

      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

  • @dr.p.sundarakumar9006
    @dr.p.sundarakumar9006 5 лет назад +1

    Sir, ph=pka+log(salt/acid) but why you used it in different

  • @mak1x3
    @mak1x3 7 месяцев назад

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

  • @ariaarmstrong2487
    @ariaarmstrong2487 5 лет назад +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.

  • @HamzaAfzal-xm7hn
    @HamzaAfzal-xm7hn Месяц назад

    hey from where you belong from which country tell plz🥰😃😃

  • @mikeyyychang8402
    @mikeyyychang8402 7 лет назад +1

    i think last one is wrong sice Ka=salt/acid (not too sure

  • @christinemukuni5608
    @christinemukuni5608 5 месяцев назад

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

  • @thatsnotme3235
    @thatsnotme3235 5 лет назад

    Do you use a mouse or finger pad to do these? Whenever I try writing with my mouse, it looks like shit

  • @awaisabdulrehman
    @awaisabdulrehman 2 года назад

    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?

  • @aestheticxgirll
    @aestheticxgirll 2 года назад

    Are we supposed to learn the dissociation constant Ka value?

  • @roseblaise7485
    @roseblaise7485 4 года назад

    Why did we not use the pOH for that last problem???????

  • @fenn8723
    @fenn8723 6 месяцев назад

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

  • @quantum_psi
    @quantum_psi 8 лет назад +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 8 лет назад

      +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 8 лет назад +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 лет назад

      +lasse skarpengland lol this cracked me up

  • @MunibaSaleheen
    @MunibaSaleheen 8 месяцев назад

    Where was pka of 9.25 given???

  • @mayaaljundi5279
    @mayaaljundi5279 3 года назад

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

  • @diver4573
    @diver4573 6 лет назад

    I would like to know the preparation of
    0.2 M phosphate buffer preparation ph 8 .
    Components we have are sodium phosphate dibasic and sodium dihydrogen phosphate
    ???????~

  • @saucemonster7137
    @saucemonster7137 Год назад

    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?

  • @viplami
    @viplami 7 лет назад

    Answer this... what is the pOH of a buffer solution if 6grams of ch3cooh and 8.2 grams of ch3coona are disolved in one liter of water ... pKa 4.75

  • @alessandrac1940
    @alessandrac1940 6 месяцев назад

    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

  • @florf619
    @florf619 2 года назад

    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?

  • @leoningman62
    @leoningman62 3 года назад

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

  • @patrickstar7693
    @patrickstar7693 2 года назад

    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

  • @MuhammadShafiq-jc7rx
    @MuhammadShafiq-jc7rx Год назад

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

  • @nancyliu5099
    @nancyliu5099 7 лет назад +1

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

  • @moristhetiger
    @moristhetiger 7 лет назад

    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 :-/

  • @fabiolamorales4306
    @fabiolamorales4306 4 года назад

    Does anyone know why for the last exercise he subtracted the log B/A instead of adding?

    • @hisyamhasbi8257
      @hisyamhasbi8257 3 года назад

      Kinda late here but if anyone is also confused, the result of log 0.18/0.26 is a negative number, that's why it's substracted

  • @SK8TESS
    @SK8TESS 7 лет назад

    Why are you using an ICE table? I thought you're supposed to use BCA tables for buffers?

  • @da-dl1cl
    @da-dl1cl 7 лет назад

    My problem is that I only have the concentration of strong base before starting the solution, and nothing else. I can't do the consommation thing because of that.

  • @paulrodriguez2901
    @paulrodriguez2901 2 года назад

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

  • @Lilliz91
    @Lilliz91 5 лет назад

    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?

  • @ا.حيدرالقيسي
    @ا.حيدرالقيسي 2 года назад

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

  • @bhaskarpandey8586
    @bhaskarpandey8586 6 лет назад

    Why the concentration of ammonium ion is equal to that of ammonium chloride, does it completely ionize in water.

  • @danicabernas2570
    @danicabernas2570 9 лет назад +1

    why did we add 0.01 in NH3?
    and why we subtracted 0.01 from NH4 and OH^- ?
    thank you!

    • @mzwenkosimbatha3329
      @mzwenkosimbatha3329 9 лет назад +1

      +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

    • @danicabernas2570
      @danicabernas2570 9 лет назад

      Thanks! :)

    • @martinfederico7269
      @martinfederico7269 7 лет назад +1

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

    • @babubassa8434
      @babubassa8434 7 лет назад

      Reactions of strong acids and strong bases go to completion.

  • @joyn.deddeh3291
    @joyn.deddeh3291 10 месяцев назад

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

  • @nurhusna1407
    @nurhusna1407 4 года назад

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

  • @abdulfattahmuhammad3923
    @abdulfattahmuhammad3923 3 года назад

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

  • @Music-zx4po
    @Music-zx4po Год назад

    What would the pH without the buffer be?

  • @helihel1323
    @helihel1323 7 лет назад +1

    How did you realize acid and base in this question ?

    • @pink_floyd1
      @pink_floyd1 6 лет назад

      NH4+--->NH3+H+ NH4+ is determined as a wake acid and we can use buffer formula for this system

    • @rivaldoalves8841
      @rivaldoalves8841 5 лет назад +2

      the acid is the one with more hydrogens typically and the base will have fewer hydrogens pr an OH

  • @sylviakazi5629
    @sylviakazi5629 3 года назад

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

  • @yejichaeryeong5545
    @yejichaeryeong5545 2 года назад

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

  • @codeonline9577
    @codeonline9577 8 лет назад

    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?

  • @nabilhaqim7757
    @nabilhaqim7757 3 года назад

    How to find Ka?
    where? is it constant?

  • @bradley7118
    @bradley7118 6 лет назад

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

  • @mithileshkumar5602
    @mithileshkumar5602 7 лет назад

    Why does salt hydrolysis doesn't take place in buffers ?

  • @tyronwood2231
    @tyronwood2231 6 лет назад

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

  • @successwokili
    @successwokili 8 лет назад

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

  • @muddassirrehman4623
    @muddassirrehman4623 7 лет назад

    How we can directly find the pOH of basic buffer?

  • @BantersM
    @BantersM 6 лет назад

    How can he use his calculator on his pc? That's neat and handy.

  • @saddamhussain9618
    @saddamhussain9618 6 лет назад

    salam sir. sir you didnt mention the volume of 0.005 M NaOH and also of acid HCl

  • @amberpawlowski7619
    @amberpawlowski7619 7 лет назад +3

    how do you know Ka??????

    • @albanaa6126
      @albanaa6126 7 лет назад +4

      Amber Stallings Ka is given,you don't have to know it.

    • @HoshinoMirai
      @HoshinoMirai 6 лет назад +2

      Ka is usually given

  • @abhinav7777777
    @abhinav7777777 8 лет назад

    please anybody help me on this.... the questions .... are about a base ... but you use pka (disociation constant for acid.... shouldnt it just have a pkb and not pka ...... which leads me to conclude .... can i compound have both pka and pkb ?

    • @darlingdarling1688
      @darlingdarling1688 8 лет назад

      you should have something to support your pkb bcz chemistry needs reasoning bt simple advice use whwt you have been told not your hypothesis...meib is still going to be discovered....

  • @Freedomboy006
    @Freedomboy006 7 лет назад +15

    pretty sure you are supposed to use moles rather than molarity for this problem.. answer should be around 7.97

    • @AVADHESHCHAUHAN01091995
      @AVADHESHCHAUHAN01091995 Год назад +7

      You can use mole or Molarity, answer will be same in both cases. Thanks 🙏🙏

  • @AVADHESHCHAUHAN01091995
    @AVADHESHCHAUHAN01091995 4 года назад

    Please calculate the ph of Tris buffer solution

  • @Hoxgene
    @Hoxgene 3 года назад

    My god. I finally get this. THANK YOU!

  • @victoroduntan
    @victoroduntan 8 месяцев назад

    Thank you for saving a live today.

  • @halidsufiyan3663
    @halidsufiyan3663 2 года назад

    Thank you for basic concept 👍👍👍🧾

  • @mhmdageeli9974
    @mhmdageeli9974 6 лет назад

    What is this program you are using in your videos

  • @alicemuyeleka2695
    @alicemuyeleka2695 Год назад

    Thank you very much this is very educative

  • @DrCrochetMadeWithLove
    @DrCrochetMadeWithLove 3 года назад

    Thank u so much

  • @iloveparamore123100
    @iloveparamore123100 6 лет назад

    i just really dont understand how to write the chemical equations by looking at the question.. If we are working with HCl i dont get why it wasnt apart of the chemical reaction

    • @mrfatboy327
      @mrfatboy327 6 лет назад

      HCl separates and combines to H2O to make H3O. I was confused about this too, but they just remove parts of the equation that dont really matter.

  • @vityamv
    @vityamv 8 лет назад

    Why in NaOH reaction we got NA + OH but in the HCl reaction we say that the H+ desolves in water and becomes H30+? thaks ahead

    • @skarpengland
      @skarpengland 8 лет назад

      +Vitya Stepanenco Think it's just to show us how the protolysis works :) Hydroxide binds a Hydrogen to form H2O, and Hydrogen in acids go out and make Oxonium, which makes sour pH solutions

  • @najma2613
    @najma2613 8 лет назад

    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.

    • @CHRISTIANZEGARRA2018
      @CHRISTIANZEGARRA2018 8 лет назад

      NH3 is a weak base. NH4 is the conjugate of NH3 because it gain a hydrogen.

  • @jennifermadrid7221
    @jennifermadrid7221 7 лет назад +1

    Extremely helpful, thank you!

  • @SnailLiquor00
    @SnailLiquor00 4 года назад

    please show harder problems

  • @Ulumanyano07
    @Ulumanyano07 10 дней назад

    Thank you

  • @ox4e
    @ox4e 10 месяцев назад

    ماكو سوادس؟ 😔

  • @AhsanKhan-eb2zb
    @AhsanKhan-eb2zb 8 лет назад +1

    Nice

  • @belle8606
    @belle8606 3 года назад

    Thaaaaaaaaaaank youuuuuuuu

  • @phumudzomamatsiari787
    @phumudzomamatsiari787 2 года назад

    brilliant...thnk u so much😇