Polyprotic Acid Base Equilibria Problems, pH Calculations Given Ka1, Ka2 & Ka3 - Ice Tables
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- Опубліковано 27 лис 2017
- This acid base equilibrium video tutorial explains how to calculate the pH of a polyprotic acid using ice tables and number lines. It discusses how to calculate the pH of a diprotic acid like H2SO4. It explains what to do when given a concentrated sulfuric acid compared to a dilute solution. In addition, it mentions how to calculate the pH of a triprotic acid like H3PO4 and how to calculate the equilibrium concentrations of H3PO4, H2PO4-, HPO4 2-, and PO4 3- given Ka1, Ka2, and Ka3. In addition, it explains how to calculate the pH of a solution that consist of a mixture of H2PO4- and HPO4 2- which is basically a buffer solution and how to calculate the pH when one of those species predominate using a simple formula. This acids and bases chemistry video contains plenty of examples and practice problems.
Acids and Bases - Introduction:
• Acids and Bases - Basi...
The 7 Strong Acids to Memorize:
• How To Memorize The St...
Conjugate Acid-Base Pairs:
• Conjugate Acid Base Pa...
pH and pOH Calculations:
• pH, pOH, H3O+, OH-, Kw...
Estimate The pH Without a Calculator:
• How To Calculate The p...
_______________________________
Autoionization of Water - Kw:
• AutoIonization of Wate...
Which Acid Is Stronger?
• Acid Base Strength - W...
Acidic, Basic, & Neutral Salts:
• Acidic, Basic, and Neu...
pH of Weak Acids:
• pH of Weak Acids and B...
Buffer Solutions:
• Buffer Solutions
_________________________________
Polyprotic Acid Base Equilibria:
• Polyprotic Acid Base E...
Acid Base Titration Curves:
• Acid Base Titration Cu...
Acids and Bases - Practice Test:
• Acids and Bases Review...
Ksp - Molar Solubility & Ice Tables:
• Ksp - Molar Solubility...
Complex Ion Equilibria:
• Complex Ion Equilibria...
___________________________________
Gibbs Free Energy, Entropy & Enthalpy:
• Gibbs Free Energy - En...
Entropy - 2nd Law of Thermodynamics:
• Entropy - 2nd Law of T...
Electrochemistry Practice Problems:
• Electrochemistry Pract...
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professor caught me off guard when he did a TRIPROTIC acid on the final exam...i'm retaking the class and not gonna let him fool me again lol
I've commented once already...but I feel like commenting once again...I just can't appreciate what you're doing for us....thanks very much and may God bless you so much...love you @organic chemistry tutor...you always makes the things clearer and clearer again😍
If I ever learned the trick that pH = (pKa1 + pKa2)/2, I had certainly forgotten it. When I watched that part of the video it really blew my mind. Thank you so much for sharing!
Why did I waste time in the last four weeks of lecture when this video exists? Thank you so much; you explained everything so much better than my professor!
Hey how'd the rest of your class go?
error at 20:11. should be ka2 = 6.2 x 10^-8 = (.1188 + x)(x) / (.1188 - x) . the error does not appear in the final calculation, because the number rounds to 6.2 x 10 ^ -8 for x. but if you are trying to follow along algebraically, this is definitely an error.
Literally spent over an hour on this problem, trying to figure it out and then once I found your video all it took was watching it only partly through, once, and I got the right answer. Thank you ;__;
Oh my god, I'm crying rn, thank you so much. Always have my back in chemistry.
Thanks, I was stuck on this for a while. I believe most of us assumed that the Ka3 value would equal [P04 3-] similarly to how Ka2 equals [HP04 2-], and missed plugging in the value for [H30+] in the Ka3 equation. (for Ionization of Triprotic acids)
You are the most beautiful person in the world right now... I learned more in the first 10 minutes of your video than I did all this semester.
You're literally my hero. I'm studying for the chem SAT 2 and you have taught me all the concepts necessary for the test while also providing effective practice problems for me to learn from. Love you fam.
I know it's been two years, but how did the SAT go!?
@@PunmasterSTP Funny story, COVID hit before my test date and it actually got cancelled. Then, the college board started phasing out the test entirely and I did not end up taking it but I was most definitely prepared.
@@benjacobson6160 Wow I’m really sorry to hear that, and I hope that your plans have gone well since then!
Thanks a lot!!! After watching a lot videos that made this simple thing a rocket science, finally found ur vid. You made my life easier as always!!! ❤❤❤💙💙💙
may God richly bless you...may he bless u with more knowledge and wisdom
CHRISTIAN MENSAH
THIS IS A CHEMISTRY CHANNEL BRING GOD OUT OF THIS!!!!!!!
so he can simply share it all with us
@@gartyqam Hey Derp, God invented chemistry, so why don't you shut your ugly, ignorant mouth.
@@benebutterbean2737 so true. God Bless
Kyle Griswold we invented god so really none of it matters.
Great video, very helpful!
This is an amazing tutorial. Thank you!
Thanks man, you helped a ton with the last part, I'm really grateful for the video
Thank you very much....may God bless you🙏🙏
thanks alot , you really helped me ,your explain is so simple and very helpful ,keep going on .
roaa wesam
dude do u know how to use commas
We in india get similar questions in JEE and we're expected to do it without a calculator... Thanks for explaining... I had difficulty there...
I'm just curious; how'd the JEE go?
Thank you so much, really helped me a lot
Extremely brilliant
Thank you for helping me do chemistry at 12am
I love you so much keep going 😃😃😃😍😍😍😍😍😘
Why did you calculated pH from the first ionization only? Why the H+ ion concentration of second n third ionization r not considered?
@15:45 The reason the answer is not exact enough (for chem classes) is because the calculated conc / initial conc > 5%. So 0.1225/2 ≈ 6% and is considered significant enough to force the need for an exact calculation. Im sure someone has pointed this out already among the huge amounts of comments, but I just wanted to make double sure.
Hi when I use ICE to find pH of 0.5M H2PO4-, the pH is 3.75. It significantly smaller than 4.67
At 21:48 why dont you plug in the ice table? Is that an estimation, or when do we know not to use the ice table and changes in concentration?
I believe it is because equilibrium has already been established so if you would plug the values into the ICE table you would end up with too many values for the concentrations.
I got a pH 3.75 not 4.67 hahahaah lol...(if estimation was not considered)
@@lumaineje i got the same thing, i think we do the estimation instead because you need to do the first removal of h+ as well for h3po4 but since we don't have the initial we don't do that.
Thanks alot ✌😊
thank you so much!
I super appreciate all your videos. I don't understand why in the first problem that you do, the H+ starts at 3M and the SO4 starts at 0 in the second equation.
I was confused about this also....
At 1:54 are we finding the pH based off of Ka2 because the value is smaller than Ka1? Or because it is where equilibrium is achieved?
Do we always use Ka2 for calculating pH for a diprotic acid?
Good explanation but why do you calculate H3O+ at the end and why do you add the value of x to 0.02 e.g please explain.
Nice
How would i be able to find the volume to get to an eq point for titration of something like H2SO4 with NaOH, like eq point for H2SO4 and for HSO4, when I know the initial molarity and volume of H2SO4 and the molarity of the NaOH. Is there a way to do that? I know how to find the first eq point where pH=pKa but how would i find the next ones?
He saved me
If I have a brain like him...
That'll be gold.
Infact...I dey feel you😍😍
At 26:00 you calculate the ph, but isn’t that the average pka and not the ph? Or is the pka the same same as the ph?
26:13 for e does concentratin of h2po4- not matter?
3rd example (from 13:30 to 22:30): why don't we add the h+ ions from h2po4^- and hpo4^2- dissociation to the total [h+] that we calculate ph from? because the h3po4 is a pretty weak acid (at least when compared to h2so4 in which case we did so)? if it was explained why in the video, respond to this comment with a timestamp please
If u make an ice table for the second dissosiation u will see that the H+ contribution is a couple orders of magnitude smaller so its contribution to ph is insignificant, in fact the H+ given by H2PO4 is the same as H2PO4's concentration and you can see 6x10^-8 is waaaaay smaller than 0.118 so u simply disestimate it. I doubt this answer will be of any use to you right now but it may help ppl with the same query.
@@Packie12 thanks i had the same question!
@@Packie12 so basically, if Ka2 is way smaller than Ka1, can we assume that the concentration of H3O+ won't be affected by the second dissociation?
@@Packie12 You're an absolute legend! Thank you so much! I was so confused as to why he didn't use the ICE table for that problem.
I thought not all species of dissociated phosphoric acid is present at the same time? As the proportion of acid species present is determined pH from the fractional composition graph, is it not?
Good
Can you upload a video on finding the [H+] of weak polyprotic acid? As this topic not given in most of the books...and it is a part of jee advanced
Could you please make a video for solving the Systematic Treatment of Equilibrium?
if the H3O concentration changed to 0.026271 as you said, then you would have to recalculate the first equilibrium as your Ka1 equilibrium equation would no longer be satisfied!
ty
Wow! ❤
What would be the first,second, and third buffer zone in the predominant species "line"
in the first equation of the first example are you simply separating the reaction or reacting H2SO4 with water?
so i think he did H2SO4 + H2O yields HSO4- + H3O+ which is the same as HSO4- + H+, for short cut
Excuse me but how do we know that the H+ in the second ICE would be = to that of the H+ in the first
Reaction 1 produces H+ ions, so it's no longer true that in reaction 2 we are starting with 0 H+ ions. That is, in reaction 2 we are starting with the amount of H+ ions that reaction 1 has produced into the solution.
you're right, it's technically [H+]= 0.1188+6.2E-8, which equals 0.118800062. The change is very insignificant and not included in sig figs
So glad y'all are asking the questions
That's cool
Isn't h2so4 a strong oxyacid?
yes. that's why it has one single arrow
what is buffer is between H3PO4 and HPO4)2- , Then which PKa will be considered or should we take average of Pka1 and Pka2?
when writing a diprotic acid reaction of sulfuric acid, H2SO4 + H2O --> SO4 + H3O, there's a missing hydrogen on the product side, so how do you write the correct equation?
Late response but there are two dissociations here. First, H2SO4 + H2O --> HSO4(-) + H3O(+), then HSO4(-) + H2O --> SO4(2-) + H3O(+).
If you wanted to balance the initial equation, you would just have to put a 2 before the hydronium since two protons are released and form H3O
in the first two problems, why did you write the molarity of the given solution under the hydronium ion but didn't do the same for the next two problems, where the concentrations of all the reactants are 0?
I am confused about (X) value when i should neglect it and when i should not do
where does the hydronium come from... i'm not sure why in the end we solved it for h3o, is it the same as using just H+?
Hi! The solution is taking place in water, so H2O + H+ = H3O+. You will often see one substituted for the other, it works the same!
why is it that in calculation 3, for ka1 the value of H+ is 0 and not 2M????
Since H3PO4 is a weak triprotic acid,so it will not completely dissociate.Therefore we will take initial concentration of H+ 0 and not 2M.
We took initial concentration of H+ as 3M in 1st question because, in that question H2SO4 is given which is a strong acid which will dissociate completely in H+ and HSO4-.
How do you calculate the Ka? Where did Ka=0.012 for HSO4- come from?
i think it wasn't calculated bc it was already given in the problem
at 13:22 why does it say H30
That is intuitively wrong at 20:27, HPO2- cant be created without a reaction, from your calculation, supposely u r still going to use the ice table to determine instead of just cancel out each other...
pH=? H2A when 1M ka1=1,0*10^-6 ka2=2,0*10^12 A) 3,0 B)3,2 C)2,6 D)2,4 E)2,8 choose the closest result which one ?
Note to self at question three:
only do Rice tabel once.
so it is possible to have a negative ph?
John Elinair Lafradez
yeah my yellow peepee discharge has a negative ph
John Elinair Lafradez yes it is
why was the pH for the first problem negative?
Late reply, but I wondered the same: pH = -log[H3O+] which is 0 when [H3O+] = 1. Any larger [H3O+] than 1 is going to produce pH less than 0. It's perfectly legit, pOH will be greater than 14,00 (=pKw) and OH-ions wil thus be very rare in the solution). Remember that pH is a logarithmic number and unlike in case of a concentration, negative values are just as legit as positive ones.
@@janikarjalainen8566 shut up nerd
why we took H concentration 0 in question 3 but in first 2 ques it wasnt 0
Because H2SO4 is a strong acid so it will dissociate almost completely in H+ and HSO4- but H3PO4 is a weak triprotic acid that's why it won't dissociate completely.That's why we will take initial concentration of H+ 0 in 3rd question.
what if i dont have ka
anyone who can give me a clue why are we getting 2,5*10-19 coz it gives a huge unbelievable pH
how can ph be negative
5:15 how do we end up with a negative ph??? is that even possible
Yeah ph scale is not only limited to 0 to 14 , it expands beyond tooo
can u pls do the Ice method , come on . I hate all these estimations done by u in this video
why am i learning this in highschool😭
to have good future
we don't have enough time to do all this shit in exam b'cz we are provided with only 1min per ques. so u better come up with smart solution which can help us getting ans. in 30 or 45 sec.....
Why in the name of god are you wasting time on all these calculations in hand in stead just writing the result.
Martin Belmont
i am bad at math so i actually find it very help ful haha