The "Assume the change in reaction 2 to be X" part really enlightened me! Thank you! Spent almost an hour looking at the answer hopelessly. You saved my day!
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!
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!
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.
@@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.
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 ;__;
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.
Professor Organic Chemistry Tutor, thank you for a classic explanation on how to calculate the PH of a Polyprotic Acid using ice tables and number lines in AP/General Chemistry. Pattern recognition and problem solving is a great way to learn this material. This is an error free video/lecture on UA-cam TV with the Organic Chemistry Tutor.
@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.
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)
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!!! ❤❤❤💙💙💙
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.
@@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.
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.
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.
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?
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.
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?
but phosphoric acid is weak acid so wont completely dissociate. so is it right to average the pKa values for this polyprotic acid and assume 100% of lets say H2PO4- in solution?
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?
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...
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?
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
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-.
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.
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.
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.....
Acids-Bases - Free Formula Sheet: bit.ly/3NsE7aP
Chapter 15 - Video Lessons: www.video-tutor.net/aqueous-equilibria.html
The "Assume the change in reaction 2 to be X" part really enlightened me! Thank you!
Spent almost an hour looking at the answer hopelessly. You saved my day!
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?
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!
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!
Oh my god, I'm crying rn, thank you so much. Always have my back in chemistry.
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 ;__;
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.
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.
Professor Organic Chemistry Tutor, thank you for a classic explanation on how to calculate the PH of a Polyprotic Acid using ice tables and number lines in AP/General Chemistry. Pattern recognition and problem solving is a great way to learn this material. This is an error free video/lecture on UA-cam TV with the Organic Chemistry Tutor.
@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.
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)
Thank you for the formation for us chemists you provide. We are very happy.
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!!! ❤❤❤💙💙💙
Thanks alot i understand this video more than the discussion of my professor
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 very much....may God bless you🙏🙏
Thanks man, you helped a ton with the last part, I'm really grateful for the video
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.
Thank you for helping me do chemistry at 12am
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.
This is an amazing tutorial. Thank 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
Thank you so much, really helped me a lot
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....
I love you so much keep going 😃😃😃😍😍😍😍😍😘
Great video, very helpful!
Infact...I dey feel you😍😍
Extremely brilliant
Why did you calculated pH from the first ionization only? Why the H+ ion concentration of second n third ionization r not considered?
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?
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?
Thanks alot ✌😊
If I have a brain like him...
That'll be gold.
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.
He saved me
thank you so much!
Hi when I use ICE to find pH of 0.5M H2PO4-, the pH is 3.75. It significantly smaller than 4.67
Could you please make a video for solving the Systematic Treatment of Equilibrium?
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
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
Wow! ❤
what is buffer is between H3PO4 and HPO4)2- , Then which PKa will be considered or should we take average of Pka1 and Pka2?
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?
Won't 3mole hydronium ion produced from the dissociation of sulfuric acid affect the dissociation of bisulfate ion (common ion effect?)
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
Isn't h2so4 a strong oxyacid?
yes. that's why it has one single arrow
26:13 for e does concentratin of h2po4- not matter?
but phosphoric acid is weak acid so wont completely dissociate. so is it right to average the pKa values for this polyprotic acid and assume 100% of lets say H2PO4- in solution?
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!
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?
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...
Good
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
What would be the first,second, and third buffer zone in the predominant species "line"
That's cool
Nice
I am confused about (X) value when i should neglect it and when i should not do
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?
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
ty
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 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-.
@@ashishagnihotri666but at 2:09, HSO4- is a weak acid so it should not dissociate completely. Why did we still consider H+ to be 3M and SO4 -2 0M ?
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
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.
at 13:22 why does it say H30
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
what if i dont have ka
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.
how can ph be negative
anyone who can give me a clue why are we getting 2,5*10-19 coz it gives a huge unbelievable pH
why am i learning this in highschool😭
to have good future
can u pls do the Ice method , come on . I hate all these estimations done by u in this video
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