Polyprotic Acid Base Equilibria Problems, pH Calculations Given Ka1, Ka2 & Ka3 - Ice Tables

[@TheOrganicChemistryTutor]

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[@sciencesynergy-g4h]

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😍

[@debbiedellacroce2612]

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!

[@PunmasterSTP]

Hey how'd the rest of your class go?

[@KidNamedLindseyy]

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

[@benjacobson6160]

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]

I know it's been two years, but how did the SAT go!?

[@benjacobson6160]

@@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.

[@PunmasterSTP]

@@benjacobson6160 Wow I’m really sorry to hear that, and I hope that your plans have gone well since then!

[@PunmasterSTP]

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!

[@jenniferobunadike6525]

Oh my god, I'm crying rn, thank you so much. Always have my back in chemistry.

[@ytb917]

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.

[@Autistotle]

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 ;__;

[@kevcon5886]

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)

[@christianmensah1452]

may God richly bless you...may he bless u with more knowledge and wisdom

[@gartyqam]

CHRISTIAN MENSAH THIS IS A CHEMISTRY CHANNEL BRING GOD OUT OF THIS!!!!!!!

[@kylegriswold9686]

so he can simply share it all with us

[@benebutterbean2737]

@@gartyqam Hey Derp, God invented chemistry, so why don't you shut your ugly, ignorant mouth.

[@michaelayeni177]

@@benebutterbean2737 so true. God Bless

[@silentislandtempest5825]

Kyle Griswold we invented god so really none of it matters.

[@captainconway2236]

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.

[@JillRhoads]

@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.

[@Riyu-jv7wf]

Thanks alot i understand this video more than the discussion of my professor

[@cynthiadent2434]

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.

[@MaureenSklaroff]

I was confused about this also....

[@harshitbalaji2892]

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...

[@PunmasterSTP]

I'm just curious; how'd the JEE go?

[@TheBest-fj4qy]

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!!! ❤❤❤💙💙💙

[@ambasing_omaygot]

Thank you for helping me do chemistry at 12am

[@sciencesynergy-g4h]

Thank you very much....may God bless you🙏🙏

[@simonlondono8059]

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?

[@Axl_Ericsson]

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 a pH 3.75 not 4.67 hahahaah lol...(if estimation was not considered)

[@nidhigeorge2411]

@@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.

[@rony12007]

Why did you calculated pH from the first ionization only? Why the H+ ion concentration of second n third ionization r not considered?

[@lukanikic4950]

Thanks man, you helped a ton with the last part, I'm really grateful for the video

[@Haraldo-Tony-Duarte]

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!

[@spencerpootis3319]

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

[@Packie12]

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.

[@rebeccacarol4439]

@@Packie12 thanks i had the same question!

[@rebeccacarol4439]

@@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?

[@Jacksonrox13]

@@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.

[@akhandpratap6379]

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

[@minhchautran507]

Hi when I use ICE to find pH of 0.5M H2PO4-, the pH is 3.75. It significantly smaller than 4.67

[@phsal5182]

This is an amazing tutorial. Thank you!

[@charitydoeses]

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.

[@roaawesam6666]

thanks alot , you really helped me ,your explain is so simple and very helpful ,keep going on .

[@gartyqam]

roaa wesam dude do u know how to use commas

[@cabangongiearnthynka.211]

If I have a brain like him... That'll be gold.

[@hivirupalihena3102]

Thank you so much, really helped me a lot

[@susantori6625]

Could you please make a video for solving the Systematic Treatment of Equilibrium?

[@sciencesynergy-g4h]

Infact...I dey feel you😍😍

[@emmataylor4328]

Great video, very helpful!

[@fardinafarha6125]

Extremely brilliant

[@Shubhagarwal_2408]

what is buffer is between H3PO4 and HPO4)2- , Then which PKa will be considered or should we take average of Pka1 and Pka2?

[@johnnyace1300]

Excuse me but how do we know that the H+ in the second ICE would be = to that of the H+ in the first

[@Bukaloom]

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.

[@lh9591]

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

[@AnthonyBecker9]

So glad y'all are asking the questions

[@snowythecat9159]

I love you so much keep going 😃😃😃😍😍😍😍😍😘

[@alihaydar8387]

He saved me

[@MikeBrown-yn7tu]

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?

[@muffinslolo8603]

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?

[@cassandrelubin1325]

Isn't h2so4 a strong oxyacid?

[@갱갱항상]

yes. that's why it has one single arrow

[@DaFireElf]

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?

[@luvdog4evr16]

in the first equation of the first example are you simply separating the reaction or reacting H2SO4 with water?

[@Anle_50]

so i think he did H2SO4 + H2O yields HSO4- + H3O+ which is the same as HSO4- + H+, for short cut

[@rizvee1968]

Won't 3mole hydronium ion produced from the dissociation of sulfuric acid affect the dissociation of bisulfate ion (common ion effect?)

[@mattrock9655]

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 ?

[@denissemaetan6918]

thank you so much!

[@preetyluli7617]

Thanks alot ✌😊

[@karlavillanueva2493]

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?

[@dewardshrewd6729]

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?

[@asiimwebrian6336]

Good

[@gerardoclaudio2041]

What would be the first,second, and third buffer zone in the predominant species "line"

[@ghaithshooter5601]

I am confused about (X) value when i should neglect it and when i should not do

[@tnt_grieferz9176]

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...

[@benjamindoku]

Wow! ❤

[@Daphtp]

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+?

[@nararabbit1]

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!

[@jhwong7943]

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?

[@rayhanrakib3113]

That's cool

[@jellolova]

How do you calculate the Ka? Where did Ka=0.012 for HSO4- come from?

[@manuelaferreras9382]

i think it wasn't calculated bc it was already given in the problem

[@nicoletsang8884]

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?

[@rumpleforeskin7032]

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

[@adiaclarke-hoyte595]

why is it that in calculation 3, for ka1 the value of H+ is 0 and not 2M????

[@ashishagnihotri666]

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-.

[@trailblazer7702]

ty

[@johnelinairlafradez9862]

so it is possible to have a negative ph?

[@gartyqam]

John Elinair Lafradez yeah my yellow peepee discharge has a negative ph

[@eddiemurphy4851]

John Elinair Lafradez yes it is

[@josemarrufo2745]

why was the pH for the first problem negative?

[@janikarjalainen8566]

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.

[@gartyqam]

@@janikarjalainen8566 shut up nerd

[@leonmozambique533]

26:13 for e does concentratin of h2po4- not matter?

[@zulanova1209]

Note to self at question three: only do Rice tabel once.

[@theiasaspirant2893]

Nice

[@RubyWilson-y1e]

❤

[@ntandoyenkosimacele1146]

anyone who can give me a clue why are we getting 2,5*10-19 coz it gives a huge unbelievable pH

[@harshanbajwa]

why we took H concentration 0 in question 3 but in first 2 ques it wasnt 0

[@ashishagnihotri666]

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.

[@chea6311]

at 13:22 why does it say H30

[@eduardax.6977]

what if i dont have ka

[@husbaansayed1697]

how can ph be negative

[@boxero415]

5:15 how do we end up with a negative ph??? is that even possible

[@dranilkumarchaudhary5899]

Yeah ph scale is not only limited to 0 to 14 , it expands beyond tooo

[@李汶灏]

can u pls do the Ice method , come on . I hate all these estimations done by u in this video

[@chen1327]

why am i learning this in highschool😭

[@ganapathib9974]

to have good future

[@aarchieverse]

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.....

[@martinbelmont]

Why in the name of god are you wasting time on all these calculations in hand in stead just writing the result.

[@gartyqam]

Martin Belmont i am bad at math so i actually find it very help ful haha