I'd like to add that at 15:14 ish he mentioned kw= 1x10^-14. recalling that for kw the concentration of H3O+ and Oh- are equal in water and that the value for each is 1x10^-7 in the Kw equation: Kw= [H3O+][OH-] Kw= [1x10^-7][1x10^-7]= 1x10^-14 this might've been obvious but it took me a while to figure out so I didn't want you guys to struggle.
I like the way you start with the basics then expand slowly on that... it helps me learn better. I feel like most professors just go from 0-100 so quick and expect you to know the material like them. Anyway, thanks for what you do.
Instead of using his complex method , just use the formula of pOH =-log[OH]- where [OH-]=3*10^-7(from NaOH)+1*10^-7 = 4*10^-7 and take negative logarithmic ull get pOH and then calculate pH by 14-pOH
Professor Organic Chemistry Tutor, thank you for a step by step and basic Introduction into the Autoionization of Water in AP/General Chemistry. Calculating hydroxide ion [OH-], hydronium ion[H3O+], Kw and PH are simple to follow and understand, however, I found Ice Table problems problematic/confusing. This is an error free video/lecture on RUclips TV with the Organic Chemistry Tutor.
For Q 2 The concentration of hydronium ions could be less than 1x10^-7 but the overall solution could still be neutral. This is because the concentrations could both be eg 1x10^-6.5 making the overall kw-1x10^-13 but because the concentrations of hydronium ions and OH- are in equal concentrations, the solution is neutral. The pH would be 6.5 but the solution would actually be neutral. pH = -log[1x10^-6.5] =pH 6.5 This was a Q on GAMSAT blue booklet Q 56 if this helps
At 9:28, if Kw= [H30+][OH-] which are considered products in this reaction - are increased, wouldn't that cause a left shift according to La Chatelier's Principle that an increase in products and a decrease in reactants cause a left shift of the reaction? If an increase in temp causes an increase in Kw = an increase in products [H30+] [OH-], I see that principle being true... unless La Chatelier's Principle doesn't apply to autoionization of water?
Yes. I think he mixed up things here. Am struggling to make sense. If Temp increases, kw also increases and also products increases and so the reaction would move to the left not right
Though it's an endothermic reaction. Consider the first explanation where if temperature is increased for an endothermic reaction, the reaction moves to the right
So I want to add a shortcut and i have a question Our teacher gave us the formula: [H+]sln = [H+]acid + [H+] water. this is for solving autoIonization for strong acids just like the last problem. (he used ICE and Quadratic formula) *note: sln = solution *note: sln, acid, water are subscripts now, I tried to solve the last problem using the formula of our teacher. The answer was pH= 7.61 it was a difference of .10. so my question is, which formula is more accurate? or is it okay if it is just a point difference?
I did not understand when we consider the intial concentration of the products is zero as question 3 and when we consider the concentration of the products is not zero as question 4 ( so my question is when we consider the concentration is zero or not ?)
It's bcos in 3 we were given the kw concentration which is 5.5×10^14 and temp so initially the conc is 0 n also water is a liquid if he had used a diff chemical den we wld hv a conc for dat chemical or compound while in que 4 we were only given the concentration of NaOH so we need to find the direction as in 3 the direction was easy since there was no conc of any subs only kw therefore leaving intial and conc 0 hope u understand
He explained that the temperature is not 25C, so you can't use pH of 7. If the concentration of H3O is equal to the concentration of OH, the solution is neutral. Divide the Kw by the concentration of H3O to get the concentration of OH [(4.4*10^-14)/(2.345*10^-7) = 2.345*10^-7] . They are both equal, and the solution is neutral.
when it is very dilute you must take into account auto ionization of water because the hydronium and hydroxide ions in water is not very small compared to when you have a high concentration. i.e. if pH is very close to 7 you must consider autoionization
The funny thing is that you study all these acids and bases and everything in chemistry and you see none of them physically, just pass the exam then forget everything.
I would demand some easy calculus explained. I am 11 years old, I am very good at math and would need a hand at differential calculus on limits. I have a less mature mindset, can you explain? Thank you!!
If this pertains to question 1, it should be the right answer. Kw=[H3O+][OH-], since we have the concentration of H3O+ and Kw (which is 1x10^-14, this number is the Kw constant at 25C or 298K). Then we get [OH-]=[1x10^-14M][2x10^-6M] => 5x10^-9M. When plugging into the calculator, make sure to put parenthesis ((1x10^-14)(2x10^-6M)).
Why do you always digress and starts talking abt something else prior to solving the problem. You make it more confusing when you do that all the time.
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I'd like to add that at 15:14 ish he mentioned kw= 1x10^-14.
recalling that for kw the concentration of H3O+ and Oh- are equal in water and that the value for each is 1x10^-7
in the Kw equation:
Kw= [H3O+][OH-]
Kw= [1x10^-7][1x10^-7]= 1x10^-14
this might've been obvious but it took me a while to figure out so I didn't want you guys to struggle.
THANK YOU SO MUCH IM DYING T-TTTTT i was so bothered by that but now i realize it was super obvious lol
Good
Damn man, thanks bro I was stuck
thank u so much!!!
Omgg!!Thank u so much!!I was juz nice struggling to figure out why is it🥹🥹
I like the way you start with the basics then expand slowly on that... it helps me learn better. I feel like most professors just go from 0-100 so quick and expect you to know the material like them. Anyway, thanks for what you do.
And I don't seem to understand why this same thing happens everywhere regardless of the country. 😥.
@@reference_realistic There must be some kind of teacher manifesto they all have 😂
@@katarina6724 definitely 😅
P1 0:09
P2 2:40
P3 3:45
short info summary 9:35
P4 11:02
15:46 if you want an inspirational speach
LOL that honestly was so straightforward i loved it
sir the last problem that of NaOH is more complex ..but anway your work is splendid.. may God bless you
Instead of using his complex method , just use the formula of pOH =-log[OH]- where [OH-]=3*10^-7(from NaOH)+1*10^-7 = 4*10^-7 and take negative logarithmic ull get pOH and then calculate pH by 14-pOH
@@manojmaheshwari4055 I didn't understand why he added 1*10^-7?
@@thatonecommentor7758 12:15 ans is there
@@manojmaheshwari4055 can't we igonore the concntraion 10^-7 oh- of water?? It makes calculation difficult..will it be a wrong way?
That last example problem has so much damn math.💀If my professor gives me an exam question like that, I'm going to explode.
😂😂😂 dude you alive
Btw if to solve the last problem as fidning [OH] and then 14-[OH] it will be 7.47. Not so far from 7.52
I finished three videos already for this topic. I will watch this tomorrow since it’s already 12 midnight here
Professor Organic Chemistry Tutor, thank you for a step by step and basic Introduction into the Autoionization of Water in AP/General Chemistry. Calculating hydroxide ion [OH-], hydronium ion[H3O+], Kw and PH are simple to follow and understand, however, I found Ice Table problems problematic/confusing. This is an error free video/lecture on RUclips TV with the Organic Chemistry Tutor.
For Q 2 The concentration of hydronium ions could be less than 1x10^-7 but the overall solution could still be neutral. This is because the concentrations could both be eg 1x10^-6.5 making the overall kw-1x10^-13 but because the concentrations of hydronium ions and OH- are in equal concentrations, the solution is neutral. The pH would be 6.5 but the solution would actually be neutral. pH = -log[1x10^-6.5] =pH 6.5 This was a Q on GAMSAT blue booklet Q 56 if this helps
You are the best af all of time i really am goig to use this on my test
thank you very much for making this video!!! Your explanations are excellent!!
Sir plz make playlist of these videos..
Wow!! This man is marvellous
Good channel mate!
At 9:28, if Kw= [H30+][OH-] which are considered products in this reaction - are increased, wouldn't that cause a left shift according to La Chatelier's Principle that an increase in products and a decrease in reactants cause a left shift of the reaction? If an increase in temp causes an increase in Kw = an increase in products [H30+] [OH-], I see that principle being true... unless La Chatelier's Principle doesn't apply to autoionization of water?
Yes. I think he mixed up things here. Am struggling to make sense. If Temp increases, kw also increases and also products increases and so the reaction would move to the left not right
I was trying hard to make sense
Though it's an endothermic reaction. Consider the first explanation where if temperature is increased for an endothermic reaction, the reaction moves to the right
hey, it would be great if you could go over the Oswald's dilution law and dependence of degree of disassociation on the different factors.
Thanks a lot for this resourceful knowledge
Thank you. From🇿🇦
where did you get 1x10^-14 from???
where did you get 4x10^-7 @ 12:58
summed up 3⋅10^-7 and 1⋅10^-7. He actually said that.
You are the best! Thanks!
17:00 Are you actually Playing Water Ionization or Algebra?🤔
how is Q greater than k in question 4?
That’s just sad most of us won’t need this in real life but now we struggle to understand it and pass the exam
So I want to add a shortcut and i have a question
Our teacher gave us the formula: [H+]sln = [H+]acid + [H+] water. this is for solving autoIonization for strong acids just like the last problem. (he used ICE and Quadratic formula)
*note: sln = solution
*note: sln, acid, water are subscripts
now, I tried to solve the last problem using the formula of our teacher. The answer was pH= 7.61
it was a difference of .10.
so my question is, which formula is more accurate? or is it okay if it is just a point difference?
I think the answer to the first Q is 0.5x10^-8 because 1x10^-14 / 2x10^-6 = 0.5x10^-8
Hi, yes that is also the answer but he just rounded off (got rid of the decimal) hence the answer 5×10^-9
I did not understand when we consider the intial concentration of the products is zero as question 3 and when we consider the concentration of the products is not zero as question 4 ( so my question is when we consider the concentration is zero or not ?)
It's bcos in 3 we were given the kw concentration which is 5.5×10^14 and temp so initially the conc is 0 n also water is a liquid if he had used a diff chemical den we wld hv a conc for dat chemical or compound while in que 4 we were only given the concentration of NaOH so we need to find the direction as in 3 the direction was easy since there was no conc of any subs only kw therefore leaving intial and conc 0 hope u understand
yo you're prob not gonna answer but I'm confused-- how can the solution be neutral for number three if the pH is 6.63?
He explained that the temperature is not 25C, so you can't use pH of 7. If the concentration of H3O is equal to the concentration of OH, the solution is neutral. Divide the Kw by the concentration of H3O to get the concentration of OH [(4.4*10^-14)/(2.345*10^-7) = 2.345*10^-7] . They are both equal, and the solution is neutral.
@@syuann appreciate it!
I am confused. my teacher said that NAOH is a strong base so we assume that it dissociates completely in water and we do not need ICE table
when it is very dilute you must take into account auto ionization of water because the hydronium and hydroxide ions in water is not very small compared to when you have a high concentration. i.e. if pH is very close to 7 you must consider autoionization
How is he getting the numbers ? Like how he’s calculating?
The funny thing is that you study all these acids and bases and everything in chemistry and you see none of them physically, just pass the exam then forget everything.
You try to give the video more brightness it will be great if you do
Please why are hydrated cations considered as bronsted-lawry acids
This topic is depressing. We discussed this today and the exam is by tomorrow 🤧
Thnks alot !!!!!
water at equilibrium
wow, that was kinda hard.
so the pH level that we learned in high school only pertains to 25 degree Celsius?
Yep 😐 they should have at least told us that it's only for 25°C
@@susa9786 they should😂
@@susa9786 well, some students whine over that
Am I the only one with a problem with his answers
U made a mistake he used sub try solving again 21:41
I would demand some easy calculus explained. I am 11 years old, I am very good at math and would need a hand at differential calculus on limits. I have a less mature mindset, can you explain? Thank you!!
Wow
keep it up ! calculus isnt too hard so you'll get it easily.
Im getting 5x10^-21 on the first problem instead of 5x10^-9. Help??
contacct
same I don't get where he gets the -9 from
If this pertains to question 1, it should be the right answer. Kw=[H3O+][OH-], since we have the concentration of H3O+ and Kw (which is 1x10^-14, this number is the Kw constant at 25C or 298K). Then we get [OH-]=[1x10^-14M][2x10^-6M] => 5x10^-9M. When plugging into the calculator, make sure to put parenthesis ((1x10^-14)(2x10^-6M)).
It just got harder I hate chemistry
The place you got 6.97 is supposed to be 1.494 thus this affects every answer you got after that
He used substruction 21:41 l tried solving it with Cal n it was correct maybe u used addition hope u understand
I dont like your page for inly one reason you dont answer questions on the comments sections
Why do you always digress and starts talking abt something else prior to solving the problem. You make it more confusing when you do that all the time.