AutoIonization of Water, Ion Product Constant - Kw, Calculating H3O+, OH-, and pH Using Ice Tables

This acids and bases chemistry video tutorial provides a basic introduction into the auto-ionization of water. It explains how to calculate the hydroxide ion [OH-] concentration given the hydronium ion [H3O+] concentration using the ion product constant of water - Kw. it explains how to determine if a solution is acidic, basic, or neutral based on the concentration of H3O+ at 25C. It explains how to calculate the pH of a neutral solution when the temperature increases and how to tell if the auto ionization of water is an endothermic or an exothermic process. Finally, it shows you how to calculate the pH of a very dilute basic solution of 3 x 10^-7 M NaOH. This tutorial contains plenty of examples and practice problems with the formulas and equations as well.
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...
Final Exams and Video Playlists:
www.video-tuto...
Full-Length Videos and Worksheets:
/ collections
Chemistry PDF Worksheets:
www.video-tuto...

[@TheOrganicChemistryTutor]

Chemistry PDF Worksheets: www.video-tutor.net/chemistry-basic-introduction.html Full-Length Videos & Exams: www.patreon.com/MathScienceTutor/collections Next Video: kzbin.info/www/bejne/hpXUd4atlLiFq8k

[@marin9207]

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.

[@user-uh9uv2yw9h]

THANK YOU SO MUCH IM DYING T-TTTTT i was so bothered by that but now i realize it was super obvious lol

[@snegan.p7685]

Good

[@austinmars2855]

Damn man, thanks bro I was stuck

[@linsdesk7240]

thank u so much!!!

[@zhiyingesther8440]

Omgg!!Thank u so much!!I was juz nice struggling to figure out why is it🥹🥹

[@katarina6724]

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.

[@reference_realistic]

And I don't seem to understand why this same thing happens everywhere regardless of the country. 😥.

[@katarina6724]

@@reference_realistic There must be some kind of teacher manifesto they all have 😂

[@reference_realistic]

@@katarina6724 definitely 😅

[@AK-op4be]

That last example problem has so much damn math.💀If my professor gives me an exam question like that, I'm going to explode.

[@ryanisber2353]

15:46 if you want an inspirational speach

[@user-uh9uv2yw9h]

LOL that honestly was so straightforward i loved it

[@siphenathitshaka6668]

sir the last problem that of NaOH is more complex ..but anway your work is splendid.. may God bless you

[@manojmaheshwari4055]

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

[@thatonecommentor7758]

@@manojmaheshwari4055 I didn't understand why he added 1*10^-7?

[@melvinketorglo309]

@@thatonecommentor7758 12:15 ans is there

[@what5493]

@@manojmaheshwari4055 can't we igonore the concntraion 10^-7 oh- of water?? It makes calculation difficult..will it be a wrong way?

[@Nemoguzapomnit]

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

[@SMM-xg7dn]

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

[@linsdesk7240]

I finished three videos already for this topic. I will watch this tomorrow since it’s already 12 midnight here

[@sarfrajansari8849]

Sir plz make playlist of these videos..

[@hadeerrashad5486]

thank you very much for making this video!!! Your explanations are excellent!!

[@AnnoyedChemistryExperime-gw1kx]

You are the best af all of time i really am goig to use this on my test

[@shreyan1607]

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.

[@tjames22123]

Good channel mate!

[@obadiahsombe6884]

Wow!! This man is marvellous

[@Oceanminded33]

where did you get 1x10^-14 from???

[@shyakafrancis6602]

Thanks a lot for this resourceful knowledge

[@leratoshasha1224]

Thank you. From🇿🇦

[@haithamh1diabete814]

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

[@AJ-ul1lt]

You are the best! Thanks!

[@saeternj]

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?

[@muhindoedith5459]

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

[@muhindoedith5459]

I was trying hard to make sense

[@muhindoedith5459]

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

[@someone8335]

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.

[@hollybidwell3754]

how is Q greater than k in question 4?

[@klodgultiano1046]

This topic is depressing. We discussed this today and the exam is by tomorrow 🤧

[@user-wp2mf8be2k]

How is he getting the numbers ? Like how he’s calculating?

[@nikolpritsky386]

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?

[@syuann]

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.

[@nikolpritsky386]

@@syuann appreciate it!

[@nastarannazari5062]

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

[@smokestack1ightning]

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

[@shukriadan8109]

Thnks alot !!!!!

[@user-Moela]

17:00 Are you actually Playing Water Ionization or Algebra?🤔

[@ykasken]

where did you get 4x10^-7 @ 12:58

[@Nemoguzapomnit]

summed up 3⋅10^-7 and 1⋅10^-7. He actually said that.

[@SMM-xg7dn]

I think the answer to the first Q is 0.5x10^-8 because 1x10^-14 / 2x10^-6 = 0.5x10^-8

[@tsuyuri_lei]

Hi, yes that is also the answer but he just rounded off (got rid of the decimal) hence the answer 5×10^-9

[@chiiko7090]

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?

[@user-oz1ex5lz8t]

You try to give the video more brightness it will be great if you do

[@user-eq9bw4ss8z]

Please why are hydrated cations considered as bronsted-lawry acids

[@kazhilasapindalo696]

water at equilibrium

[@zenamohammed4524]

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 ?)

[@melvinketorglo309]

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

[@linsdesk7240]

wow, that was kinda hard.

[@bunny8352]

Am I the only one with a problem with his answers

[@melvinketorglo309]

U made a mistake he used sub try solving again 21:41

[@rosiedianne1599]

so the pH level that we learned in high school only pertains to 25 degree Celsius?

[@susa9786]

Yep 😐 they should have at least told us that it's only for 25°C

[@don_fabrice14]

@@susa9786 they should😂

[@lastpro6244]

It just got harder I hate chemistry

[@agentcarranzo]

Im getting 5x10^-21 on the first problem instead of 5x10^-9. Help??

[@darlingbamford8791]

contacct

[@AngelGarcia-yr9bp]

same I don't get where he gets the -9 from

[@Berns8]

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)).

[@astrochemistry6007]

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!!

[@carlamariedelcarmen5944]

Wow

[@user-uh9uv2yw9h]

keep it up ! calculus isnt too hard so you'll get it easily.

[@OmevwerhoElijah]

I dont like your page for inly one reason you dont answer questions on the comments sections

[@bunny8352]

The place you got 6.97 is supposed to be 1.494 thus this affects every answer you got after that

[@melvinketorglo309]

He used substruction 21:41 l tried solving it with Cal n it was correct maybe u used addition hope u understand

[@jenn5816]

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.