Common ion effect and buffers | Chemistry | Khan Academy
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- čas přidán 15. 08. 2014
- Example of calculating the pH of solution that is 1.00 M acetic acid and 1.00 M sodium acetate using ICE table. Another example of calculating pH of a solution that is 0.15 M ammonia and 0.35 M ammonium nitrate.
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Thank you. Really learn more here than lessons.
thank you so much khan academy! this helped a great deal! :-))
thank you, this video was so helpful ! :)
I swear you and organic chem tutor have the lowest volume videos on youtube.
I guess it depends on the speakers on your device. It's loud on mine.
yess I agreed. But they are the best tutors
Organic chemistry is the best
Wooow thanks Khan Academy. I have always had troubles in this topic. Now am happy I have followed
Excellent, thanks!
Thank you!! never knew it was this easy
Yaaaas, thank you!
Thx! Much help
Check this out.
[anything] = n x 10^(m)
p[anything] = (n-1).(10-m)
So in the video,
Ka= 1.8 x 10^(-5) = [H3O+]
p[H3O+]= (5-1).(10-2) [don't include negative & round).
pH= 4.8 ~4.74
Learned this through a MCAT Kaplan Video, so the credit goes to them! You can use this for anything...so if you're given Ka and need to get the pKa, or the pOH and you know the [OH]. Very handy for fast calculations. You can always to the long-log method for an accurate answer.
You're a god.
Thank you so much
Y'all *demanding* a solution without calculator should really be ashamed not to appreciate the work this man is doing. If you want to know how to approximate the value of pH, that's not something that the chemistry video must necessarily cover. Go search CZcams for a video on logarithms.
Thank you...
Thanks!
thanks...
thank for sharing
thank you
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best explanation
gless u man saved my grade
Thank you so much sir! I 🥹🥹
good
What if Ka and Kb are not given? How would we find the pH buffer?
10:56 Can we assume every time that the x is of a very less value?
THank u for the video!
yah +hunter sikari
hunter sikari No
we can only assume x is a small value only when ka < 1
for example : 1.83x10^-4, we ignore x since the degree of ionization is very small ( less than 5%) and ka < 1
where as if ka = 6.73x10^3 , we have to include x in the equation since ka > 1 and degree of ionization is greater than 5%
That is obviosly idiot
Sayam Bandyopadhyay obvious, idiot.*
Idiot.
lol
OK I'm sorry. I was wrong Kh
= Kw/Ka is in case of hydrolysis of salt. Thanks. For video
Do we have to always assume that x is a very small number?
Only a very small amount of a weak acid/base dissociates.
If initial concentration/ka >100 then we can ignore x
why is the x small??
Why is the volume so low?
go to a doctor
Or become one
ughhh i agree... its getting on my nerves
it's good for me dude. bruh
How can we equate Ka = [CH3COO-][H3O+]/[CH3COOH]
what I have learnt Is this [CH3COO-][H3O+]/[CH3COOH] = Kh (i.e hydrolysis constant) and Kh = Kw/Ka. And from here we calculate x and then [H3O+] = x. And then pH = - log (x).
i know this is old but another way to get [H3O+] is using:
[OH-][H3O+] = Kw
so,
[H3O+] = Kw/[OH-]
Kw is constant: 1.01 x 10^-14
now u can use the -log[H+]
Henderson Hassleback equation will be your friend with these type problems. PH= Pka+log ([conjugate base]/[acid])
when I did CA/A I did NH3/NH4 and got the wrong answer but when I did the reverse got the correct. Anyone know why?
What does happen to Nh4NO3??
it makes the concentration of NH4 increase, this is all happing at once
Please fix the sound
Arrigato
not me abusing the Henderson-Hasselbalch equation to get the same answers
whose watching after 6years
You cannot use a calculator on the MCAT. You should show people how to best approximate these values. Example: -log(m x 10^-n) = approx. n - 0.m. So for this video -log (7.7 x 10^ -6) = 6- 0.77 = 5.23
Can u explain this elaboratly
Why is this not a buffer system ? why didn't we take the -log of the Ka and then the log of the ratio of conjugate base over acid?
that works too. For example the second problem whose PH was 8.89. You could have use kw/kb to find Ka and then the solution is simply PH = -log(5.56 x 10^10) + log (0.15/0.35) will give you the 8.89 PH he found without having to go through all of those steps . It is good to know the ICE method solution however
Plz also translate ur lectures into urdu...........
advance thanks :-)
These videos are meant for MCAT preparation, an exam tested in English. Go look for videos in your language else.
HKUST anyone?
All your videos are calculate pH, the EASIEST. Why not do an example on calculate strong base conc. if the pH of acetic acid buffer is _____________.
Why is the volume always so low in khanacademy videos?
These are AAMC approved videos and given that use of calculator is not permitted in the exam, is it the time you showed your viewers how to solve these problems WITHOUT a calculator?
how can pH be greater than 7???
+Parth Ranawat pH is a measure of how acidic/basic water is. The range goes from 0 - 14, with 7 being neutral. pHs of less than 7 indicate acidity, whereas a pH of greater than 7 indicates a base. Got this from google.
Wtf??
Of cause, otherwise why do we have pH up to 14?
chemistry sucks but thx for the video anyway
Marry me professor
Am i the only one who dint get this lecture??
There are a lot of terms that you have to learn before watching this video. Maybe you are confused because you don't know the definition of some of the terms (le Chatlier's principle, equilibrium, shift, etc.)
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rubbish
Thank you so so so muchhh , I have my chemistry exam tomorrow ( 6 units included ) & you're being a lifesaver rn. 🤍