Titration of a weak acid with a strong base (continued) | Khan Academy
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- čas přidán 31. 08. 2014
- Calculating the pH for titration of acetic acid with strong base NaOH at equivalence point and past the equivalence point. Created by Jay.
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I have been banging my head against the wall for 3 days until this video came on my recommended. Thanks my g.
Best video on the web teaching titration. I needed a step by step walk through and they delivered. God bless Khan Academy.
Great set of titration videos! Thank you so much!!!
Thank you! :D
I thought question d was incorrect. Why would you assume that the number of moles of OH remaining from the addition of OH is the actual number of moles of OH in solution. Adding OH causes the formation of CH3COO- which is going to react with water and produce more OH like it did in question C. I know the surplus of OH will cause the reaction to favor the left side (CH3COO-) much more heavily but I don't know if that means that CH3COO- will no longer react with h2o to form any more OH which is what this answer seems to be assuming.
I think you'd need another ICE box with a starting concentration of 0.014 for OH and everything else like the last question
Edit: trying the solution above gives an answer that barely differs from the original so this answer in the video is correct. Just wish the instructor emphasized this point
The reason that the narrator of the video disregarded the CH3COO- (and it's potential reaction to produce even more additional OH-) is because the concentration of OH- produced from CH3COO- is VERY negligible in comparison to the concentration of OH- anions from the NaOH. Actually, these are the following concentrations of OH- anions from NaOH dissociation: 0.0143 Molarity of OH-. and from CH3COO- ionization: 4x10^-6 Molarity. Hence, when we add 0.0143+4x10^-6 the number that we get is still effectively 0.0143 Molarity - Hence the narrator disregarded that additional step of calculating the ionization of CH3COO- since the OH- we get out of it is measly and negligible.
@@stavshmueli6932 thank you my friend, this was the answer i have been searching and i found it just before my exams thanks to you.
Why is it that in this video you consider the acetate ion reacting with water, but in the previous video you just ignore it and go straight to Hendersen-Hasselbach?
Because the previous video was something called a buffer, which means it was a weak acid and its conjugate base. If you can recognize that it is a buffer titration, you are able to use the Henderson-Hasselbach equation.
Thank you teacher!
thank you so much!
Thanks a lot for sharing and being awesome!
There was a lot of useful theory information needed to solve the problems in these two videos. Would be nice to have a quick written explanation after or before to better wrap my mind around it; went back and forth many times to keep track of it all.
great day..... can any one help me with this Calculate the ph of 1.0*10^-8 M of Koh
Why did you flip the equation around 5:30? Is it because you don't want the anion to be a product? Like you initially want the product side to be zero?
Are you still Alive?
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