Ahhh I was confused about the Kb of the conjugate base of a strong acid ... I figured they should be the same because dissociation is dissociation. However, I did not realize that the Kb expression assumes the proton is extracted from water, not hydronium. That clears it up.
Question, The secound part doesn't make sense, if it takes back the proton shoudn't it logicaly take it from hydronium and form water rather than OH? As a reversable reaction?
Thank you for your great educational videos...iam confused and have two questions please : 1) About the HF example , in the F- 2nd reaction equation at equilibrium , shouldnt the F- reacts with H3O+ instead of H2O?. 2) why the concentrations of F- and HF are equal in both reactions inspite that they are two different reactions ? . Am i missing something i the concept ?...i am totally lost :( ..thank you in advance
Before I explain, note that HF and F- in both equations are represented as acids and conjugate bases for each equation where HF is the acid in the first one but a conjugate base in the second. Same follows for F-, but flipped for its case. 1. If you look at the first HF example next to the F- examples, we see that the HF and the water molecule has reacted to produce the product of F and H30, so therefore the F- wouldn't react with the h30 (they're both products of the same reactants) H from h2o was given another h from HF which then led for F to be left alone in the products. 2. Folowing from the above explanation, the reactions might not look the same, but it's from the same interactions between the F and the HF of the two equations. In the first equation F- is produced and as a lone molecule, it seeks other H2O molecules to attach to form another equation to yield the HF and the OH molecule. I hope this helps :)
Dude this needs more views. Thank you for opening another door of chemistry for me to jump through!
THANK YOU. You're the only one I could find who explained this and made it so SIMPLE.
thank you. finally some damn intuition. too many videos out there just repeating the formula
You saved a life today. It was awesome.
Thanks. This really helps
5:28 what do you mean here?
Paid 18K yet have the worst chemistry teacher at school. Thank you ChemistNATE for giving us the opportunity to learn.
You are great...... Great teaching.......this really helped
Loved it, helped me a lot
Bro that was precise and clear
Super clear. Super helpful. Thank you
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Ahhh I was confused about the Kb of the conjugate base of a strong acid ... I figured they should be the same because dissociation is dissociation. However, I did not realize that the Kb expression assumes the proton is extracted from water, not hydronium. That clears it up.
Question, The secound part doesn't make sense, if it takes back the proton shoudn't it logicaly take it from hydronium and form water rather than OH? As a reversable reaction?
thanks dada(Elder brother)iam from india !thankss for sharing.
When you find the (x). And put the negative log of x. Do you get the ph or poh?
How does this video not have a 1000 likes already?
I would give 1000, but I can only give 1 haha.
Thank you for your great educational videos...iam confused and have two questions please :
1) About the HF example , in the F- 2nd reaction equation at equilibrium , shouldnt the F- reacts with H3O+ instead of H2O?.
2) why the concentrations of F- and HF are equal in both reactions inspite that they are two different reactions ? .
Am i missing something i the concept ?...i am totally lost :(
..thank you in advance
Before I explain, note that HF and F- in both equations are represented as acids and conjugate bases for each equation where HF is the acid in the first one but a conjugate base in the second. Same follows for F-, but flipped for its case.
1. If you look at the first HF example next to the F- examples, we see that the HF and the water molecule has reacted to produce the product of F and H30, so therefore the F- wouldn't react with the h30 (they're both products of the same reactants) H from h2o was given another h from HF which then led for F to be left alone in the products.
2. Folowing from the above explanation, the reactions might not look the same, but it's from the same interactions between the F and the HF of the two equations. In the first equation F- is produced and as a lone molecule, it seeks other H2O molecules to attach to form another equation to yield the HF and the OH molecule.
I hope this helps :)
@@yanic-p162 you helped me, thank you!!
thx man .
مين عربي هون😅😅
Thanks it helps me a lot
My question is ...what is Ka and Kb??😅😅
The rate at which the Acid (Ka) or Base (kb) dissociate in water
This taught me where my chem teacher didn't