Titration,Henderson-Hasselbach

[Glycogen]

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Is there anyone out there who can explain to me this concept in any word other that chemistry!!!!!
I understand,equivalent point and curve and all that.Also the fact that at equivalent point the pH=pka but if there will be a Q with all that,all my knowledge is useless.I can not answer it.
Thank you.

 

[bumpski20]

just remember the equivalence point is when the # of MOLES of the acid equal the # of MOLES of base.

the pH = the pka when the CONCENTRATIONS of the acid and base equal. the equation is pH = pka + log [conj base]/[acid] (THIS IS ONLY FOR BUFFERS). if the concentrations are equal then technically you are adding the log of 1 (which is zero) to the pka.

ex (you can do this in your head): pka = 4.5. concentration of conj base = 1M and conc of acid = 0.1M whats the pH of this buffer?

well if you divide 1 by .1 you get 10. the log of 10 is 1. 4.5 plus 1 is 5.5.

 

[Glycogen]

just remember the equivalence point is when the # of MOLES of the acid equal the # of MOLES of base.

the pH = the pka when the CONCENTRATIONS of the acid and base equal. the equation is pH = pka + log [conj base]/[acid] (THIS IS ONLY FOR BUFFERS). if the concentrations are equal then technically you are adding the log of 1 (which is zero) to the pka.

ex (you can do this in your head): pka = 4.5. concentration of conj base = 1M and conc of acid = 0.1M whats the pH of this buffer?

well if you divide 1 by .1 you get 10. the log of 10 is 1. 4.5 plus 1 is 5.5.

Thank you very much.This surely make sense now.In this case even in the last step, we don't even need to get the log of 10,we know that we will increase pH by 1 point for each 10 ka.Right!?
So,4.5-->5.5

 

[bumpski20]

no problemo! the reasoning behind that last sentence sounds right but i never thought of it like that so i cant confirm anything. does any1 else no?