You are saying that 0.1M of CH3COOH will fully disproportionate to give 0.1M of H3O+, but... isnt it a weak acid? so it doesnt 100% give 0.1M of H3O+? i am getting the idea that its like ur doing strong acid + strong base, but ur doing weak acid with base, so do you consider the weak acid a strong 1? thanks btw
@Sulfhur1k No, I don't say that this weak acid will dissociate 100% but it will react with OH- 100%. There's a difference. OH- is such a strong base that it will completely remove the H+ from the weak acid.
OH! so that's why we do that little subtraction!!!....to find the excess moles of stuff in solution which determines the ph....wow, the only chemistry teacher i have is on the internet...
The excess hydroxide will have such a higher concentration than what the conjugate base ion will produce of hydroxide in solution that you can disregard it.
But is it one of those conceptual things that I can take for granted 100%, or is it a circumstantial thing? Just so that I dont get an MCAT question that tries to trick me.
I do know that if it was the acid that was in excess (like an example you present later), then we factor the conjugate base ion into the pH via the Henderson-H equation--because it would constitute a buffer.
I'll be very content to just remember that the conjugate base matters in pH when there is excess acid.
When you say disregard 'it' you mean we disregard the conjugate base of the reaction and just get the ph of the excess NaOH which, because of it's higher concentration, will dominate the ph of the solution at the end of the reaction?
So the reaction itself only serves the purpose of using up an amount of the NaOH in order to reduce it's total amount and also dilute what's left in more solvent?
i dont get why the products side is not factored into the pH of the solution. We looked at the excess of NaOH, and found the concentration of that, but the reaction still proceeded in the 1:1 ratio and a base, CH3COO- was formed in addition to our excess NaOH.
You are fantastic!
like, big time.
shmily018 1 year ago
You are saying that 0.1M of CH3COOH will fully disproportionate to give 0.1M of H3O+, but... isnt it a weak acid? so it doesnt 100% give 0.1M of H3O+? i am getting the idea that its like ur doing strong acid + strong base, but ur doing weak acid with base, so do you consider the weak acid a strong 1? thanks btw
Sulfhur1k 1 year ago
@Sulfhur1k No, I don't say that this weak acid will dissociate 100% but it will react with OH- 100%. There's a difference. OH- is such a strong base that it will completely remove the H+ from the weak acid.
Chemguy
bannanaiscool 1 year ago
@bannanaiscool s this all that i need to knoe in acids and bases???......m a 12th std student!
soumyasayujya 3 weeks ago
in XS i am using that from now on
BLJohnson10 1 year ago
fooled ya this side! :) HAHA. chemguy you're like a billion times better than my chem teacher, she's horrible :( depending on your lessons now!
DarthMaul913 2 years ago
chemguy i love u !!!!!!!!! i really do///
priyakoko 2 years ago
OH! so that's why we do that little subtraction!!!....to find the excess moles of stuff in solution which determines the ph....wow, the only chemistry teacher i have is on the internet...
sodr2 2 years ago 4
haha! INXS
Mortalvis 2 years ago 4
The excess hydroxide will have such a higher concentration than what the conjugate base ion will produce of hydroxide in solution that you can disregard it.
Glad you we pondering this. It shows you get it!
Chemguy
bannanaiscool 2 years ago
ok I see.
But is it one of those conceptual things that I can take for granted 100%, or is it a circumstantial thing? Just so that I dont get an MCAT question that tries to trick me.
I do know that if it was the acid that was in excess (like an example you present later), then we factor the conjugate base ion into the pH via the Henderson-H equation--because it would constitute a buffer.
I'll be very content to just remember that the conjugate base matters in pH when there is excess acid.
hookup83 2 years ago
When you say disregard 'it' you mean we disregard the conjugate base of the reaction and just get the ph of the excess NaOH which, because of it's higher concentration, will dominate the ph of the solution at the end of the reaction?
So the reaction itself only serves the purpose of using up an amount of the NaOH in order to reduce it's total amount and also dilute what's left in more solvent?
Monkmaney 1 year ago
Pretty much right! The excess NaOH will easily produce more OH- ion in solution than that very weak conjugate base ion of low concentration!
Chemguy
bannanaiscool 1 year ago
i dont get why the products side is not factored into the pH of the solution. We looked at the excess of NaOH, and found the concentration of that, but the reaction still proceeded in the 1:1 ratio and a base, CH3COO- was formed in addition to our excess NaOH.
I'm confused....
hookup83 2 years ago
YES! THANK YOU for explaining this to me.
sLiniss 3 years ago
I LOVE YOU
monkeyfacefoot 3 years ago 2
thanks
im not good at calculation , i feel upset
but thanks again
ddrose272 3 years ago 2