Multiple choice questions
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Original post by charco
0.1M ethanoic acid does not have a pH of 1
ka of ethanoic acid is 1.78 x 10-5
ka of ethanoic acid is 1.78 x 10-5
Ahh i was wondering why there was no Ka given - am i supposed to know it?
Original post by kiiten
Ahh i was wondering why there was no Ka given - am i supposed to know it?
you ARE supposed to know that ethanoic acid is a weak acid ...
Original post by charco
you ARE supposed to know that ethanoic acid is a weak acid ...
Yeah i know that, just didnt take any notice for this question
I dont need to know what Ka for ethanoic acid is though?
Original post by kiiten
Yeah i know that, just didnt take any notice for this question
I dont need to know what Ka for ethanoic acid is though?
I dont need to know what Ka for ethanoic acid is though?
no
Original post by kiiten
Yeah i know that, just didnt take any notice for this question
I dont need to know what Ka for ethanoic acid is though?
I dont need to know what Ka for ethanoic acid is though?
A GCSE student would not know about Kc, but would know that EtOOH is a weak acid and therefore only partially ionises, i.e. [H+] =/= [EtOOH], (like wot i sed b4).
Original post by Pigster
A GCSE student would not know about Kc, but would know that EtOOH is a weak acid and therefore only partially ionises, i.e. [H+] =/= [EtOOH], (like wot i sed b4).
This is for something else but when you write ionic equations. E.g. barium hydroxide and sodium carbonate - why isnt the product separated into its ions.
Ba2+ + 2OH- +2Na+ + CO32- ------> BaCO3 + 2NaOH
Original post by kiiten
This is for something else but when you write ionic equations. E.g. barium hydroxide and sodium carbonate - why isnt the product separated into its ions.
Ba2+ + 2OH- +2Na+ + CO32- ------> BaCO3 + 2NaOH
Ba2+ + 2OH- +2Na+ + CO32- ------> BaCO3 + 2NaOH
BaCO3 is really rather insoluble, whereas NaOH is very soluble. You'd therefore show BaCO3 as a product, but Na+ and OH- remain in solution.
The ionic equation is: Ba2+(aq) + CO32-(aq) -> BaCO3(s)
Original post by Pigster
BaCO3 is really rather insoluble, whereas NaOH is very soluble. You'd therefore show BaCO3 as a product, but Na+ and OH- remain in solution.
The ionic equation is: Ba2+(aq) + CO32-(aq) -> BaCO3(s)
The ionic equation is: Ba2+(aq) + CO32-(aq) -> BaCO3(s)
Ohh so if its insoluble you write it as a product and not ions
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