Chemical equilibrium system and where my ions went

NO3)2 and crO4(s), it precipitates barium chromate (possibly ba(OH)2) which then dissolves back into the aqueous solution. the equilibrium constant for the reaction is Ksp of barium chromate is 2.2 x 10-10.
  • #1
sorry if this has been asked before, i did a search and couldn't find any results.
Explain fully what happens when [tex]H^+[/tex] is added to a mixture containing [tex]BaCrO_4(s)[/tex] and [tex]CrO_4^2^-[/tex].

The overall equation of ions in this experiment is [tex]H^+(aq)~+~2CrO_4^2^-\rightleftharpoons~Cr_2O_7^2^-(aq)~+~OH^-(aq)[/tex]

when HCl is added to a solution of K2CrO4 and NaOH and Ba(NO3)2, the precipitate dissolves and the solution becomes orange. I am pretty stumped, but I am assuming K2CrO7 is present because it is orange in colour and that the precipitate from before becomes aqueous again but I am not sure what the percipitate is (possibly Ba(OH)2??) or what is involved in the reaction. any push in the right direction is very appreciated!
 
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  • #2
Ksp of barium hydroxide is 5 X 10-3 and the Ksp of barium chromate is 2.2 X 10-10.

Hope it helps.
 
  • #3
chemisttree said:
Ksp of barium hydroxide is 5 X 10-3 and the Ksp of barium chromate is 2.2 X 10-10.

Hope it helps.

i just realized i even identified barium chromate as a solid in my first post, hahaha :P
thank you anyway

so now my equation is looking more like HCl(aq) + BaCrO4 -> Cr2O7 + BaCl(aq) + H+
 
  • #4
allywallyrus said:
[tex]H^+(aq)~+~2CrO_4^2^-\rightleftharpoons~Cr_2O_7^2^-(aq)~+~OH^-(aq)[/tex]

It won't work this way. Think what will happen to produced OH- in the presence of H+.
 
  • #5
Borek said:
It won't work this way. Think what will happen to produced OH- in the presence of H+.

so we'd have H2O in the products instead, which makes more sense because this also balances nicely as well
i think this clears up where H+ ions were going, thanks :)

i don't suppose it would be a problem if i recycled this thread for questions on the same subject? i will probably have some more
 
  • #6
No, you better start a new thread. Give it more informative subject.
 
  • #7
alright, thanks again
 

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