1. Write the equilibrium constant expression for the third ionization of phosphoric acid (H3PO4) in water.
@emmynimmy @Hayhayz @byronichero @xlovely_Lizardx @Zarkon @hperez777 @timo86m
imquerty, any idea?
:) its imqwerty but you can call me that too :D okay so we have \(H_3PO_4\) do you know the stepwise ionization reactions of this acid? :)
i can tell you in case you forgot :)
@CreatingMyself u there? (:
YesYes I'm here!
And no I only know the first ionization :(
okay so these are the three reactions in sequential order- http://prntscr.com/9h0ah5
so this how you do this question- 1st wee write the 3rd ionization reaction of \(H_3PO_4\) in water-> \(H_3PO_4 ^{-2}\)\(_{(aq)}\)---------->\(H^+ _{(aq)}~~+~~PO_4 ^-3_{(aq)}\) the ionic concentration is given like this-> ionic concentration =\(\Large \frac{[Products]}{[Reactants]}\) so in out case we will have this-> ionic concentration=\(\Large \frac{[H^+][PO_4 ^-3]}{[HPO_4 ^{-2}]}\) and this value comes equal to-> \(4.8 \times 10^{-13}\)
Ahhhhh!! How did you calculate the ionic concentration though? what values did you substitute in place of the reactants and products?
well thats a general value that i remember :) i wrote that just for info you don't have enough data in this problem to calculate that so in you case the answer is just this-> ionic concentration=\(\Large \frac{[H^+][PO_4 ^-3]}{[HPO_4 ^{-2}]}\)
Ah so my answer ends up as Ka=(that equation)
Thank you! I have 3 more questions if you wouldn't mind answering them, in separate open questions of course. More medals i can give ya!
oops not Ka Ka is acid equilibrium
yeah :) i am really sorry but i gtg now x-x i have a class and i got few hours to sleep
Well thanks anyway man! Medal!
lol np (; buh bye :D
get some sleep!!
lol i really must go but its really difficult to log off x-x os is kinda addicting
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