Initial Concentrations of ICE Box and 12.79 (6th Edition)

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Brandon Mo 4K
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Initial Concentrations of ICE Box and 12.79 (6th Edition)

Postby Brandon Mo 4K » Fri Jan 18, 2019 2:39 am

Are there ever any instances where the initial concentrations of the products are not 0?

In the answer key of 12.79 (6th Edition), why is it that the initial molarity of H30+ is not 0?

deniise_garciia
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Re: Initial Concentrations of ICE Box and 12.79 (6th Edition)

Postby deniise_garciia » Fri Jan 18, 2019 9:36 am

I had this same question, but I assume that the products in an instance that they are not zero is when the question first asks you to find a concentration at equilibrium and there is a follow up question that states that a certain product was added towards the end. As, for the question in the book I am also confused, I'm not sure as to why they used the same molarity for H3O+ and H2SO4.

Dimitri Speron 1C
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Re: Initial Concentrations of ICE Box and 12.79 (6th Edition)

Postby Dimitri Speron 1C » Fri Jan 18, 2019 12:20 pm

If you have a reaction with a polyprotic acid, (acid capable of donating more than one proton, of which H2SO4 is an example) then you can express it as two equilibrium reactions in which the second one the concentration of H3O+ is more than 0. For the vast majority of polyprotic acids the second or third deprotonation adds an insignificant amount to the concentration (of H3O+) and can therefore be disregarded, but H2SO4 is a special case. The first deprotonation is nearly total, and the second is that of a weak acid. I do not have the 6th edition, but I hope this helps answer your question!

Kenan Kherallah 2C
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Re: Initial Concentrations of ICE Box and 12.79 (6th Edition)

Postby Kenan Kherallah 2C » Fri Jan 18, 2019 12:27 pm

So just adding on to what Dimitri said, sulfuric acid is a diprotic acid and it is a strong acid for the first dissociation and a weak acid for the second. In the first dissociation the equation is H2SO4 +H20 --> HSO4- + H30+ . this would mean that the 0.15 M of H2SO4 has completely dissociated and become 0.15 M of HSO4- and 0.15 M of hydronium ions.
For the second dissociation we have HSO4- + H20 <--> SO4 2- + H30+
now the initial concentration of HSO4- and H3O+ are both 0.15 M once setting up the ice box.
you should then see that the equilibrium constant would be written as such= (0.15+x) (x)/ 0.15-x
solving this for x should get you the correct answer
the equilibrium constant for the second dissociation equation in found in table 12.9


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