6E #1

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marlene 1K
Posts: 98
Joined: Wed Sep 30, 2020 9:53 pm

6E #1

Postby marlene 1K » Wed Jan 20, 2021 10:10 am

Calculate the pH of 0.15M H2SO4(aq) at 25C.
I'm a little confused on how to set this problem up especially in regards to the Ka and what the reaction itself would be for H2SO4 + H2O.

Marisa Gaitan 2D
Posts: 102
Joined: Wed Sep 30, 2020 9:47 pm
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Re: 6E #1

Postby Marisa Gaitan 2D » Wed Jan 20, 2021 10:17 am

Since H2SO4 is a diprotic acid, it has two reactions that give off H3O+, the first: H2SO4 + H2O --> H3O+ + HSO4- and the second: HSO4- +H2O --> H3O+ + SO42-. Because H2SO4 is a strong acid, it dissociates completely. However, the second reaction needs an ICE table. Make sure to use the right concentration of H3O+ that results from the first reaction. Solve like a regular ICE table, using the Ka2 given and you can find the pH from there! Hope this helps.

Kiran Singh 3A
Posts: 95
Joined: Wed Sep 30, 2020 9:41 pm

Re: 6E #1

Postby Kiran Singh 3A » Wed Jan 20, 2021 10:24 am

I believe the reactions would be : (1) H2SO4 + H2O <--> HSO4(-) + H3O(+) (2) HSO4(-) + H2O <--> SO4(2-) + H3O(+). As for the Ka value, only the Ka2 (the Ka value for the second reaction) is needed to determine the concentration of hydronium and pH because the concentration calculated for the first reaction would be very similar, if not equal, to the initial concentration. Hope this helps!


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