6E.1

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David Perlman 1H
Posts: 106
Joined: Sat Feb 04, 2023 8:17 am

6E.1

Postby David Perlman 1H » Mon Jan 29, 2024 5:29 pm

"Calculate the pH of 0.15 M H2SO4(aq) at 25 °C."

How do we know to solve for the pH of the 2nd dissociation? In previous problems in the textbook we solved a strong acid for the 1st dissociation so how do we know to do it differently for this one?

Sydni Bednarek 3I
Posts: 163
Joined: Fri Sep 29, 2023 12:00 pm
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Re: 6E.1

Postby Sydni Bednarek 3I » Mon Jan 29, 2024 5:40 pm

For the first dissociation, we are starting with 0.15 M H2SO4, which is a strong acid that will dissociate completely. Therefore, the first dissociation will produce 0.15 M of H3O and 0.15 M of HSO4-. The second dissociation will then start with 0.15 M HSO4-, which is a weak acid. To calculate how much H3O is produced from this dissociation you will need to set up an ICE table and solve for the concentration of H3O, which will be added to the 0.15 M H3O produced by the first dissociation. You can then use the concentration of H3O to find the pH.


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