6D.15 b

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Gabriella Bates 2L
Posts: 113
Joined: Thu Jul 11, 2019 12:15 am

6D.15 b

Postby Gabriella Bates 2L » Thu Jan 23, 2020 6:27 pm

Calculate the pH of (b) 0.055 m AlCl3(aq).

In the solutions manual, the chemical equation they used was
Al(H2O)63+ +H2O <--> H3O+ + Al(H2O)5OH2+

Can someone explain how they got this equation? I'm confused.

Connie Chen 1E
Posts: 51
Joined: Mon Jun 17, 2019 7:24 am

Re: 6D.15 b

Postby Connie Chen 1E » Fri Jan 24, 2020 10:04 am

In water, AlCl3 dissociates, and Al3+ acts as an acid and takes the form Al(H2O)63+, as shown in the table 6D.1. In this reaction, Al(H2O)63+ will donate an H+ because it is acting as an acid, which results in H3O+ and Al(H2O)5OH2+.


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