Textbook 6D.15 b) Setting up equation for AlCl3 + H2O

[Vienna 1G]

Hi there! I am currently working through some of the acid/base equilibrium questions in section 6D of the textbook. Question 15 b) asks what the pH of a solution made from 0.055 M AlCl_{3 (aq)}. I understand that Al³⁺ will act as an acid, but how would I set up mu ICE table for this question? I am confused how I would write out the equation for the deprotonation of AlCl3.

Would I write it like this (because I'm using the net ionic and the Cl- does not change the pH of the solution)? This seems wrong:

Al³⁺_(aq) + H₂O_(l) <--> H₃O⁺_(aq) + Al(OH)_{3 (aq)}


In the textbook answer key, they do:
Al(H₂O)₆³⁺_(aq) + H₂O_(l) <--> H₃O⁺_(aq) + Al(H₂O)₅OH²⁺_(aq)

But I don't understand that at all, so if someone could explain it that would be great!

[Soumya 2D]

When the AlCl3 dissolves, the Al3+ forms a coordination compound with water which is what leads to the Al(H2O)63+. This complex ion now acts as a weak acid as shown in the book. Now that you have this new sequence, use an ICE table with the Ka from 6D.1 (1.4 x 10^-5) to find the concentration of H30+. Hope this helps!