6D.15 part b

[Amy Luu 2G]

Calculate the pH of (b) 0.055 M AlCl3(aq)
I am really confused on how to solve this problem. I looked at the solutions to see how they solved it and the chemical reaction was
Al(H20)₆ + H2O H30+ + Al(H2O)₅OH2²⁺
How did they get this? Where did H2O come from and what happened to Cl3?

[Isha_Maniyar_Dis2E]

This is an example of a hydrated complex. We haven't learned this in class, so I wouldn't worry about it. It is a topic that combines coordination complexes being hydrated in water (recall that it typically has 6 binding sites, which is why 6 H2O's can bind).

Hope this helped!

[Hope Hyland 2D]

The Cl is not included in the equation because it's not involved in the reaction, so it would just cancel out anyway.

[Sebastian Lee 1L]

If you took Lavelle for 14A, we did cover the transition metals acting as acids in salts briefly. Remember that transition metals like aluminum can form coordination compounds with ligands including water when in solution. There are 6 available bonding sites in an octahedral arrangement where the aluminum accepts the oxygen's lone pair to form a covalent bond. This is how we get Al(H₂O)₆. In solution, this complex can be acidic because one of the water molecules can give off a proton (H⁺) to form H₃O⁺ as seen in the solution's equation. You will end up with the aluminum bonded to the remaining 5 water molecules as well as a deprotonated OH^-.

Remember that Cl will have no effect on the pH when bonded to the Al salt. It is the conjugate base of a strong acid HCl so, if we assume 100% deprotonation for HCl, Cl^- will not undergo the reverse reaction to form HCl.