6D.15 part b 0.055 m AlCl3(aq)

[Natalie Do 3F]

Calculate the pH of 0.055 mAlCl3(aq).

In the solutions manual, it replaces Cl3 with H2O6. I understand the rest of the problem but I don't know where we learned to do that and where we can find the Ka/pKa for the new compound it makes. Anything helps!

[Najia Saleem 2G]

Hello,

I was confused by this problem as well, but the textbook explains how AlCl3 is the same as Al(H2O)6 ^+3, and the Ka value and pKa value for this compound are given in table 6D.1. I do not recall this information being emphasized during lecture, but you can find it under Focus 6D.3.

Now, for the new compound's Ka and pKa values that can be determined by setting the Ka value you found equal to the acidity constant you can make using an ICE table. It should look like Ka = (x^2)/(0.055-x). Then, you can approximate, solve for x, take the -log of x and that is your pH value.

Hope that helps!

[Kiyoka Kim 3C]

I believe it's because Cl- is the conjugate base of the strong acid HCl that it would have no significant effect on the pH. And so you would only consider the ionization Al3+ when calculating the pH.

[Silvi_Lybbert_3A]

In this problem, there is technically two events that happen: AlCl3 dissolves in water and then the coordination compound gives off protons, acting as an acid. When AlCl₃ dissolves in water the Al⁺³ ions are surrounded by six water molecules each, as the Al⁺³ can accept the lone pair of electrons on the oxygen of a water molecule for each of the six water molecules. So, the aluminum ion forms an octahedral shaped coordination complex as it is surrounded by water molecules when dissolved in water. The Cl^-1 ions are the anions of the strong acid HCl, and because the anions/cations of strong acids/bases do not affect pH, the Cl ions act as spectator ions and can be ignored when writing out the equation for how the dissociation of Al(H₂O)₆⁺³ affects the pH: Al(H₂O)₆⁺³ + H₂O <--> H₃O⁺ + Al(H₂O)₅OH⁺². As you can see, one of the hydrogen bonds between a water molecule attached to the aluminum ion and another free-floating water molecule in solution becomes a covalent bond as the proton is given off to this water molecule becoming hydronium ion (while the water molecule apart of the aluminum coordination complex becomes a hydroxide group).

[Kyla Roche 2K]

Since Cl is the conjugate base of HCL, a strong acid, it doesn't affect the solution yet AL does because it doesn't come from a strong base.