Textbook problem 6D.15

[Jazlyn Romero 1I]

Calculate the pH of (a) 0.19 M NH4Cl (aq); (b) 0.055 M AlCl3 (aq)

Hi, For part b, I am having trouble understanding how to calculate the pH since I am not sure how to find the Ka value. I cannot seem to find it in the table. If anyone can help explain how to approach this, that would be great, thanks!

[Sahiti Annadata 3D]

I believe you can look up the Ka online for [Al(H2O)6]3+ because that table is not comprehensive.

[Yun Su Choi 3G]

Why did Al3+ all of a sudden become Al(H2O)6 3+?

[Megan Lu 3D]

Hi! I also had some trouble finding the Ka value in the textbook at first, but it is available in Table 6D.1 (page 477).
I approached this problem by writing out the equation Al(H2O)3+(aq) + H2O(l) <—> H3O+(aq) + Al(H2O)2(OH-)2+(aq) , where Al3+ is a small, highly-charged metal cation that can act as an acid in water. Then, I set up an ICE table and solved for x as usual using the Ka value in the textbook (1.4 * 10^-5) to get the molar concentration of H3O+. I then took the negative log of [H3O+] to get a pH of 3.06. Hope this helps!