Textbook 6D.15 part(b)

[Qinyan Feng 1H]

Hello,

I'm having trouble calculating the pH of 0.055 M AlCl3(aq).
What will the equilibrium equation look like for the highly charged cations that act as an acid?

Thank you

[LavieTran2B]

Hi Qinyan,

Just from the structure of other problems, the Ka equation will look like x^2/(0.055-x) = 1.4 x 10^-5. The equation will be Al(H2O)6 3+ (aq) + H2O (l) -> <- H3O+ (aq) + Al(H2O)5OH2+. You technically don't have to write out the equation you can assume its structure looks like [HA] + H2O -> H3O+ + [A-]. Once you find x, take the -log(x) to find the pH. Hope this helps.