[THC]

NaOH (s) is added to 1.0 L 0.05 M Al₂(SO₄)₃. The pH of the solution is 9.0 at equilibrium.
L(Al(OH)₃) = 1.3*10^-33 and K for the following reaction
Al³⁺ + 4 OH^- <--> [Al(OH)₄]^-
is K = 1.1*10^33.


a) Is Al³⁺ found as [Al(OH)₄]^- or Al(OH)₃ in the solution?

b) At what pH are all the Al³⁺ found as  [Al(OH)₄]^-?

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a) I have no idea how to start this. I know that [OH-] = 10^-5 M...

[AWK]

Use K formation of [Al(OH)₄]^-. Assume this is the only important equilibrium in this solution.

[THC]

Is it something like this:

K = (0.1 M - x)/(x*(10^-5)^4) => x =9.09*10^-15 M

?

[THC]

Ok, I guess that's wrong, since the [OH-] I'm using already is at equilibrium while the other concentrations aren't.

Also, if I calculate
[Al3+][OH-]^3 = 9,09*10^-15 M*10^-15 M^3 = 9,09*10^-30 M
which is less than L, and the Al(OH)4- is formed after Al(OH)3 is formed, by adding more OH-.

[AWK]

Is it something like this:

K = (0.1 M - x)/(x*(10^-5)^4) => x =9.09*10^-15 M

?

Do the same with Ksp treating conc. of OH- as an excess cocentration.
You will get another values of Al3+ but these are limits between for the real concentarion of Al3+. Further calculations should include activity coefficients and iteration procedures.