Question #0be3f

So, the aluminium cation acts as a weak acid, its acid dissociation constant, `K_a`, being equal to `1 * 10^(-5)`.

You can use the value of its acid dissociation constant and the molarity of the solution to determine the concentration of hydronium ions, `H_3O^(+)`, present in solution.

For a weak acid, you have

`[H_3O^(+)] = sqrt(K_a * C)`

`[H_3O^(+)] = sqrt( 1 * 10^(-5) * 2.5) = 0.005`

Now that you have the concentration of the hydronium ions, you can determine the concentration of the hydroxide ions by

`[OH^(-)] = 10^(-14)/([H_3O^(+)])`

`[OH^(-)] = 10^(-14)/(5 * 10^(-3)) = 2 * 10^(-12)`

Now, in order to determine whether or not a precipitate will form, you need the solubility product constant, `K_(sp)`, for aluminium hydroxide, which is listed as being equal to `1.3 * 10^(-33)`.

`Al(OH)_(3(s)) rightleftharpoons Al_text((aq])^(3+) + color(red)(3)OH_((aq))^(-)`

By definition, `K_(sp)` is

`K_(sp) = [Al^(3+)] * [OH^(-)]^(color(red)(3))`

The minimum concentration of hydroxide ions needed to precipitate aluminium hydroxide, given that the concentration of `Al^(3+)` is 2.5 M, will be

`[OH^(-)] = root(3)(K_(sp)/([Al^(3+)])) = root(3)((1.3 * 10^(-33))/2.5) = 8.0 * 10^(-12)`

Compare this value to the one you've calculated for `[OH^(-)]`, and you'll see that you don't have enough hydroxide ions present to form a precipitate

`2 * 10^(-12) < 8 * 10^(-12)` `=>` NO precipitate is formed.