# Question #0be3f

May 2, 2015

So, the aluminium cation acts as a weak acid, its acid dissociation constant, ${K}_{a}$, being equal to $1 \cdot {10}^{- 5}$.

You can use the value of its acid dissociation constant and the molarity of the solution to determine the concentration of hydronium ions, ${H}_{3} {O}^{+}$, present in solution.

For a weak acid, you have

$\left[{H}_{3} {O}^{+}\right] = \sqrt{{K}_{a} \cdot C}$

$\left[{H}_{3} {O}^{+}\right] = \sqrt{1 \cdot {10}^{- 5} \cdot 2.5} = 0.005$

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

$\left[O {H}^{-}\right] = {10}^{- 14} / \left(\left[{H}_{3} {O}^{+}\right]\right)$

$\left[O {H}^{-}\right] = {10}^{- 14} / \left(5 \cdot {10}^{- 3}\right) = 2 \cdot {10}^{- 12}$

Now, in order to determine whether or not a precipitate will form, you need the solubility product constant, ${K}_{s p}$, for aluminium hydroxide, which is listed as being equal to $1.3 \cdot {10}^{- 33}$.

$A l {\left(O H\right)}_{3 \left(s\right)} r i g h t \le f t h a r p \infty n s A {l}_{\textrm{\left(a q\right]}}^{3 +} + \textcolor{red}{3} O {H}_{\left(a q\right)}^{-}$

By definition, ${K}_{s p}$ is

${K}_{s p} = \left[A {l}^{3 +}\right] \cdot {\left[O {H}^{-}\right]}^{\textcolor{red}{3}}$

The minimum concentration of hydroxide ions needed to precipitate aluminium hydroxide, given that the concentration of $A {l}^{3 +}$ is 2.5 M, will be

$\left[O {H}^{-}\right] = \sqrt[3]{{K}_{s p} / \left(\left[A {l}^{3 +}\right]\right)} = \sqrt[3]{\frac{1.3 \cdot {10}^{- 33}}{2.5}} = 8.0 \cdot {10}^{- 12}$

Compare this value to the one you've calculated for $\left[O {H}^{-}\right]$, and you'll see that you don't have enough hydroxide ions present to form a precipitate

$2 \cdot {10}^{- 12} < 8 \cdot {10}^{- 12}$ $\implies$ NO precipitate is formed.