In a solution with the composition implied by your question, the least
soluble Al-containing solid phase will be aluminum hydroxide, Al(OH)3.
The solubility product of Al(OH)3 is defined as:
K_sp = [Al3+][OH-]^3
where the brackets indicate the concentration of the species between
the brackets. At saturation, the product of the concentrations is
equal to the value of the solubility product, which is a constant at a
given temperature and assuming dilute solutions.
You are interested in how the solubility of Al(OH)3 varies as a
function of pH, which is defined as –log[H+]. Because this is an
aqueous solution, the dissociation constant of water provides us a way
to relate the concentration of H+ to that of OH- ions:
K_w = 10^-14 = [OH-][H+]
[OH-] = 10^-14/[H+]
Substituting the above expression for [OH-] into the first equation
gives:
K_sp = [Al3+] * {10^-14/[H+]}^3
Taking the common logarithm of both sides gives:
log(K_sp) = log[Al3+] + 3*log(10^-14) – 3*log[H+]
(log(K_sp))/3 – (log[Al3+])/3 + 14 = -log[H+] = pH
In your solution, [Al3+] = 0.05M. Tables of solubility products
available on the web give values of K_sp for Al(OH)3 ranging from
2*10^-32 to 3*10^-34 (Google search terms: solubility product
al(oh)3). Assume something in the middle of the range like 10^-33,
then we have:
(log(10^-33))/3 – (log(0.05))/3 + 14 = pH
-11 – (-1.30)/3 + 14 = pH
3.43 = pH
This says that if the solubility product of Al(OH)3 is, indeed,
10^-33, then this phase will precipitate at a pH higher than ~3.43.
Using a smaller K_sp would predict precipitation at a lower pH, and
conversely, a larger pH would predict precipitation at a higher pH.
Note that solubility is a function of temperature. These calculations
use values for K_w and K_sp of Al(OH)3 appropriate for ~25 degrees C. |
