At what ph will Al(OH)3 begin to preciptate from a .10 M solution of AlCl3?
To determine the pH at which Al(OH)3 will begin to precipitate from a 0.10 M solution of AlCl3, we first need to understand the solubility equilibrium of Al(OH)3.
The solubility product constant (Ksp) expression for Al(OH)3 is:
Al(OH)3(s) ⇌ Al3+(aq) + 3OH-(aq)
The Ksp value for Al(OH)3 is approximately 3 x 10^(-34).
When the concentration of Al3+ and OH- ions in the solution exceeds the solubility product constant, precipitation of Al(OH)3 occurs.
At equilibrium, the concentration of Al3+ is equal to the concentration of OH- ions, which is denoted as [OH-]. Therefore, we can rewrite the Ksp expression as:
Ksp = [Al3+][OH-]^3
Since [Al3+] = 0.10 M (given in the question), we substitute this value:
3 x 10^(-34) = (0.10 M)([OH-]^3)
Now, rearrange the equation to solve for [OH-]:
[OH-]^3 = (3 x 10^(-34))/(0.10 M)
[OH-] = ∛((3 x 10^(-34))/(0.10 M))
Taking the cube root of the right-hand side gives us the concentration of OH- ions in the solution. To convert this concentration to pH, we need to calculate the pOH and then convert it to pH using the relation:
pOH = -log[OH-]
pH = 14 - pOH
Substitute the value of [OH-] into the equation, calculate the pOH, and then convert it to pH using the given relation.