At what ph will Al(OH)3 begin to precipitate from a .10 M solution of AlCl3?
Work this problem the same as your Ca(OH)2 problem except that Al is the common ion here instead of OH as in the NaOH in that problem. Solve for OH and convert to pH. Post your work if you get stuck.
To determine the pH at which Al(OH)3 will begin to precipitate from a 0.10 M solution of AlCl3, we need to consider the solubility product constant (Ksp) of Al(OH)3 and the hydrolysis reaction of Al3+.
First, let's write the balanced equation for the hydrolysis of Al3+ in water:
Al3+ + 3H2O ⇌ Al(OH)3 + 3H+
The equilibrium constant expression for this reaction is:
K = [Al(OH)3] [H+]^3 / [Al3+]
Solubility product constant (Ksp) is the product of the concentrations of the ions involved in the equilibrium of a sparingly soluble salt. For Al(OH)3, the Ksp is given as:
Ksp = [Al3+] [OH^-]^3
Since Al(OH)3 is a weak base, it reacts with water to produce OH- ions. And since [OH^-]^3 = [H+]^3, we can set up the equilibrium:
Ksp = [Al3+] [H+]^3
Now, by substituting the concentration of Al3+ in the solution (0.10 M) and rearranging the equation, we get:
[H+]^3 = Ksp / [Al3+]
[H+] = (Ksp / [Al3+])^(1/3)
To find the pH, we need to convert [H+] to pH using the equation:
pH = -log[H+]
Therefore, the steps to calculate the pH at which Al(OH)3 will begin to precipitate are:
1. Calculate [H+] using the expression [H+] = (Ksp / [Al3+])^(1/3)
2. Convert [H+] to pH using the equation pH = -log[H+]
However, we still need the value of the Ksp of Al(OH)3 to proceed with the calculations.