The Ksp of aluminum hydroxide is 5 x 10 -33.At what ph foes Al(OH)3 just begin to precipitate in a 0.15M solution of AlCl3?
Al(OH)3 ==> Al^+3 + 3OH^-
Ksp = (Al^+3)(OH^-)^3
Plub in Ksp and Al^+ from the problem, solve for OH^-
To determine the pH at which Al(OH)3 just begins to precipitate in a 0.15M solution of AlCl3, you need to compare the concentrations of the hydroxide ions (OH-) and the aluminum ions (Al3+). When the concentration of OH- exceeds the solubility product constant (Ksp) of Al(OH)3, precipitation of Al(OH)3 occurs.
First, let's write the balanced equation for the dissociation of AlCl3 in water:
AlCl3 → Al3+ + 3 Cl-
Since AlCl3 is a strong electrolyte, it will completely dissociate in water, resulting in a 0.15M concentration of Al3+ ions.
Next, let's write the equation for the solubility equilibrium of Al(OH)3:
Al(OH)3 ⇌ Al3+ + 3 OH-
The Ksp expression is given as:
Ksp = [Al3+] [OH-]^3
We know the Ksp value for Al(OH)3, which is 5 x 10^-33. We also know that the concentration of Al3+ is 0.15M. Let's represent the concentration of OH- as "x" since we don't know it yet.
Therefore, we can rewrite the Ksp expression as:
5 x 10^-33 = (0.15) (x)^3
Now, solve for "x". Take the cube root of (5 x 10^-33 / 0.15) to find the value of "x". This will give you the concentration of OH- ions.
Once you have the concentration of OH-, you can calculate the pOH using the formula:
pOH = -log [OH-]
Finally, determine the pH using the relation:
pH = 14 - pOH
So, by plugging in the value of pOH into the above equation, you can find the pH at which Al(OH)3 just begins to precipitate in a 0.15M solution of AlCl3.