The K_sp of Al(OH)₃ is 2.0 × 10^-32. At what pH will a 0.2 M Al³⁺ solution begin to show precipitation of Al(OH)₃?

Question

Quizplus · Accepted Answer

The balanced equation for the dissolution of Al(OH)3 in water is:

Al(OH)3(s) ⇌ Al3+(aq) + 3OH-(aq)

The solubility product constant (Ksp) expression for this equilibrium is:

Ksp = [Al3+][OH-]^3

The value of Ksp for Al(OH)3 is given as 2.0 × 10^-32.

Assuming complete dissociation of AlCl3 in water,

AlCl3 → Al3+ + 3Cl-

The initial concentration of Al3+ in solution is therefore 0.2 M.

Let x mol/L of OH- be produced.
The equilibrium concentration of Al3+ is [Al3+]=0.2 mol/L.
The equilibrium concentration of OH- is [OH-]=x mol/L.
Using the Ksp expression, we have:

Ksp = [Al3+][OH-]^3

2.0 × 10^-32 = (0.2)(x)^3

x= (2.0 × 10^-32/0.2)^1/3

pOH = -log[OH-] = -log(x) = -log((2.0 × 10^-32/0.2)^1/3)=11+pH

pH+pOH= 14
pH= 14-11=3

Therefore, the pH at which a 0.2 M Al3+ solution begins to show precipitation of Al(OH)3 is 3.7. Choice C is the best answer.

The Ksp of Al(OH)3 Is 2

The K_sp of Al(OH)₃ Is 2