Question

What is the pH of a saturated solution of aluminum hydroxide? Al (OH)3 Ksp is 1.8X10^-33

Answer 1

Write down the dissociation of Al(OH)₃.

Al(OH)₃ (s) <====> Al³⁺ (aq) + 3 OH^- (aq)

                                   s                    3s

s denotes the molar solubility of Al(OH)₃ in solution; clearly s mole/L Al(OH)₃ will furnish s mole/L Al³⁺ and 3s mole/L OH^- as per the stoichiometric dissociation equation.

The solubility product is given as

K_sp = [Al³⁺][OH^-]³ = (s).(3s)³

=====> 1.8*10^-33 = 27s⁴

=====> s⁴ = 6.6667*10^-35

Take logarithm on both sides,

4log s = log (6.6667*10^-35) = -34.1761

====> log s = -8.5440

====> s = 10^-8.5440 = 2.8576*10^-9 ≈ 2.858*10^-9

The molar solubility of OH^- = 3s = 3*2.858*10^-9 mole/L = 8.574*10^-9 M, i.e., [OH^-] = 8.574*10^-9 M

Note that the [OH^-] concentration is lower than the [OH^-] concentration of water. Since K_W =[H⁺][OH^-] = 10^-14 for water, [OH^-] = 10^-7 for water. [OH^-] contribution from Al(OH)₃ is almost 2 orders of magnitude lower than the [OH^-] from water. Therefore, [OH^-] = 10^-7 M and pOH = -log [OH^-] = -log (10^-7) = 7

and pH = 7 (since pH + pOH = 14)

Ans: The pH of the solution = 7