Question

Determine the molar solubility for Al(OH)3 in pure water. Ksp for Al(OH)3 = 1.3 x 10-33

2.6 x 10-9 M is the answer, but how? Please explain.

Answer 1

Use Al(OH)₃(s) which has a K_sp = 1.3 x 10^-33 M⁴

1. Write the chemical reaction.

Al(OH)₃(s) <---> Al³⁺(aq) + 3OH^-(aq)

2. Write the K_sp expression.

K_sp = [Al³⁺] [OH^-]³

3. Let "s" equal the molar solubility of Al(OH)₃(s).

From the reaction stoichiometry:

[Al³⁺] = s and [OH^-] = 3s

Substitute into the K_sp expression:

Ksp = (s)(3s)3 = 1.3 x 10-33 M4

                      (s)(27s3) = 1.3 x 10-33 M4

                          27s4  = 1.3 x 10-33 M4

                            s4 = 4.8 x 10-35 M4

                             s = 2.6 x 10-9 M

So   [Al3+] = 2.6 x 10-9 M

and

[OH^-] = 3(2.6 x 10^-9 M) = 7.8 x 10^-9 M