Question

What is the molar solubility of Al(OH)₃ in a solution containing 1.2 ✕ 10⁻³M NaOH?

Al(OH)_3 Ksp= 4.6 * 10^(-33)

______ M

Answer 1

Let S be the molar solubility of Al(OH)₃

Al(OH)₃ Al³⁺ + 3OH^-

NaOH Na⁺ + OH^-

[Al³⁺] = S     

[OH^-] = 3S ( from Al(OH)₃ ) + (1.2x10^-3M) ( from NaOH )

Solubility product constant , K_sp = [Al³⁺][OH^-]³

                              4.6x10^-33 = S x(3S+(1.2x10^-3))³

                                           S = 3.6x10^-9 M

Therefore the molar solubility is 3.6x10^-9 M