Question

Obtain relation between solubility product and its solubility for Al(OH)₃

Answer 1

Dissociation Equation

\[\ce{Al(OH)3 < = > Al^3+ + 3OH^-}\]

Let the solubility of Al(OH)3 be S mol/L.

[Al³⁺] = S

[OH⁻] = 3S (because 3 moles of OH⁻ are released per mole of Al (OH)3)

Expression for Solubility Product

K_sp​ = [Al³⁺] [OH⁻]³

K_sp​ = (S) (3S)³ = S⋅27S³ = 27S⁴

Ksp = 27S⁴ or S = `root4(K_(sp)/27)`