Question

Consider the equilibrium reaction: AX3(s) ⇌ A3+(aq) + 3X-(aq). What is the solubility of AX₃ (MM = 78.0 g/mol) in mg/L at 25⁰C if its Ksp is 1.0 x 10⁻³³?

Answer 1

$AX_{3(s)} ⇌ A^{3+}_{(aq)} + 3X^-_{(aq)}$

K_sp = 1.0 x 10⁻³³

Let us denote the solubility of AX₃ as S molL^–1.

[A³⁺] = S

[X^-] = 3S

K_sp = [A³⁺][X^-]³ = [S][3S]³ = 27S⁴

1.0 x 10⁻³³ = 27S⁴

S = 2.47×10^-⁹ M

Molar mass of AX₃ = 78.0 g/mol

S = 2.47 × 10^-9 moldm^-3 × 78.0 g/mol = 1.93 × 10^-7 gdm^-3 = 0.193 mg per 1L

Therefore, the the solubility of AX₃ at 25⁰C is 0.193 mg/L