Answer:
Conc of Q at Equilibrium 2.1 × 10⁻³ mol/dm³
Ksp = 1.57 × 10⁻¹² mol/dm³
Explanation:
The Solubility of the Ionic compound is given by s
where s = 7 × 10⁻⁴ mol/dm³
solubility equation is given as follows
XQ₃(s) ⇄ X³⁺(aq) + 3Q⁻(aq)
s s 3s
let the concentration of each be s
The concentration of Q = 3s
⇒ [Q] = 3 × 7 × 10⁻⁴ mol/dm³
⇒ [Q] = 2.1 × 10⁻³ mol/dm³
hence,
Ksp = [X³⁺][Q⁻]³
Ksp = [s][3s]³
Ksp = [7 × 10⁻⁴ ][2.1 × 10⁻³]³
Ksp = 1.57 × 10⁻¹² mol/dm³
