Question

calculate the solubility of A2X3 in pure water, assuming that nither kind ofion reacts with water. the solubiity product of A2X3, Ksp=1.1 X 10 to the power of -23

Answer 1

Here,

A₂X₃ → 2A³⁺ + 3X^2-

Now, Ksp = [A³⁺ ]² [X^2-]³ = 1.1x 10^-23

If the solubility of A₂X₃ is y then , [A³⁺ ]= 2y and [X^2-] = 3y

or, (2y)² (3y)³ = 108 y⁵ = 1.1x 10^-23

Hence, y⁵ = 1.1x 10^-23 / 108 = 1.0 x 10^-25

Or, y = 1.0 x 10^-5 mol/L

Hence solubility of A₂X₃ in pure water is 1.0 x 10^-5 mol/L